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© 2023 Mathspace Pty Ltd Copyright Notice This Work is copyright. All rights are reserved. Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. For permission to use material from this text or product, please email hello@mathspace.com.au For our full digital offering, visit mathspace.co Mathspace Pty Ltd Level 11, 477 Pitt St Sydney 2000 The Victorian Curriculum F-10 content elements are © VCAA, reproduced by permission. The VCAA does not endorse or make any warranties regarding this resource. The Victorian Curriculum F-10 and related content can be accessed directly at the VCAA website. Title: Mathspace Victorian Curriculum 2.0 - Year 10 (Pre-General): 2024 Edition Print Run: 1 ISBN: 978-1-963022-45-2 Editors and lead authors: Erin Gallagher, Elisha Templin, Sarah Butterworth, Julia Reid, Anjelica Sieger, Jacqueline Cichello, Stephen George, Grace Peake, Jonathan Kim Sing, Toni Smythe, Dot Yung Further reviewers: Melissa Archey, Warren Bardsley, Lianna Beeching, Kathryn Bennett, Benjamin Bui, Eric Chan, Cory Duker, Bahija Elmir, Wayne Gillie, Jeremy Hunter, Shahad Odeh, Helen Soar, Patricia Toutzaris, Paul Unkles, Jie Yang Writing and development team: Adam Humphries, Neil Lopez, Ma. Lourdes Ona, Bernadette Barillo, Sheramie Castillo, Abegail Fontanilla, Shaira Llanita, Lesley Lopez, Kerby Mendoza, Regine Jara, Albert Jara, Rolly Apdo, Michaela Belen, Valerie Baja, Marielle Belen, Allen Calabia, Adrian Doctolero, Dhave Fernandez, Janine Francisco, Rose Francisco, Rona Josue, Crystel Lontoc Images and Design team: Alex Duckmanton, Chastine Marquez, Keith Gimeno

Contents 1

Indices and surds

1.01

Index laws

1.02

Negative indices

1.03

Optional: Simplify surds

1.04

Optional: Add and subtract surds

1.05

Optional: Multiply and divide surds

1.06

Optional: Binomial expansions with surds

1.07

Optional: Rationalise the denominator

1.08

Optional: Fractional indices Chapter 1 review

Expand and factorise

2.01

Factorise algebraic expressions

2.02

Expand binomial products

2.03

Factorise binomial products

2.04

Factorise monic quadratic trinomials

2.05

Optional: Factorise non-monic quadratic trinomials

2.06

Complete the square

Chapter 2 review

Linear equations

3.01

Linear graphs

3.02

Equations with more than 3 steps

3.03

Equations involving fractions

3.04

Rearrange equations

3.05

Parallel and perpendicular lines

3.06

Linear models

104

3.07

Graph linear inequalities

117

Chapter 3 review

133

Contents mathspace.co

iii

Quadratics 136 4.01

Quadratic equations

138

4.02

Quadratic graphs

146

4.03

Solve by factorisation

170

4.04

Solve by completing the square

181

Investigation: Use bisection to solve quadratics Investigation: Derive the quadratic formula 4.05

The quadratic formula

4.06

The discriminant

4.07

Optional: Further quadratic equations Chapter 4 review

Simultaneous equations

188

190

5.01

Solve graphically

192

5.02

Simultaneous inequalities

200

5.03

Solve using substitution

210

5.04

Solve using elimination

218

5.05

Applications of simultaneous equations

225

Investigation: Intersections of curves 5.06

Optional: Intersections of curves Chapter 5 review

Trigonometry 242 6.01

Pythagoras’ theorem

244

6.02

Trigonometric ratios

253

6.03

Angles of elevation and depression

266

6.04

Bearings

276

6.05

Optional: Pythagoras in 3D

6.06

Optional: Trigonometry in 3D

6.07

Optional: Sine and cosine rules Chapter 6 review

238

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

292

Measurement 296 7.01

Surface area of prisms and cylinders

298

7.02

Volume of prisms and cylinders

308

7.03

Surface area and volume of composite solids

318

Investigation: Rainfall collection

331

Estimation and accuracy

335

Investigation: Instruments and methods

344

Investigation: Scale models

347

7.04

7.05

Optional: Spheres

7.06

Optional: Cones

7.07

Optional: Pyramids Chapter 7 review

349

Probability 354 8.01

Review Venn diagrams and two-way tables

356

8.02

Independent and dependent events

368

8.03

Conditional probability

382

8.04

Mixed probability concepts

394

Investigation: Simulation

410

8.05

Optional: Geometric probability Chapter 8 review

Univariate statistics

413

416

9.01

Representations of data

418

9.02

Quartiles

436

9.03

Boxplots

449

9.04

Cumulative frequency

464

9.05

Compare data sets

476

9.06

Samples and populations

491

9.07

Optional: Standard deviation

9.08

Optional: Comparisons using standard deviation Chapter 9 review

501

Contents mathspace.co

Bivariate statistics 10.01

Two-way tables

510

10.02

Associations between categorical variables

522

10.03

Bivariate numerical data

541

10.04

Associations between numerical variables

550

10.05

Line of good fit

565

10.06

Interpolation and extrapolation

576

10.07

Misrepresentations of data

590

Investigation: Misrepresentations in the media

605

Investigation: Conduct a statistical investigation

611

10.08

Optional: Correlation coefficient Chapter 10 review

11.01

Introduction to networks

622

11.02

Network representations

636

11.03

Planar graphs

648

11.04

Walks, paths, and trails

658

11.05

Optional - Traversable networks

Algebraic fractions 12.01

Add and subtract fractions with numeric denominators

12.02

Add and subtract fractions with algebraic denominators

12.03

Optional: Add and subtract further algebraic fractions

12.04

Multiply and divide algebraic fractions with numeric denominators

12.05

ptional: Multiply and divide algebraic fractions with O algebraic denominators Chapter 12 review

615

Networks 620

Chapter 11 review

508

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

674

Introduction to functions (Optional) 13.01

Optional: Define relations and functions

13.02

Optional: Domain and range

13.03

Optional: Graphs of xn

13.04

Optional: Transformation of functions Chapter 13 review

Exponentials 14.01

Exponential relationships

14.02

Characteristics of exponential graphs

14.03

Optional: Transformations of exponentials

14.04

Exponential equations

14.05

Compound interest Investigation: Half life

14.06

Exponential models Chapter 14 review

Logarithms 15.01

Logarithms

15.02

Order of magnitude and logarithmic scales

15.03

Optional: Other bases Investigation: Logarithms

15.04

Optional: Logarithm laws

15.05

Optional: Connect exponentials and logarithms

15.06

Optional: Solve exponential equations with logarithms Chapter 15 review

Circles Investigation: Circles 16.01

Circles at origin

16.02

Translations of circles Investigation: Equation of any circle

16.03

Optional: Transform circle equations Chapter 16 review Contents mathspace.co

vii

Hyperbolas Investigation: Hyperbolas 17.01

Direct and indirect proportion

17.02

Hyperbolas

17.03

Optional: Transformations of hyperbolas Chapter 17 review

Geometry and proofs 18.01

Deductive reasoning

18.02

Proofs with parallel lines

18.03

Congruent triangles

18.04

Similar triangles

18.05

Further geometric proofs Investigation: Shortest path

18.06

Optional: Circles and chords

18.07

Optional: Angles in circles

18.08

Optional: Tangents Investigation: Algorithms in geometry

18.09

Optional: Transformations on the plane using algorithms Chapter 18 review

Functions and polynomials (Optional) 19.01

Optional: Operate with polynomials

19.02

Optional: Polynomial division

19.03

Optional: Factor and remainder theorem

19.04

Optional: Graph polynomials in factored forms

19.05

Optional: Rates of change Chapter 19 review

viii

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Further trigonometry (Optional) 20.01

Optional: Angles with exact values

20.02

Optional: The unit circle

20.03

Optional: Graphs of sine and cosine

20.04

Optional: Trigonometric equations Chapter 20 review

Counting Methods (Optional) 21.01

Optional: Counting principles

21.02

Optional: Factorial notation

21.03

Optional: Permutations

21.04

Optional: Combinations

21.05

Optional: Applications of counting methods to probability Chapter 21 review

Answers

678

Contents mathspace.co

Big ideas • Understanding and applying index laws enable us to manipulate and simplify expressions involving powers, facilitating problem-solving and exploring mathematical concepts. • Exploring irrational numbers, surds and their operations reveals the underlying principles of manipulating expressions involving roots. • Mathematical expressions can be transformed and represented in different forms using principles of equivalence and algebraic properties.

Curriculum Outcomes VC2M10A01 factorise algebraic expressions by taking out a common algebraic factor VC2M10A02 simplify algebraic products and quotients using exponent laws VC2M10AN01 define rational and irrational numbers and perform operations with surds and fractional indices VC2M10AN02 perform operations on numbers involving fractional exponents and surds

Indices and surds Chapter outline 1.01 Index laws 1.02 Negative indices 1.03 Optional: Simplify surds 1.04 Optional: Add and subtract surds 1.05 Optional: Multiply and divide surds 1.06 Optional: Binomial expansions with surds 1.07 Optional: Rationalise the denominator 1.08 Optional: Fractional indices Chapter 1 review

4 13

A “surd” is a term used to describe an irrational number. The name originally meant “deaf” or “unheard” because they cannot be written exactly using fractions and decimals.

1.01

Index laws After this lesson, you will be able to… • apply the multiplication law to simplify expressions. • apply the division law to simplify expressions. • apply the power of a power law to simplify expressions. • apply the zero power law to simplify expressions. • solve problems using a combination of index laws.

Multiply terms with indices

For the expression a5 × a3, since the factors have like bases we can use the multiplication law. a5 × a3 = a5 + 3 = a8 We can remind ourselves of the multiplication law by writing out the original expression in expanded form.

a× a× a× a× a × a× a× a For any base number a and any indices, also called exponents, m and n the multiplication law is: am × an = am+n When multiplying terms with a common base: • Keep the same base. • Find the sum of the indices.

Example 1 Simplify 6x2 × 4x6 using index laws.

Create a strategy Use the multiplication law.

Apply the idea 6x2 × 4x6 = 6 × 4 × x2+6 = 24x

Multiply the coeﬃcients and add the indices of x Evaluate and simplify

Reﬂect and check Notice that the coeﬃcients are multiplied separately when we use the multiplication law on the powers.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Idea summary The multiplication law: am × an = am+n

Divide terms with indices Similarly, we can avoid having to write each expression in expanded form by using the division law, which is also known as the quotient law.

a × a ×a× a× a a×a

To simplify

we can use the fact that

to get:

For any non-zero base number a, and any indices m and n we get,

When dividing terms with a common base: • Keep the same base. • Find the difference between the powers. We can also write the division law in the form: am ÷ an = am−n

Example 2 Simplify

Create a strategy Use the division law.

Apply the idea Subtract the indices of x Evaluate the subtraction

Idea summary The division law:

where a ≠ 0.

1.01 Index laws mathspace.co

Power of a power law For any base a, and indices m and n, we also have the power of a power law: (am)n = am×n To see how this law works, let’s work through an example of (52)3. (52)3 = (5 × 5)3 = (5 × 5) × (5 × 5) × (5 × 5) = 56 This gets generalised to (am)n = am×n When simplifying a term with a power that itself has a power: • Keep the same base. • Find the product of the powers. Brackets are used to show which powers are applied to any coefficient. Be careful when raising negative numbers to a power. A negative number raised to an odd power gives a negative result, but raised to an even power gives a positive result. For example: (−2x2)3 = −8x6

Example 3 Simplify (−2x2)2

Create a strategy Use the power of a power law.

Apply the idea (−2x2)2 = (−2)2 x2×2 = 4x

Multiply the powers of the variable Evaluate the multiplication and coeﬃcient

Idea summary The power of a power law: (am)n = am×n

Zero power law We also have the zero power law which says that for any non-zero base a, a0 = 1 We know using the division law that

= a m – n.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

= an – n = a0.

If m = n, then Also

= 1.

This means that a0 = 1. That is, taking the zeroth power of any number will always result in 1.

Example 4 Simplify 18a0

Create a strategy Use the zero power law.

Apply the idea Notice that only a is raised to the power of 0, but not the coeﬃcient 18. 18a0 = 18 × a0

Separate the two parts of the term

= 18 × 1

Evaluate a0

= 18

Evaluate

Idea summary The zero power law: a0 = 1 where a ≠ 0.

Mixed operations Now we are going to look at questions that can be solved using a combination of these rules. It’s important to remember the order of operations when solving such questions. • The multiplication law: am × an = am+n • The division law: am ÷ an = am−n, a ≠ 0 • The power of a power law: (am)n = am×n • The zero power law: a0 = 1, a ≠ 0 We may also come across expressions such as (am × bn)p, and we can use the power of a power law on each variable to see that: (am × bn)p = am×p × bn×p

1.01 Index laws mathspace.co

Example 5 Simplify (x5 y4)4

Create a strategy Use the power of a power law.

Apply the idea (x5 y4)4 = (x5)4 × ( y4)4

Raise each variable to the power of 4

× y Multiply the powers

y16 Simplify

=x =x

Example 6 Simplify a

Create a strategy We can use the zero power law, followed by the power of a power law, on the numerator. Then we can use the division law to simplify the expression, since there will be a power of x in both the numerator and denominator.

Apply the idea Use the zero power law

Use the power of a power law

Simplify the numerator

Use the division law

Simplify the power

Create a strategy Use the division law then the power of a power law.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Apply the idea Use the division law Simplify Use the power of a power law Simplify

Idea summary Some expressions require a combination of the power laws to simplify them. It’s also important to remember the order of operations when simplifying these expressions. The multiplication law: am × an = am+n The division law: am ÷ an = am−n, a ≠ 0 The power of a power law: (am)n = am×n The zero power law: a0 = 1, a ≠ 0

1.01 Practice questions Understanding

3, 4

Explorer

1–4

Adventurer

1–4

Trailblazer

When simplifying a division of terms with powers of the same base, what operation is performed on the indices?

Rewrite this expression in expanded form and demonstrate how the division law for indices works.

1.01 Index laws mathspace.co

Consider the expression ( p3)2. a

Are these expressions equivalent to ( p3)2:

( p × p × p) × ( p × p × p)

iii

( p × p × p)2

( p × p × p) × ( p × p)

p3 × p3

p3 × p2

Determine whether these equations are true or false: ( p3)2 = p3+2

i c

( p3)2 = p3×2

Complete the equation: ( p3)2 = p⬚

Complete each statement: a

x⬚ ÷ x3 = x7

(b8 × b2) ⬚ = b20

5a–f, 7a–g, 8a–c, 9a–g, 10a–c, 11a–d, 12a–h, 13a–c

Fluency

Ex 1

Simplify: a

a3 × a5

6y × y

i 6

Ex 2

( y⬚ × y17)1 = y17

(3m2 × ⬚)4 = 81m24

5c–j, 7c–i, 8b–e, 9d–j, 10b–f, 11b–e, 12e–n, 13d–f

a2 × a8

b × b4

3y × 4y

a × −2a

2w4 × w10 × w7

5z5 × 3z8 × z9 6bc4 × 8b6a × 0

Simplify: a

5g4h2 × 8g3h4

8m2n4 × 7m6n

5g–l, 6, 7e–l, 8c–f, 9g–l, 10d–h, 11d–h, 12g–t, 13e–h

d h l

y3 × y7 a ma n

4y8 × 6y6 × 3y3

Simplify: a

a10 ÷ a5

c e

c m5 ÷ m2

4y4 × (−2y6) × 4y5 2s2 t × (−5r2 s3) × 3r3 t4

(−20b4) ÷ 2b4

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

j11 ÷ j4

Simplify: a

u2 × u6 × u3

i 8

b d f

h l

(−4y7) × (−6y2) × (−4y2) 9y9 × 8 (−y)8 × 7y7 (−9y9) ÷ (−3y5)

9r8 ÷ (4r2)

Ex 3

Simplify: a

(j4)6

(w2)11

(5h7)2

(xy2)5

(−x9)4

(−5x4)3

i 10

Ex 4

Ex 5

Ex 6

Simplify:

( y4)3

(3a2)4

(w9 v6)4

(k5)3 × (k2)8

6 (r6)5 × 4 (r4)7

(x9 y)7 × (xy2)4

10 (r7)5 × (2rs)4

(a5)9 × (b6)10

( p10)6 × ( p4)3

((x3)4)5

(g0)7

(j8)0

(7st)0

6k0 ÷ (4x2)0

Simplify: a

24a0

10r0 + (2q3)0

Simplify: a

(x7 y2)4

(11x ) − (12x )

(4u5 v2)2

m8 ÷ m2 × m5

m ((x ) ) o

p5 × ( p4)3

n p

(−2x )

(3y5)2 × (5y2)3

(c10)11 ÷ (c8)3

(−240u32) ÷ (−8u9) ÷ (−5u12)

265

p9 ÷ p5 ÷ p2 (3a2 b5 c)4

(15x3)0 + 150 − 15y0

(4h0)2 + 18 ÷ (3g0)

(10x2 × 10x5)0 − 10x0 (4y4)3

Simplify:

1.01 Index laws mathspace.co

Reasoning 14

15, 16ace

When simplifying the expression

15, 16

14, 16

, Clarice’s ﬁrst line of work was a3 + a2.

Is Clarice’s work correct? Explain. 15

When simplifying the expression

• Indira’s first line of work was a3 × a4 • Chad’s first line of work was a3 × a2 • Ernestino’s first line of work was Who is correct? Explain. 16

Explain why each statement is incorrect, then give the correct solution: a

3s6 × 3s2 = 3s12

c e

(2x3)5 = 2x15

(4a)0 = 4a

Problem-solving 17–19

18, 19, 21

18–20, 22

A scientist is modelling the growth of a bacteria colony. The colony doubles in size every hour. If she starts with a single bacterium, how many bacteria will there be after 6 hours? Express your answer using indices.

Will any values of x make 3x be equal to x3? Use examples to justify.

Find the value of a and b in the equation:

Determine whether each statement is true or false. a c

−xn = (−x)n for all x, only if n is odd −xn = (−x)n for all x, only if n is even

Rewrite 16p with a prime base.

Simplify: a

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−xn = (−x)n for all x always

1.02

Negative indices After this lesson, you will be able to… • apply the negative index law to simplify expressions. • extend the concept of negative indices to include fractional bases.

Negative indices By simplifying

using the division law, we subtract the powers and get a−2. By using expansion

we see this is the same as

a×a×a 1 = a × a × a × a× a a× a This gives us the negative index law: for any non-zero base a, base to a negative power: • Take the reciprocal of the expression. • The power becomes positive.

. That is, when raising a

This also means that:

Example 1 Fill in the box to make the statement true:

Create a strategy Use the negative index law.

Apply the idea

Write the denominator as a power of 5 Apply the negative index law. −2 should be placed in the box.

1.02 Negative indices mathspace.co

Example 2 Simplify

Create a strategy Use the division law.

Apply the idea Subtract the powers and divide the coeﬃcients Simplify the power Apply the negative index law

Example 3 Simplify 20m6 ÷ 5m13 × 9m2

Create a strategy Use the division law then the multiplication law.

Apply the idea Use the division law Multiply the coefficients then use the multiplication law Use the negative index law

Idea summary The negative index law:

where a ≠ 0.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Fractional bases with negative indices When raising a fractional base to a negative power we can combine the individual rules we have seen so far.

Exploration We may see expressions in the form

We can simplify these expressions by starting with

Power law

Rewrite as division

Use negative index law

Multiply and take the reciprocal

Simplify the product

1. How could we rewrite

using positive indices?

2. Are there any values that a or b cannot have?

Example 4 Simplify: a

Create a strategy Find the reciprocal of the fraction inside the bracket then change the power to positive.

Apply the idea Find the reciprocal of the fraction then change the power to positive

Apply the power to the fraction

Evaluate the powers

1.02 Negative indices mathspace.co

Create a strategy Use the negative index law, remembering we take the reciprocal of the fraction.

Apply the idea Take the reciprocal of the fraction and change the power to positive Use Use (a × b)n = an × bn Simplify using index laws Use the negative index law again to end up with positive indices

Idea summary Negative index law with fractions:

where a, b ≠ 0.

1.02 Practice questions Understanding 1

1–3

Explorer

1–3

True or false: a

Is 4−2 less than or greater than 1?

Write

1–3

Adventurer

using a negative index?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

a−4 = −a4

Trailblazer

4b, 5a–h, 6a–h, 7ac, 8ab, 9a–c, 11a–c, 12a–f, 13ab

Fluency 4

4a, 5c–k, 6g–n, 7bd, 8bc, 9d–g, 10a–d, 11c–e, 12f–k, 13b–d

4c, 5d–l, 6i–p, 7cd, 8cd, 9gh, 10e–h, 11d–h, 12k–p, 13d–f

Complete each table: a

2−4

2−3

10−4

10−3

2−2

2−1

10−2

10−1

100

101

102

103

104

100 c

3−4

3−3

3−2

3−1

27 Ex 1

Fill in each box. Evaluate if possible: a

Rewrite with positive indices: a

a−9

p−2

3x−4

8p−3

2x−8 y3

m 3a4 b−5 c−7

4−2 k−3 l7

m−5 n−4 p4

Simplify: a

p−2 q3

Rewrite as terms with negative indices: a

Ex 2

7x−9

Simplify: a

5y9 × 4y−3

7a4 × 4a−2

5x4 × (−3x−8)

−4y × (−4y )

−4

× 4h

−2

× 2y

−5

3y−2 × 4y−3

(5mp)2 × mp−2

1.02 Negative indices mathspace.co

Evaluate: a

2−2 × 24

2 × 3−2

10 ÷ 2 + 3

20 × 2−2 + 6

3−1 + 4−1

3 × 4−2 + 5 × 2−4

g Ex 3

2y6 × 4y7 × 4y−5

6y7 × 2y−5 × 5y3

4y × 3y ÷ 2y

−1

40x−2 ÷ 5x8 ÷ 4x−9

12x−9 × 2x4 × 3x−2 8y10 ÷ 2y−4 ÷ y3

14x−8 ÷ 2x × 3x6

(4m−6)4

Simplify: (2m)−3

i m

(3p−4)−2

Simplify: a

(10x3 y−4)2

(3x−4 × 2x2)2

(4x3)−1 × ( y−2)−3

(10x3)3 × (2x4 y3)−2

Reasoning 14

(7 + 32) × 2−2

2x−2 × 5x5 × 4x−5

c g 12

102 × 5−2

Simplify: a

Ex 4

−1

14, 16

14, 16, 18

Oreste and Donovan are writing an equivalent expression for Oreste writes down

(3y2)−2

15–18

as a single term.

and Donovan writes down x−10.

Who is correct and why? 15

We can evaluate 03 = 0, but not 0−3. Explain why 0−3 is undeﬁned.

A student wanted to express 5a−1 without negative indices and wrote their answer is incorrect, and write the correct answer.

For x > 1, what happens when we raise x to a very large negative index? That is, what happens to x−a as a gets closer to 0? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

. Explain why

For variable p, calculate the following expressions for p = −2, 0, 4, 10: • 2p • 4p • 2−p + 4p • 2−p × 4p

What do you predict will happen to the value of each expression as p increases signiﬁcantly?

Problem-solving 21 19

Write

Complete:

Solve for n:

19, 21, 23

19–24

at least three different ways.

(x3 y−5)n = x−12y20

(a−5 b3)n = a15 b−9

Why is 2−2 < 2 but

It is estimated there were 7 × 108 jellyﬁsh in the largest bloom ever observed. If each jellyﬁsh can consume 5 grams of plankton in a day, how long would it take the bloom to consume 10 tonnes (1 tonne = 1 000 000 g) of plankton?

In electrical engineering, the formula used for calculating the capacitance of a capacitor is

? Give another example where this pattern applies.

, where C is the capacitance measured in farads, Q is the charge on the capacitor measured in coulombs and V is the voltage across the capacitor measured in volts. Use this formula to answer the following questions. a

A capacitor has a charge (Q) of 15 × 10−6 C (coulombs) and a voltage (V ) of 5 volts. What is the capacitance (C) of the capacitor?

1 microcoulomb is equivalent to 1 × 10−6 coulombs. A capacitor has a charge of 20 microcoulombs, and a voltage of 4 volts, what is the capacitance of the capacitor? Remember to ﬁrst convert the charge to coulombs.

A capacitor has a capacitance (C) of 10 × 10−4 F ( farads), and a charge of 30 microcoulombs. What voltage was applied across the capacitor?

1.02 Negative indices mathspace.co

1 Chapter review 1

Simplify: a

x3 × 6x2

a4 × −3a5

Find the value of a and b in the equation:

Simplify: a

4y10 × 5y−4

(4a)−3

5x4 × (−3x−8)

3y−2 × 4y−3

(6x−2 × 2x3)4

Given a = m3n2, b = m−1n4, and c = m5n−3, find: a

Fill in the blanks: a

Simplify, giving your answers with positive exponents: a

(−2x3)4

Simplify, giving your answers with positive exponents: a

6m3 n7 × 9m6 n

Fill in each box. Evaluate if possible. a

−2y7 × 2y2 × −2y6 d

Simplify: a

a×b

a÷c

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

c×b a

(5mp)2 × mp−2

“The greatest shortcoming of the human race is our inability to understand the exponential function.” –Albert A. Bartlett

Big ideas • Algebraic expressions can be manipulated and transformed through different ways allowing for a deeper understanding of their structure and relationships. • Algebraic fractions can be manipulated by ensuring common denominators, performing operations and simplifying through factorisation.

Curriculum Outcomes VC2M10A01 factorise algebraic expressions by taking out a common algebraic factor VC2M10A04 expand binomial products and factorise monic quadratic expressions using a variety of strategies VC2M10AA07 factorise monic and non-monic quadratic expressions and solve a wide range of quadratic equations derived from a variety of contexts

Expand and factorise Chapter outline 2.01 Factorise algebraic expressions 2.02 Expand binomial products 2.03 Factorise binomial products 2.04 Factorise monic quadratic trinomials 2.05 Optional: Factorise non-monic quadratic trinomials 2.06 Complete the square Chapter 2 review

24 29 35 40 46 52 58

Engineers use algebraic factorisation and expansion to simplify complex equations representing the motion of robotic arms, spacecraft, and drones.

2.01

Factorise algebraic expressions After this lesson, you will be able to… • identify the highest common factor in algebraic expressions. • factorise the highest common factor out from algebraic expressions. • apply the distributive law to factorise algebraic expressions with common factors.

Highest common factor Any factorising problem should start with an inspection of the terms to see if there is a common factor. If there are common factors, other than 1, these should be factored out before any other work is done.

Example 1 Find the highest common factor for the following pairs of terms: a 9ab and 12ac

Create a strategy Break down the terms into factors. Note, the HCF of 9 and 12 is 3.

Apply the idea 9ab = 3 × 3 × a × b

Factorise 9ab

12ac = 3 × 4 × a × c

Factorise 12ac

HCF = 3 × a Select the common factors = 3a Simplify b 16c3d2 and 20c2d3

Create a strategy Break down the terms into factors. Note, the HCF of 16 and 20 is 4.

Apply the idea 16c3d2 = 4 × 4 × c × c × c × d × d

Factorise 16c3d2

20c2d3 = 4 × 5 × c × c × d × d × d

Factorise 20c2d3

HCF = 4 × c × c × d × d 2 2

= 4c d

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Select the common factors Simplify

c 4(x − 3) and y(x − 3)

Create a strategy Note that a common factor can be an expression in brackets.

Apply the idea 4(x − 3) = 4 × (x − 3)

Factorise 4(x − 3)

y(x − 3) = y × (x − 3)

Factorise y(x − 3)

HCF = (x − 3)

Select the common factor

Idea summary The ﬁrst step in any factorising problem is to identify and factor out the highest common factor. It can help to break each term down into factors and compare.

Factorise algebraic expressions Consider 9ab + 12ac We can write each term in factors: 3×3×a×b+3×4×a×c We can see that the common factors are 3 and a, so we factor 3a out, and what is left goes into a bracket. 9ab + 12ac = 3a(3b + 4c) We can view this as factorising using the law, AB + AC = A(B + C) A is the highest common factor, which is taken out. This extends to three or more terms. AB + AC + AD = A(B + C + D)

Example 2 Factorise the expression by taking out the highest common factor: 8 + 16x − 4x2

Create a strategy The terms have a highest common factor of 4, which we can factor out using AB + AC + AD = A (B + C + D)

2.01 Factorise algebraic expressions mathspace.co

Apply the idea 8 + 16x − 4x2 = 4 × 2 + 4 × 4x − 4 × x2 2

= 4(2 + 4x − x )

Write the terms as products of 4 Factorise

Example 3 Factorise the expression 4(x − 3) − y(x − 3).

Create a strategy Factorise using the law, AB + AC = A (B + C). Here the common factor is the binomial expression (x − 3).

Apply the idea 4(x − 3) − y(x − 3) = 4 × (x − 3) − y × (x − 3) Write the terms as products of (x − 3) = (x − 3) (4 − y)

Factorise

Example 4 Factorise the expression 2t2k7 + 18t9k9.

Create a strategy Factorise using the law, AB + AC = A (B + C).

Apply the idea The coeﬃcients have a common factor of 2, and both terms have factors of t2 and k7, so we can factorise 2t2k7 out. 2t2k7 + 18t9k9 = 2t2k7 × 1 + 2t2k7 (9t7k2) = 2t2k7 (1 + 9t7k2)

Write the terms as products of 2t2 k7 Factorise

Idea summary We can use the distributive law to factorise an algebraic expression like AB + AC. This means writing the expression with any common factors between the terms taken outside of the brackets: AB + AC = A(B + C) Factorising is the reverse of expanding.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2.01 Practice questions Understanding Ex 1

15n2m and 8nm2

16abnx and 12abmy

Ex 4

20x and 15x2n

6abmn and 8bnxy

12a7bm3y2 and 9ab5n2x3

x(x + 1) + 3(x + 1)

2x(4 − x) − (4 − x)

5y3 + 15y2 + 10y = 5y( y2 + ⬚ + ⬚)

Fill in the boxes to complete the factorisation: 2x3 + 4x2 + 2x = ⬚(x2 + 2x + 1)

ax2 − 3ax + 6a = ⬚(x2 − 3x + 6)

4kx2 + 12kx + 16k = 4k(⬚ + ⬚ + 4)

Find the highest common factor of the terms: 12p2r2t3, 9prt3, and 18p3r3. 4a–f, 5a–d, 6, 7

4c–h, 5c–h, 6, 7

Factorise: a

3ax2 + 6ax − 6a

2ax + 2ax + 4ax

4b3 − 16b2 + 20b

3xyz + 21xyz + 9xz

i Ex 3

16a2bx3y6 and 12b4m2n6x

Fluency 4

1–3

Trailblazer

Ex 2

1–3

Adventurer

Find the highest common algebraic factor of each pair of terms:

1–3

Explorer

−6efgh2 − 30efgh − 12ehg

f h

4m (n − 3) + 6(n − 3)

( y + 4) ( y + 7) + x( y + 7)

6qr2 − 36qr − 12q

−4abc2 + 8abc + 4ac

3m2(n + 2) + 5(n + 2)

2y2(x + 2) − 3(x + 2) 2p3q(r + 4) + 5q(r + 4)

5xy2 + 10xy + 15x

10rstu2 + 70rstu − 40rtu

a e

2x3 − 16x2 + 2x

Factorise: 2

4e–j, 5g–l, 6, 7

f h

5y3(z + 5) + 7(z + 5)

6a2b(c − 2) − 4b(c − 2) 2f(g + h) + (g + h)2

4(7 − x2) − 2y(7 − x2) + 5(x2 − 7)

3(8 − c2) − 5d(8 − c2) − 4(c2 − 8)

(3m − 2) (4m + 7) − (2m + 3) (4m + 7)

(6z − 5) (2z + 4) + (4z + 2) (2z + 4)

A farmer wants to fence a rectangular field with an area of 48xy − 24y. If the length of the field is 24y, find an expression for the dimension of the width of the field.

Factorise: a

20j − 15jk − 55j2k

2yz − 10xy + 12xy2z

−hf − hj + hg

3p2q2 + 4p4q4 + 5p6q6

pqr + p2q2r + p3q3r

25m − 15mn − 10m2n

2.01 Factorise algebraic expressions mathspace.co

Reasoning

8, 9, 11

9–11

12x4 + 18x3 − 24x2 can be factorised into the form 6x(2x3 + 3x2 − 4x). Has the expression been fully factorised? Explain your answer.

Xander was asked to factorise the expression 35x2y + 10xy2 − 5xy. He stated that the highest common factor was 5xy so the factored form of the expression is 5xy(7x + 2y). Identify and correct Xander’s error.

Explain why the highest common factor of the variables in any expression has the least possible index of any of the terms.

When factorising an algebraic expression, explain how you can check your answer.

Problem-solving 12, 14, 15 12

13–15

A rectangle’s length is three times its width. Its length can be represented by the expression: 3x2 + 6x + 3

Factorise this expression by taking out a common factor.

Interpret what the factors represent in the context of this problem.

A farmer wants to create a set of adjacent fields which all have the same width. He plans to create the smallest field in the shape of a square, with the largest field 9 times the size of the smallest field, and the middle field to have a length that is 5 units more than the smallest field. a

Write expressions for the area of each ﬁeld.

Use factorising to determine the dimensions of the entire ﬁeld.

A rectangle has area 4x2 + 8x cm2 and length 4x cm. For what values of x is it a square? For what values is it just a rectangle?

Using the digits 0 to 9, at most once each, place a digit in each box to create a true statement. ⬚x + ⬚⬚ = ⬚(⬚x + ⬚⬚)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2.02

Expand binomial products

After this lesson, you will be able to… • understand and apply the rule for expanding binomial products. • visualise the expansion of binomial products using area models. • expand perfect squares. • understand and apply the difference of two squares identity. • solve problems requiring the expansion of binomial products.

Expansion of binomial products A binomial is a type of algebraic expression that consists of two terms. A binomial product refers to the result obtained when multiplying two binomials. A C

+ D

Consider the expression (A + B) (C + D). We can represent it visually using this area model. This diagram shows that (A + B) (C + D) = AC + AD + BC + BD. In other words, the area of the whole shape is the sum of the smaller areas.

Let’s look at using this model to expand (x + 3) (x + 5). Here we can see that:

x × x = x2

(x + 3) (x + 5) = x2 + 3x + 5x + 15

5 × x = 5x

We can simplify this to: x2 + 8x + 15

+ 3

3 × x = 3x

5 × 3 = 15

Example 1 Expand and simplify: (x + 10) (x + 5)

Apply the idea Use the rule (A + B) (C + D) = AC + AD + BC + BD. (x + 10) (x + 5) = x × x + x × 10 + 5 × x + 10 × 5 Apply the rule = x2 + 10x + 5x + 50 2

= x + 15x + 50

Simplify the products Combine like terms

2.02 Expand binomial products mathspace.co

Idea summary We can expand the product of two binomial expressions using the rule: (A + B) (C + D) = AC + AD + BC + BD

Expand perfect squares Exploration Consider the expansions of the binomials of the form (a + b) (a + b) = (a + b)2: • (x + 3) (x + 3) = x² + 6x + 9 • (x + 5) (x + 5) = x2 + 10x + 25 • (x + 6)2 = x² + 12x + 36 • (2r + 3s)2 = 4r2 + 12rs + 9s2 • (5s − 3)2 = 25s² − 30s + 9 1. What do you notice about the linear coefficient of the product? 2. What do you notice about the constant of the product? 3. Is there a general rule for this type of product?

A perfect square is something of the form (A + B)2.

We can use the same area model to determine the expansion. A

+ B

Here we can see that the expansion is: (A + B)2 = A2 + 2AB + B2 In a similar way: (A − B)2 = A2 − 2AB + B2 We can use perfect square expansions whenever we have to expand a perfect square.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 2 Expand and simplify: (x + 2)2

Apply the idea Use the perfect square expansion: (A + B)2 = A2 + 2AB + B2 (x + 2)2 = x2 + 2 × 2x + 22

Use perfect squares

= x + 4x + 4

Evaluate

Idea summary A special case of expanding binomials is for perfect squares: • (A + B)2 = A2 + 2AB + B2 • (A − B)2 = A2 − 2AB + B2

Difference of two squares Exploration Consider the expansions of the binomials of the form (a + b) (a − b): • (x + 3) (x − 3) = x² − 9 • (x − 5) (x + 5) = x5 − 25 • (x + 6) (x − 6) = x² − 36 • (2r + 3s) (2r − 3s) = 4r2 − 9s2 • (5s − 3) (5s + 3) = 25s² − 9 1. What do you notice about the linear coefficient of the product? 2. What do you notice about the constant of the product? 3. Is there a general rule for this type of product?

The difference of two squares identity is very useful in many contexts in mathematics. Consider expanding (A − B) (A + B), using the distributive law: (A − B) (A + B) = A2 + AB − AB − B2 As the middle terms, AB − AB = 0, the right hand simplifies and we have (A − B) (A + B) = A2 − B2 The right-hand side, A2 − B2, is a difference of two squares, which is where the identity gets its name.

2.02 Expand binomial products mathspace.co

Example 3 Expand and simplify: (m + 3) (m − 3)

Create a strategy Expand using the identity for difference of two squares: (A − B) (A + B) = A2 − B2

Apply the idea We are given: A = m and B = 3. (m + 3) (m − 3) = m2 − 32 2

=m –9

Apply the identity Evaluate the square

Idea summary A special case of expanding binomials is the difference of two squares: (A + B) (A − B) = A2 − B2

2.02 Practice questions Understanding 1

(2x) (3x)

(4y) (5y)

2–4

Trailblazer

(−3z) (7z)

(x + 2) (x + 1) = x2 + ⬚ + 2

(x − 3) (x + 2) = x2 − ⬚x − 6

(x + 2) (x − 5) = x2 − 3x − ⬚

(x + 3)2 = x2 + ⬚ + 9

(5p) (−2p)

(x − 3) (x − 4) = x2 − ⬚x + ⬚ (x − 3)2 = x2 − ⬚x + ⬚

(x + ⬚)2 = x2 + ⬚x + 36

(3x − 8)2 = 9x2 − ⬚ + 64

(x + 5) (x − 5) = x2 − ⬚

(x + 2) (x − 2) = ⬚2 − 4

Fill in the missing boxes. a

Adventurer

Fill in the missing boxes. a

2–4

Fill in the missing boxes. a

Explorer

Multiply the terms. a

1–3

(x − ⬚) (x + ⬚) = x2⬚64

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

(x + 14) (x − ⬚) = x2 − ⬚

5a–d, 6a–ce, 8a–d, 9a–d, 11ab, 12a

Fluency Ex 1

(x − 1) (x − 8)

(x + 1) (x + 4)

(x − 3) (x − 6)

(x − 4) (x + 3)

(x − 7) (x − 5)

(x − 5) (x − 1)

(x + 5) (x − 6)

(x − 9) (x + 2)

Expand and simplify: a

(2x + 3) (3x − 4)

(3x − 2) (4x + 5)

(4x + 1) (x − 2)

(3x − 4) (4x + 5)

2(x + 3) (x−2)

3(2x−5) (x + 4)

5(2x + 1) (x − 2)

−3(3x + 2) (2x − 1)

Expand and simplify: b

Expand and simplify: (x + 7)2

(c + 6)2

(7 − z)

Expand and simplify: (3x + 7)2

5(m − 5)2

(e + 5)2

(−6 − a)

(a + b)

(a − b)2

(2y − 5)2

(2y − 3)2

(x2 + 1)2

6(n + 6)2

( y − 5)2

8(q + 3)2

3x(7x − 5y)2

Expand and simplify: a

(x − 7) (x + 7)

(8 − m) (m + 8)

(7 − 6p) (7 + 6p)

(−9y − 1) (−9y + 1)

(9x − 2y) (9x + 2y)

(0.2x + 3) (0.2x − 3)

Expand and simplify: a

3x(5x − 6y) (5x + 6y)

3(9x − 8y) (9x + 8y)

a + (9 − a) (9 + a) − 10

[(5t + 9) − 6r] [(5t + 9) + 6r]

6(8x − 9y) (8x + 9y)

5x(2x − 9y) (2x + 9y)

Reasoning 14

Expand and simplify: a

Ex 3

5f–h, 6df–h, 7, 8c–h, 9d–h, 10, 11c–f, 12, 13

Expand and simplify:

a Ex 2

5d–f, 6d–f, 7, 8c–g, 9b–f, 10, 11b–d, 12ab, 13ab

14, 17

16, 17

15–18

Is (3x + 4)2 equal to 9x2 + 16? Explain your answer. 2.02 Expand binomial products mathspace.co

Show and explain how 36 × 24 may be multiplied using the structure of a difference of two squares.

Given two expressions (x + 3) (x + 3) and 13 × 13, perform the multiplication and describe the similarities between them.

A student wrote (a + b)2 = a2 + b2. Explain why they are incorrect and write the correct answer.

Consider the expression a

Explain how to expand and simplify the expression.

Is the result a trinomial?

Problem-solving 19, 20 19

19–22

21–24

The length of a rectangular ﬁeld is (x + 10) metres and the width of the ﬁeld is (x − 2) metres. a

Find the simpliﬁed expression for the perimeter of the ﬁeld.

Find the simpliﬁed expression for the area of the ﬁeld.

If x = 6, ﬁnd the length, width, perimeter, and area of the ﬁeld.

A cube has side length x + 3: a

Find the surface area of the cube in expanded form.

Find the volume of the cube in expanded form.

x+3

A square of side lengths measuring (x − 1) centimetres has each side enlarged by a factor of 2. Write an expression, in expanded form, for the area of the new square.

Form an expression in expanded form for the shaded area in the given ﬁgures. a

5x − 7

x 8x − 5y

8x + 5y 5x + 7

If the smallest of three consecutive integers is n, ﬁnd the product of the three integers in terms of n. Write your answer in expanded form.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

A rectangular prism has side lengths y, 12 − y, and y + 8. a

Write an expression for the volume of the prism in expanded form.

Determine which of these are viable options for the length y and explain why or why not in relation to both the side lengths and volume.

y = −10

y = 20

y=0

iii

y=2

y y+8

12 − y

y = 12

By trial and error or by using technology, determine the maximum volume of the prism.

2.03

Factorise binomial products

After this lesson, you will be able to… • understand the process of factorising binomial products. • factorise expressions by identifying and grouping common factors. • understand and apply perfect square factorisation. • understand and apply the factorisation of the difference of two squares. • apply the knowledge of factorising binomial products to solve problems.

Factorise by grouping We’ve seen the distributive law: A(B + C) = AB + AC We have also seen how the distributive law applies when the ﬁrst term is a binomial expression: (A + B) (C + D) = A(C + D) + B(C + D) This means that if we have an expression such as: y(x + 4) + 5(x + 4) we can write it as a product of two brackets: ( y + 5) (x + 4)

Example 1 Fully factorise 5(a + b) + v(a + b)

Create a strategy Identify the common factor between the two terms, then factorise using the rule: A (B + C) + D (B + C) = (A + D) (B + C)

2.03 Factorise binomial products mathspace.co

Apply the idea We can see that (a + b) is a common factor of both terms. 5 (a + b) + v (a + b) = (5 + v) (a + b)

Use the rule for factoring

Example 2 Factorise this expression by grouping in pairs: 2x + xz − 40y − 20yz

Create a strategy To factorise an expression with four terms, group the expression in pairs so that each pair has a common factor.

Apply the idea We can then factorise both pairs, since x is a common factor of (2x + xz) and 20y is a common factor of (40y − 20yz). 2x + xz − 40y − 20yz = x (2 + z) − 20y (2 + z) = (2 + z) (x − 20y)

Factorise by x and 20y Simplify

Idea summary We can use the rule (A + B) (C + D) = A(C + D) + B(C + D) = AC + AD + BC + BD. to factorise expressions in terms of the product of two binomial expressions.

Factorise using perfect squares We’ve seen how to expand perfect squares and use a model to visualise it. (A + B)2 = A2 + 2AB + B2 A

+ B

When factorising an expression of the type A2 + 2AB + B2: • Look for a common factor to both terms first as this can simplify the expression for the remaining steps. • Identify values for A and B that satisfy this form. • Write in factorised form using the rule (A + B)2 = A2 + 2AB + B2.

In our work in quadratics, the first term A, is often the variable x, so for the expression x2 + 4x + 4, we would identify A = x and B = 2, because 22 = 4. The middle term 2AB holds true as 2 × x × 2 = 4x.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 3 Factorise x2 − 12x + 36

Create a strategy Using the perfect squares rule (A + B)2 = A2 + 2AB + B2.

Apply the idea We can use the perfect squares rule because half of −12 squared is (−6)2 = 36. Using the perfect squares rule, A = x and B = −6. x2 − 12x + 36 = (x − 6)2 Use the rule

Idea summary We can factorise the product of two binomial expressions using the rule (A + B)2 = A2 + 2AB + B2 (called a perfect square)

Factorise using the difference of two squares We know that the difference of two squares looks like this: A2 − B2 = (A − B) (A + B) When factorising an expression of the type A2 − B2: • Look for a common factor to both terms first as this can simplify the expression for the remaining steps. • Identify the terms that are the squares A and B. • Write in factorised form using the rule A2 − B2 = (A − B) (A + B).

Example 4 Factorise: 121m2 – 64

Create a strategy Factorise using the difference of two squares, A2 − B2 = (A + B) (A − B)

Apply the idea We can tell that this is a difference of two squares expression because there are 2 terms that are both perfect squares, and the sign between them is a minus (−). Since (11m)2 = 121m2 and 82 = 64, we can use A2 − B2 = (A + B) (A − B), where A = 11m and B = 8.

121m2 − 64 = (11m + 8) (11m − 8)

2.03 Factorise binomial products mathspace.co

Idea summary We can factorise expressions of the form A2 − B2 using the rule: A2 − B2 = (A + B) (A − B) (called the difference of two squares)

2.03 Practice questions Understanding 1

1–5

Identify the highest common factor: b

32(c + 5) − x(c + 5)

6d4 + 8d2

⬚ = 42

Fill in the box to make the equation true: 81 =⬚2

100 =⬚2

One linear factor of x2 − 16 is x − 4. What is the other factor?

One linear factor of n2 + 12n + 36 is n + 6. What is the other factor?

Are these expressions a difference of two squares? a2 – 16

6a–d, 7a–d, 8a–d, 9a–d

Fluency 6

4w2 – 25a2

6b–e, 7c–g, 8a–f, 9c–g, 10a–d

25 = ⬚2

100 + 25y2

6e–h, 7e–j, 8f–j, 9f–l, 10

Factorise: a

m(x + y) − 6(x + y)

x( y − z) − w( y − z)

5y(4w + 3x) − z(4w + 3x)

8y( y − 4) + 3(4 − y)

3f(g + h) + (g + h)

(2c − d) (c + 5d) − 3(d − 2c)

3x2(x + 4y) − 5y(x + 4y)

x2 + 2x + 5x + 10

x2 − 3x + 8x − 24

a + 8a + a + 8

8x + xz − 16y − 2yz

24 + 3y + 8x + xy

7xy + wx + 7yz + wz

2mp + 6 + 3p + 4m

4x + 24yz + 32xy + 3z

2x + 18yz + 12xy + 3z

Factorise: a

b2 – 9b – 81

1–5

Trailblazer

10xy + 25x2

2a + 18b

Ex 2

Adventurer

Ex 1

Explorer 1–3, 5

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2y(2x2 + 3z) − (2x2 + 3z)

3df 2 − 11ydf + 3rzf − 11yrz

Ex 3

Factorise: a

x2 + 6x + 9

49 − 14p + p

e g

6c2 − 24c + 24 16w2 − 40w + 25

i Ex 4

Factorise: a

x2 − y2

e i 10

n2 − 25

f 25m2 − 49

81x2 − 16y2

3x2y − 6y4 + 9x2y3

121c2 − 49d2

8a b + 12bc

g i

2 4

b2 − 2br + r2

v2 − 1

121 − v2

16 − 9y2

x2y2 − 49

7x2 − 63

45t2 − 20

16x2y2 + 24xyz + 9z2

14h3p2r + 21h5p2 + 49hp4

16q4 + 24q2r + 9r2

Reasoning

11, 12

5q2 + 10qt + 5t2 64r2 + 48r + 9

u2 − 2uq + q2

Factorise: a

64 + 16e + e2

3x2y + x2z + 3y2 + yz (b + 4)2 − (3 − b)2

16w2 − 64wt2 + 6wt − 24t3 48df 2 − 3dh2

12–14

Does the sum of two squares, a2 + b2, equal (a + b) (a + b)? Explain your reasoning.

Identify and correct the error in each statement. a

x2 − 4x + 4 = (x + 2) (x + 2)

What strategy or combination of strategies would you use to factorise? Explain your choice.

Factorise the expression fully.

x2 + 2x + 1

2x − 6x + 5x − 15

(x + 16)2 − y2

n3 − 121n

9x2 − 12xy + 4y2

4x3 + 16x2 − x − 4

Explain why the trinomial 4a2−72a + 9 cannot be factorised using perfect squares.

Problem-solving 15ac, 16 15

For each expression:

e 14

x2 + 4x + 3 = (x + 2) (x + 2)

15bd, 16–18

Using the difference of two squares, evaluate: a

652 − 632

992 − 972

15d, 16–20

452 − 422

582 − 522

2.03 Factorise binomial products mathspace.co

A square has area t2 + 14t + 49 units2:

Find the side length of the square.

Find the perimeter of the square.

Using four of the digits 1 to 9 with no repeats, ﬁll in the blanks to create a perfect square trinomial. x2 + ⬚⬚x + ⬚⬚

Factorise: a

x4 + 8x2 + 16

a4 − 18a2 + 81

Factorise x4 − 1.

The area of a quilt can be expressed as x2 + 6x + 4x + 24. a

Label the area model so that it represents the area of the quilt.

Express the area of the quilt in factorised form.

2.04

Area I

Area II

Area III

Area IV

actorise monic quadratic F trinomials

After this lesson, you will be able to… • find two numbers that satisfy the conditions of the trinomial’s coefficients. • factorise monic quadratic trinomials.

Factorise monic quadratic trinomials A monic quadratic trinomial is an expression with three terms (trinomial), whose highest power is 2 (quadratic) and whose coefficient of the x2 term is a 1 (monic). This means that quadratics of the form ax2 + bx + c, when a = 1 are monic quadratic trinomials. Let’s look at how we can approach trying to factorise an expression in the form x2 + bx + c. If we can ﬁnd numbers p and q so that b = p + q and c = pq then we can write x2 + bx + c = x2 + ( p + q)x + pq 2

Since b = p + q and c = pq

= x + px + qx + pq

By expanding the middle term we have four terms

= x(x + p) + q(x + p)

Factorise the ﬁrst two and the last two terms

= (x + q) (x + p)

Finish the factorisation

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

We can see that we have just used AC + AD + BC + BD = (A + B) (C + D) so once we have identiﬁed p and q, we can write (x + p) (x + q) To factorise a monic quadratic trinomial of the form x2 + bx + c, we want two numbers p and q so that when you add them you get the coefficient of the middle term (b) and when you multiply them you get the constant term (c). For example: • To factorise x2 + 8x + 12, we want two numbers that sum to 8 and multiply together to give 12. The two numbers would be 6 and 2. • To factorise x2 + 4x + 3, we want two numbers that sum to 4 and multiply together to give 3. The two numbers would be 3 and 1. • To factorise x2 − x − 6, we want two numbers that sum to −1, and multiply together to give −6. The two numbers would be −3 and 2.

Example 1 Factorise: x2 + 6x + 8

Create a strategy

Apply the idea

Find two numbers p and q that add to 6 and multiply to 8, then factorise using (x + p) (x + q).

Notice that 6 and 8 are positive, then both p and q must be positive. We need factors of 8 that add to 6, which are 2 and 4. x2 + 6x + 8 = (x + 2) (x + 4)

Factorise

Example 2 Factorise: x2 − 17x + 60

Create a strategy

Apply the idea

Find two numbers that add to −17 and multiply to give 60, then factorise.

We need to choose factors of 60 that have sum of −17. The factors are −5 and −12. x2 − 17x + 60 = (x − 5) (x − 12)

Factorise

Example 3 Factorise: 3x2 + 6x − 45

Create a strategy There is a common factor of 3 to factor out. Factorise the remaining quadratic.

2.04 Factorise monic quadratic trinomials mathspace.co

Apply the idea 3x2 + 6x − 45 = 3 (x2 + 2x − 15)

Factor out the common factor

If we now consider the trinomial in the bracket, we need to choose factors of −15 that have sum of 2. The factors are −3 and 5. 3x2 + 6x − 45 = 3 (x − 3) (x + 5)

Factorise the quadratic trinomial

Idea summary An expression of the form x2 + bx + c is a monic quadratic trinomial. To factorise expressions like this, we ﬁnd a pair of numbers p and q such that p + q = b and pq = c. Then the factorisation is x2 + bx + c = (x + p) (x + q).

2.04 Practice questions Understanding 1

1–3

Explorer

Sum: 19

Product: 90

x2 + 16x + 64

(m + 4) (m + ⬚) = m2 + 10m + 24 2

( y − 14) ( y + ⬚) = y − 12y − 28

5ab, 6a–f, 7a–f, 8a–f, 9ab

x2 + 3x + 2x + 6

Sum: 3 Product: −88

x2 + x − 30

(m + 8) (⬚ + ⬚) = m2 + 18m + 80

(x − 2) (⬚ − ⬚) = x2 − 6x + 8

5bc, 6c–h, 7c–h, 8c–h, 9a–h

t2 + 7t + 8t + 56

Factorise: a

Product: 24

Factorise: c

Sum: −10

Find the values of a and b that will make the equation true: y2 + ay + 35 = ( y + 5) ( y + b).

Ex 1

Product: −63

x2 + 11x + 18

Fluency 5

Complete the missing values in each equation. a

Sum: −2

For each of the quadratic expressions, list all factor pairs of the constant term. a

1–3

Trailblazer 1, 3, 4

Find a pair of numbers with the given sum and product. a

Adventurer

x2 + 11x + 24

t2 + 7t + 10

x + 7x + 12

x2 + 8x + 16

t2 + 8t + 7

y + 9y + 14

x2 + 4x + 5x + 20

y2 + 6y + 12y + 72

x2 + 17x + 72

y2 + 9y + 20

40 + 13x + x

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5cd, 6g–l, 7g–l, 8g–l, 9e–l

d h l

x2 + 16x + 64 x2 + 8x + 15

36 + 12x + x2

Ex 2

Factorise: a

x2 − 4x + 3

x2 − 16x + 60

x2 − 19x + 84

x − 8x + 7

y − 10y + 16

t − 13t + 42

i 8

y2 − 18y + 77

t2 − 20t + 96

x2 − 18x + 45 x2 − 3x − 54

x2 − x − 30

x2 + x − 20

x − 3x − 70

x + 4x − 117

y − 13y − 48

y2 − 12y − 45

x2 + 10x − 56

4x2 + 24x + 32

Factorise: e i

2x2 + 14x + 20 5x2 + 5x − 30 −8 − 6x − x2

Reasoning

f j

−2x2 − 18x + 44

10–12

x2 + 10x − 24

4x2 − 4x − 288

−12 + 7x − x2

x2 − 20x + 51

x2 − 12x + 11

x2 − x − 6 2

x2 − 12x + 35

Factorise:

i Ex 3

−4x2 + 12x + 40 x3 + 7x2 + 12x

10–12, 14

t2 + 8t − 84

x2 + 3x − 28

3x2 − 21x + 30

h l

−3x2 + 12x − 12

2x3 + 16x2 + 30x

12–16

Octavia and Rajeev are factorising −3x2 − 15x − 18. • Rajeev states that the first step in factorising the trinomial is to find two numbers that multiply to −18 (the constant term) and add to −15 (the coefficient of the middle term). • Octavia states that the first step is factorising the highest common factor, which is −3. Who is correct? Explain and factorise −3x2 − 15x − 18.

Omar and Dima both factorise the expression x2 + 24x + 144. Their work is shown. Omar:

x2 + 24x + 144 = x2 + 2(x) (12) + (12)2 = (x + 12)2

Dima:

⬚ + ⬚ = 24

⬚ × ⬚ = 144

x + 24x + 144 = (x + 12) (x + 12) = (x + 12)2

Explain and compare each of their strategies. 12

Identify and correct the error in each statement. a

x2 + 7x − 18 = (x + 2) (x + 9)

x2 − 11x + 30 = (x − 5) (x + 6)

Suppose we want to write the trinomial x2 + bx + c in the form (x + p) (x + q). For each condition, explain what must be true about the signs of p and q. a

c is negative.

c and b are both positive.

c is positive and b is negative.

2.04 Factorise monic quadratic trinomials mathspace.co

A square table of side length x has one of its dimensions decreased by 4. This can be expressed visually by the area model shown.

x 4

Determine if the following expressions are equivalent to the area x (x − 4). If so, explain why. If not, show what they are equivalent to. a b

x2− Area II

Area I

d 15

Maximises the value of b.

(x + ⬚) (x + ⬚)

Minimises the value of c. (x + ⬚) (x + ⬚)

Minimises the value of b. (x + ⬚) (x − ⬚)

A square ﬁeld with side length x will have its length extended by a and its width extended by b. Both a and b are integers. a

What values of a and b would maximise the length of the entire ﬁeld? Explain.

What values of a and b would minimise the difference in area of the bottom left and top right ﬁeld? Explain.

17–20, 21ac

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

10m

Find an expression for the total area of the rectangles in factorised form.

19, 20, 21bd, 22, 23

One expression for the area of the rectangle is m2 + 14m + 45. The rectangle is made up of four smaller rectangles. Use the diagram to express the area of the large rectangle in factorised form.

Area II

x2− Area I

Problem-solving 17–19 17

Area I

Find two digits (1 to 9) to ﬁll in the blanks and create a trinomial in the form x2 + bx + c that

Area I + Area II

a b

Factorise u2 − 11u + 24.

Factorise (x2 + 2x)2−11(x2 + 2x) + 24.

The area of a patchwork artwork can be expressed as x2 + 8x + 15. a

Express the area of the quilt in factored form (x + p) (x + q).

What is the difference in the side lengths?

If both side lengths are tripled, what is the new expanded expression for the area of the artwork?

Factorise: a

x4 − x2 − 12

(a + 3) + 3(a + 3) − 10

y4 − 5y2 + 4

(d − 1)2 − 12(d − 1) + 36

A square with side length y metres has c metres added to its length and d metres added to its width. The resulting rectangle has an area of ( y2 + 8y + 15) m2. Find the values of c and d.

Students decide to make rectangular appreciation cards for their family members. The total area of each card is equal to x2 + 2x − 3 cm2. If the shorter sides of a card are 6 cm in length and the longer sides are 10 cm, ﬁnd the value of x.

2.04 Factorise monic quadratic trinomials mathspace.co

2.05

Optional: Factorise non-monic quadratic trinomials

After this lesson, you will be able to… • understand the difference between monic and non-monic quadratic trinomials. • understand and apply the method of factorising non-monic quadratic trinomials by grouping. • apply the splitting of the middle term to factorise non-monic quadratic trinomials. • factorise non-monic quadratic trinomials.

Factorise non-monic quadratic trinomials A non-monic quadratic trinomial is an expression of the form: where a ≠ 1.

ax2 + bx + c

The general approach to factorising a quadratic trinomial of the form ax2 + bx + c: 1. Factor out any HCF. (If a is negative it is advised to factor out -1 at this stage). 2. Find two numbers, p and q that add to give b and multiply to give ac. 3. Rewrite the trinomial with four terms in the form ax2 + px + qx + c. 4. Factorise by grouping. 5. Check that the answer will not factorise further, and verify the factorised form by expanding. We will explore this approach in the following examples.

Example 1 Factorise: 2x2 + 11x + 12

Create a strategy Find numbers that multiply to give 2 × 12 = 24 and add to give 11. Use those numbers to split the middle term and factorise by grouping.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Apply the idea 8 × 3 = 24 and 8 + 3 = 11, so split the middle term into 8x and 3x. 2x2 + 11x + 12 = 2x2 + 8x + 3x + 12

Split the middle term

= 2x(x + 4) + 3(x + 4)

Factorise the ﬁrst two and the last two terms

= (2x + 3) (x + 4)

Finish the factorisation

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: (2x + 3) (x + 4) = 2x2 + 8x + 3x + 12 2

= 2x + 11x + 12

Expand the brackets Collect like terms

Example 2 Factorise: 6x2 − 26x + 24

Create a strategy There is a common factor of 2 to factor out. Look for numbers p and q to split the middle term, then factorise by grouping.

Apply the idea 6x2 − 26x + 24 = 2 (3x2 − 13x + 12)

Take out the common factor of 2

The trinomial in the bracket is 3x2 − 13x + 12 so we need to ﬁnd numbers p and q so that pq = 36 and p + q = −13. p = −9 and q = −4 will do this. 6x2 − 26x + 24 = 2 (3x2 − 9x − 4x + 12)

Use p = −9 and q = −4 to split the middle term

= 2 (3x(x − 3) − 4(x − 3))

Factorise the ﬁrst two and last two terms

= 2 (3x − 4) (x − 3)

Complete the factorisation by grouping

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: 2 (3x − 4) (x − 3) = 2(3x2 − 13x + 12) 2

= 6x − 26x + 24

Complete the binomial expansion Do the remaining expansion

2.05 Optional: Factorise non-monic quadratic trinomials mathspace.co

Example 3 Factorise the expression by ﬁrst taking out a common factor: −6x2 − 21x + 45

Create a strategy There is a common factor of −3 to factor out. Look for numbers p and q to split the middle term, then factorise by grouping.

Apply the idea −6x2 − 21x + 45 = −3(2x2 + 7x − 15)

Take out the common factor of −3

The trinomial in the bracket is 2x2 + 7x − 15 so we need to ﬁnd numbers p and q so that pq = −30 and p + q = 7. p = 10 and q = −3 will do this. −6x2 − 21x + 45 = −3(2x2 + 10x − 3x − 15)

Use p = 10 and q = −3 to split the middle term

= −3(2x(x + 5) − 3(x + 5))

Factorise the ﬁrst two and last two terms

= −3(2x − 3) (x + 5)

Complete the factorisation by grouping

Reﬂect and check We can check our factorisation by expanding to make sure we get the original expression: −3 (2x − 3) (x + 5) = −3(2x2 + 7x − 15) 2

= −6x − 21x + 45

Complete the binomial expansion Do the remaining expansion

Idea summary The steps in factorising a quadratic trinomial of the form ax2 + bx + c 1. Factor out any HCF. (If a is negative it is advised to factor out −1 at this stage). 2. Find two numbers, p and q that add to give b and multiply to give ac. 3. Rewrite the trinomial with four terms in the form ax2 + px + qx + c. 4. Factorise by grouping. 5. Check that the answer will not factorise further, and verify the factorised form by expanding.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2.05 Practice questions Understanding

To factorise −8x2 − 40x − 48, what is a possible ﬁrst step?

Fill in the boxes to complete each factorisation. c

8x2 + 11x + 3 = (8x + 3) (x + ⬚)

Ex 3

4a–d, 5a–d, 6a, 7, 11

Factorise fully:

2x2 + 3x − 20 = (2x − 5) (x + ⬚)

10x2 + 29x + 21 = (⬚ + 3) (5x + ⬚)

4c–f, 5b–e, 6a–e, 8, 9, 11

3x2

18x

4e–h, 5e–h, 6d–h, 8–10

3y2 + 28y + 9

2x2 − 11x − 40

3x2 − 25x + 28

12t − 13t − 4

81x + 72x + 16

12x + 7x − 10

6x2 + 41x − 56 9x2 − 42x − 15

Factorise fully:

8x2 − 21x − 9

3x2 − 21x − 54

4x2 + 40x + 100

8s2 + 6s − 54

8x − 20x + 8

72x − 6x − 6

105x − 50x + 5

168x2 − 88x − 16

−6x2 + 25x − 14

−12x2 − 12x + 45

−6x2 + 22x + 40

Factorise fully: e

1–3

Trailblazer

Find an expression for the total area of the rectangle, 3x2 + 19x + 6, in factorised form.

2x2 − 19x + 45 = (⬚ − 9) (⬚ − 5)

Fluency

Ex 2

1–3

Adventurer

Ex 1

1–3

Explorer

−6x2 + 5x + 6

−4x2 + 11x − 6

−9x2 + 24x − 16

−5x2 − 14x − 8

The length of a rectangle is L = −56y + 11 and the width is W = 5y2. a

What is the perimeter of the rectangle in terms of y in expanded form?

What is the perimeter in fully factorised form?

−3x2 − 10x + 8 W

The side length of a regular pentagon is S = 2x2 + 21x + 49. a

What is the perimeter of the pentagon in terms of x in expanded form?

What is the perimeter in fully factorised form?

2.05 Optional: Factorise non-monic quadratic trinomials mathspace.co

A cube has a surface area of (6x2 + 36x + 54) square units, where x > 0. a

Factorise 6x2 + 36x + 54 fully.

What is an expression for the length of a side of the cube?

A rectangle has an area of 6x2 + 23x + 20. If the length and width are linear factors of 6x2 + 23x + 20, what are the dimensions of the rectangle in terms of x?

A ball is thrown from the top of a 140 m tall cliff, with an initial velocity of 50 m/s. The height of the ball after t seconds is approximated by the quadratic expression −10t2 + 50t + 140. Factorise this quadratic.

Reasoning 12

12, 13

13, 14

Consider each quadratic expression. The factorised form of the expression is (Ax + C) (Bx + D). Determine: i

the product of A and B

the signs for A and B: opposite or the same

iii

the product of C and D

the signs for C and D: opposite or the same

15x2 + 14x − 16

−45x2 + 29x − 4

Sheldon and Gabriella each factorised 16x2 + 48x + 36. Whose work is incorrect? Identify and explain the error. Sheldon: 2

Gabriella: 2

16x + 48x + 36 = 16x + 24x + 24x + 36

16x2 + 48x + 36 = (4x + 6)2

= 8x(2x + 3) + 12(2x + 3)

= 2 (2x + 3)2

= (8x + 12) (2x + 3) = 4(2x + 3) (2x + 3) = 4(2x + 3)2 14

Consider these polynomials: 2x2 + x + 1,

3x2 − 2x − 4,

6x2 + 10x − 5,

12x2 − 12x + 1

What do the polynomials have in common?

How can you show that a polynomial where a ≠ 1 cannot be factorised?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Problem-solving 15, 16, 20 15

15–17, 20

17–21

Tara is practicing diving. She springs up off a board 10 metres high, and after t seconds, her height in metres above the water is described by the quadratic: −5t2 + 5t + 10

Fully factorise the quadratic.

Substitute t = 2 into the factorised quadratic and ﬁnd the value of the expression.

Substitute t = 2 into the original quadratic and ﬁnd the value of the expression.

What is happening 2 seconds after Tara jumps off the board?

Consider the ﬁgure. a

Write an expression in expanded form for the area of the shaded region.

Write the expression for the area in factorised form.

The area of a rectangular ﬁeld is given by the general expression (6x2 + 17x + 12) m2 where x is a positive whole number. a

Find the length and width of the ﬁeld in terms of x.

Write an expression for the perimeter of the ﬁeld. Describe how to determine the expression for the perimeter.

If the perimeter of a particular ﬁeld is 114 metres, ﬁnd x.

A photographer wants to put a border around her photo. a

Why might the photographer use a variable to represent the width of the border?

What are the dimensions, length and width, of the total artwork if the total area of the photograph including the border is 4x2 + 24x + 35 square units?

What is the area of the photograph? Explain how you reached your solution.

Using the digits 1 to 9 with no repeats, ﬁll in the blanks to create a trinomial that can be factorised, and provide the factorised form: 4x2 + ⬚⬚x + ⬚.

Find the value of k that will make 16x2 − 24x + k a perfect square trinomial.

Quadratic trinomials can be factorised using the identity:

where m + n = b and mn = ac. Find the values m and n for the quadratic 4x2 − 14x + 12.

2.05 Optional: Factorise non-monic quadratic trinomials mathspace.co

2.06

Complete the square

After this lesson, you will be able to… • understand how to complete the square using visual representation. • understand and apply the process of completing the square in transforming quadratic expressions. • factorise expressions from the completed square form.

Complete the square We have seen the perfect square identity (A + B)2 = A2 + 2AB + B2 and we have seen how we can use this to factorise an expression such as x2 + 4x + 4 as (x + 2)2. All quadratics of the form y = ax2 + bx + c can be transformed into the form y = a(x − h)2 + k through the process of completing the square. This form is very useful for us in graphing, solving and identifying key features of quadratics, for now we will just focus on the algebraic skills associated with completing the square. From earlier in this chapter you will remember that Perfect squares : (x + a)2 = x2 + 2ax + a2

(x − a)2 = x2 − 2ax + a2

Difference of two squares : (x − a) (x + b) = x2 + b2 We also used a visual to help us understand the expansion of a perfect square. Let’s use the same ideas and look at how to complete the square, and how it got that name. Let’s start with the expression x2 − 4x + 23. We will start with the ﬁrst two terms and build a perfect square using our visual. x x

We know that x × x is x2, so we can put that in.

−2

−2x

−2 −2x

To show the −4x, we will have two rectangles each with an area of −2x. This is now a square, with a side length of (x − 2).

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−2

−2x

−2 −2x

To complete the square we will use −2 × −2 = 4. If the constant is 23, and we have used 4, there is 19 remaining.

So we can rewrite x2 − 4x + 23 = (x − 2)2 + 19. Algebraically the same example would look like this: x2 − 4x + 23 = x2 − 4x + (−2)2 + 23 − (−2)2 Add the square of half the coeﬃcient of x, but also subtract that same value so the expression’s value is not altered = (x − 2)2 + 23 − (−2)2

= (x − 2)2 + 19

Simplify the constant term

Identify the perfect square we created by adding the square of half the coeﬃcient of x

Example 1 Using the method of completing the square, rewrite x2 + 4x in the form a(x − h)2 + k.

Create a strategy Form a perfect square from the terms and see what the required constant is.

Apply the idea Use x and half of the coeﬃcient of x to make a perfect square, (x + 2)2 (x + 2)2 = x2 + 4x + 4 2

Expand the perfect square 2

It follows that x + 4x = (x + 2) − 4

Example 2 Write 2x2 + 4x + 7 in the form a(x − h)2 + k.

Create a strategy Factorise 2 from the ﬁrst two terms. Complete the square inside the bracket, re-expand and simplify.

Apply the idea 2x2 + 4x + 7 = 2(x2 + 2x) + 7 2

Factor 2 out from the ﬁrst two terms

= 2((x + 1) − 1) + 7

Complete the square on x2 + 2x

= 2(x + 1)2 − 2 + 7

Expand the bracket

= 2(x + 1) + 5

Simplify

2.06 Complete the square mathspace.co

Idea summary Completing the square is a method that allows us to write a quadratic trinomial ax2 + bx + c in the form a(x + h)2 + k where

and

Factorise from completed square form We have met the difference of two squares identity: (A − B) (A + B) = A2 − B2 We can use this to factorise an expression such as x2 − 9 x2 − 9 = (x − 3) (x + 3) or an expression such as x2 − 5

If we have a completed square expression such as (x + 2)2 − 25, we can factorise as: (x + 2)2 − 25 = ((x + 2) − 5) ((x + 2) + 5)

Use difference of two squares

= (x + 2 − 5) (x + 2 + 5)

Remove the brackets

= (x − 3) (x + 7)

Simplify

Example 3 Factorise x2 + 4x + 3 using the method of completing the square.

Create a strategy Complete the square. Use the difference of two squares identity (A − B) (A + B) = A2 − B2

Apply the idea x2 + 4x + 3 = (x + 2)2 − 4 + 3 2

= (x + 2) – 1 2

Complete the square on the ﬁrst two terms Simplify

= (x + 2) − 1

Write as difference of 2 squares

= ((x + 2) + 1) ((x + 2) − 1)

Factorise by difference of 2 squares

= (x + 3) (x + 1)

Simplify each bracket

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 4 Factorise 5(x + 3)2 − 15.

Create a strategy Factor out the common factor of 5 Use the difference of two squares identity (A − B) (A + B) = A2 − B2

Apply the idea Factor out the common factor Use the difference of two squares identity Remove the unnecessary brackets

Idea summary If completing the square leaves us with an expression of the form (x + h)2 − k we can use the difference of two squares identity A2 − B2 = (A − B) (A + B) to factorise the completed square form to

2.06 Practice questions Understanding 1

Explorer

1, 2

Adventurer

1, 2

Trailblazer

For each quadratic expression, calculate the square of half the coefficient of x: a

1, 2

x2 + 6x + 36

x2 + 8x + 12

x2 − 14x − 72

x2 − 10x + 24

Identify if the correct value for ‘the square of half the coeﬃcient of x’ has been applied in each expression, ensuring that the expression value is not altered. a b c d

x2 + 4x + 18 = x2 + 4x + 22 + 18 − 22 x2 − 10x − 21 = x2 − 10x + 52 − 21

x2 + 30x − 64 = x2 + 30x + 152 − 64 + 152

x2 − 24x + 80 = x2 − 24x + 122 + 80 − 122

2.06 Complete the square mathspace.co

3ac, 4, 5a–d, 6ac, 7a–h, 8a–d

Fluency 3

Complete each equation: a c

Ex 4

5x2 − 10x + 4 = ⬚(x2 − 2x) + 4

2x2 − 10x + 1 = ⬚(x2 − 5x) + 1

6x2 + 24x − 7 = ⬚(x2 + ⬚) − 7 6x2 + 36x − 1 = ⬚(x2 + ⬚) − 1

= ⬚(x2 − 2x + ⬚) + 4 − ⬚

= ⬚(x2 + 6x + ⬚) − 1 − ⬚ = ⬚(x + ⬚)2 − ⬚

x2 + 18x

x2 − 10x + 30 (x − 3)2 − 1

x2 − 8x

x2 + 10x + 31

x2 − 18x + 77

x2 + 9x + 16

x2 − 7x + 15

(x + 4)2 − 9

(x + 5)2 − 49

(x − 1)2 – 3

2x2 + 8x − 7

5x2 + 20x − 9

4x2 − 16x + 3

7x − 14x + 1

9x + 54x − 2

3x + 33x + 88

4x2 − 11x + 7

3x2 + 9x + 8

2x2 − 7x + 6

10x2 + 20x − 11 5x2 + 5x + 1

4x2 + 20x + 16 x2 + 11x + 10

Factorise fully using the method of completing the square: a

x2 + 6x + 4

x2 − 8x + 11

x2 − 6x + 5

x − 11x + 30

x + 24x + 63

x − 20x + 19

x2 − 28x + 115 −2(x − 9)2 + 32

Factorise fully:

3(x − 1)2 − 12

2(x + 4)2 − 50

4(x + 2)2 − 1

−4(x + 2) + 16

9(x + 2) − 4

4(x + 3) − 5

Reasoning

11, 12, 14a

10–12, 14ab

13, 14

Explain why we cannot factorise x2 + 6x + 13.

Alessandro is completing the square on x2 + 6x + 12. Here is his work:

x2 + 6x + 12 = (x2 + 6x + 9) + 9 + 12

= (x + 3)2 + 9 + 12

= (x + 3)2 + 21

What mistake has Alessandro made?

x2 + 14x + 47

Rewrite the following in the form a(x − h)2 + k by completing the square:

i Ex 3

Factorise fully: a

4x2 + 12x − 6 = ⬚(x2 + ⬚) − 6

3bd, 4, 5e–h, 6cd, 7e–l, 8e–h, 9c–h

Rewrite each quadratic in the form a(x − h)2 + k using the method of completing the square: e

Ex 2

= ⬚(x − ⬚)2 − 1

3x2 + 6x − 8 = ⬚(x2 + 2x) − 8

Complete each solution for completing the square: a

Ex 1

3bd, 4, 5c–f, 6bc, 7c–j, 8b–e, 9a–e

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2(x − 1)2 − 3

Beatrix is completing the square on 3x2 + 18x − 7. Here is her work:

3x2 + 18x − 7 = 3(x2 + 6x) − 7

= 3[(x + 3)2 − 9] − 7

= 3(x + 3)2 − 9 − 7

= 3(x + 3)2 − 16

What mistake has Beatrix made? 13

An equation is written as y = 3(x − 4)2 + 2. Explain why the minimum value of y is 2. For what value of x does y take the value 2?

Factorise each quadratic expression in 2 different ways. Show your working. a

x2 − 18x + 81

4c2 − 12c + 9

Problem-solving 15, 17 15

3a2 + 14a + 8

15–17

2x2 + 5x − 3

16–18

A square has an area of x2 − 10x + 25 cm2. a

What is the length of one of the sides?

What are the restrictions on the value of x?

A cube has a surface area of 6x2 − 48x + 96. What is the length of one of the sides?

Complete the square for the equation y = −x2 − 8x + 20.

What is the maximum value of y and at what value of x does this occur?

Using four of the digits 1 to 9 with no repeats, ﬁll in the blanks to complete the expression: x2 + ⬚x + ⬚ = (x + ⬚)2 − ⬚

2.06 Complete the square mathspace.co

2 Chapter review 1

Find the highest common factor of each pair of terms. a

20x and 15x2 n

16a2 bx3 y6 and 12b4 m2 n6 x

5y2 + 10y + 15

Factorise: a c

k2 − 8k −4bc2 + 8bc + 4c

3p2 q2 + 4p4 q4 + 5p6 q6

A student was asked to factorise the expression 35x2 y + 10xy2 − 5xy. They stated that the highest common factor is 5xy, so the factored form of the expression is 5xy (7x + 2y). Identify and correct the student’s error.

Expand and simplify: a

(x + 1) (x + 3)

(x + 5) (x − 6)

(x − 9) (x + 4)

(4y + 3)2 = 16y2 + 24y + 9

(y − 3)2 = y2 − 9y + 9

(3x − 0.7) (3x + 0.7) d

Expand and simplify: a

Are the expansions correct or not? If not, write the correct expansion. a

(x − 2) (x − 8)

(m + 4) (m − 4)

(8 − 5p) (8 + 5p)

Expand and simplify: a

−2 (3x−4) (4x + 3)

A square of side lengths measuring (x − 4) centimetres has each side enlarged by a factor of 3. Write an expression, in expanded form, for the area of the new square.

Factorise:

k3 + 8k2 + k + 8

35m2 − 49

5 (a + b) + x (a + b)

5x2 (x + 4y) − 5y (x + 4y)

Factorise:

5r2 + 10rs + 5s2

x2 − 10x + 25

x2 + 16x + 64

1 − 2x + x2

x + 4x − 117

y − 13y − 48

x + 10x − 24

x2 + x − 20

6x2 + 41x − 56

81x2 − 90x + 25

−6x2 + 22x + 40

Factorise fully: a

12t2 − 13t − 4

x2 + 17x + 72

A rectangle has an area of 20x2 + 39x + 18. If the length and width are linear factors of 20x2 + 39x + 18, what are the dimensions of the rectangle in terms of x?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The side length of the regular pentagon is S = 2x2 + 15x + 25.

What is the perimeter of the pentagon in terms of x in expanded form?

What is the perimeter in fully factorised form?

Find the value of k that will make 4x2 − 12x + k a perfect square trinomial.

Rewrite each quadratic in the form a (x − h)2 + k using the method of completing the square:

x2 − 8x

x2 − 4x + 6

x2 + 10x + 21

10x + 20x − 11

2x + 8x − 7

4x + 20x + 28

5x2 + 5x + 1

x2 − 11x + 30

Factorise fully using the method of completing the square: a

x2 + 6x + 4

Complete the square for the equation y = 20 − 8x − x2.

What is the maximum value of y and at what value of x does this occur?

x2 + 14x + 33

x2 + 42x + 185

x2 + 3x + 3

Did you know?

“Completing the square” might sound super fancy, but it’s actually an ancient maths trick! Way back in 2000 BCE, Babylonian mathematicians were already onto it, using geometry to solve equations by filling in squares to complete rectangles—kind of like finishing a puzzle. Fast forward to the 9th century, and the legendary Al-Khwarizmi took things up a notch, perfecting the method to solve quadratic equations and paving the way for modern algebra. Chapter 2 review mathspace.co

Big ideas • Graphs and models represent relationships between variables, with linear functions exhibiting a constant rate of change and non-linear functions demonstrating a varying rate of change. • Graphical representations of inequalities provide a visual tool to depict solution regions in the Cartesian plane. • Equations can be solved more easily by manipulating algebraic expressions and applying a series of organised steps and techniques. • Parallel and perpendicular lines have special rules that make them different from other lines. Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10A07 solve problems involving linear equations, including those derived from formulas VC2M10A08 solve linear inequalities and graph their solutions on a number line VC2M10A10 solve problems involving gradients of parallel and perpendicular lines VC2M10A12 solve linear equations involving simple algebraic fractions

Chapter outline

Linear equations Chapter outline 3.01 Linear graphs 3.02 Equations with more than 3 steps 3.03 Equations involving fractions 3.04 Rearrange equations 3.05 Parallel and perpendicular lines 3.06 Linear models 3.07 Graph linear inequalities Chapter 3 review

62 72 79 87 94 104 117 133

If you bought one less ice cream per week, how much would you save in a year? You can use a linear model to measure it!

3.01

Linear graphs After this lesson, you will be able to… • plot a series of points from a table of values on the Cartesian plane and connect them to form a linear graph. • plot linear graphs using the x-intercept and y-intercept as primary points. • plot a linear graph using its gradient and a specified point on the line.

Sketch from a table of values Each pair of values in a table of values represents the x-coordinate and y-coordinate for a point on the line. To plot from a table of values we identify the points from the table. x y

1 −2

2 1

3 4

4 7

The x and y-value in each column of the table can be grouped together to form the coordinates of a single point, (x, y).

(1, -2), (2, 1), (3, 4), (4, 7)

Example 1 For y = 2x − 4: a Complete the table of values. x y

Create a strategy

Apply the idea

Substitute each x-value into the given equation.

For x = 0:

For x = 1:

For x = 2:

For x = 3:

x y

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

y=2×0−4 = −4

Substitute x = 0 Evaluate

y=2×1−4 = −2

Substitute x = 1 Evaluate

y=2×2−4 =0

Substitute x = 2 Evaluate

y=2×3−4 =2

Substitute x = 3 Evaluate

0 −4

1 −2

2 0

3 2

b List the 4 coordinates from the table of values.

Apply the idea The coordinates are (0, −4), (1, −2), (2, 0) and (3, 2).

c By choosing any two coordinates from part (b), plot the line on the Cartesian plane.

Create a strategy Connect the plotted points.

Apply the idea Choosing points (0, −4) and (3, 2), we plot them on the Cartesian plane and connect them with a straight line. y 4 3

y = 2x − 4

2 1 −2 −1

−1

x 1

−2 −3 −4

Idea summary To plot a point, (a, b), on a Cartesian plane, we ﬁrst identify where x = a lies along the x-axis, and where y = b lies along the y-axis. If a series of points lie on a straight line, we can join them to represent all points that satisfy the relationship y = mx + c.

3.01 Linear graphs mathspace.co

Sketch using the gradient and a point We can graph a line by identifying the gradient and any point on the line.

Example 2 Sketch the line that has a gradient of −5 and an x-intercept of −4.

Create a strategy Use the gradient to ﬁnd another point starting from the x-intercept.

Apply the idea 6 5 4 3 2 1 −5 −4 −3 −2 −1

−5 −4 −3 −2 −1

The gradient is

−1 −2 −3 −4 −5 −6

so the rise is −5 and the run

is 1. We can use this to ﬁnd another point on the line. x 1

−1 −2 −3 −4 −5 −6

6 5 4 3 2 1

The x-intercept has coordinates of (−4, 0).

By moving right 1 unit and down 5 units we get to the point (−3, −5).

So now we can draw our line through these two points.

x 1

Example 3 Sketch the line that has a gradient of 2 and passes through the point (3, −1).

Create a strategy Plot the point (3, −1).

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

A gradient of 2 means that for every 1 unit we go to the right, we go 2 units up. A positive gradient means that the line will go upwards from left to right.

Apply the idea 8 7 6 5 4 3 2 1 −1

−1

−1 −2 −3 −4

Produce a series of points from (3, −1) using the gradient.

A gradient of 2 means that for each 1 unit we move to the right, we also move 2 units up. Alternatively, it means that for each 1 unit we move to the left, we also move 2 units down. x 1

−1 −2 −3 −4

8 7 6 5 4 3 2 1 −1

x 1

−1 −2 −3 −4

8 7 6 5 4 3 2 1

Plot the point (3, −1).

Draw the line through the points.

x 1

3.01 Linear graphs mathspace.co

Idea summary To sketch a line given the gradient and a point on the line: 1. Plot the point on the Cartesian plane. 2. Use the gradient to find one other point that lies on the line. 3. Draw a line through the two points that spans the plane. If we have the choice, the y-intercept is a good point to use.

Sketch by using the intercepts The x-intercept of a graph occurs where y = 0. To ﬁnd the x-intercept, set y = 0 and then solve the equation. The y-intercept of a graph occurs where x = 0. To ﬁnd the y-intercept, set x = 0 and then solve the equation. This method is often useful when the equation is not in gradient-intercept ( y = mx + c) form but in the two intercept form (ax + by = c).

Example 4 Sketch the graph of 2x + 3y = 12 by finding the x and y-intercepts.

Create a strategy First, ﬁnd the x-intercept by setting y to zero and solving for x. Next, ﬁnd the y-intercept by setting x to zero and solving for y. Then, plot the intercepts and sketch the line.

Apply the idea Solve for the x-intercept: 2x + 3y = 12 2x + 3(0) = 12 2x = 12 x=6

Write the equation Substitute y = 0 into the equation Simplify Divide both sides by 2 7 6 5 4 3 2 1

−7−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7

Plot the coordinate (6, 0).

x 1 2 3 4 5 6 7

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Solve for the y-intercept: 2x + 3y = 12

Write the equation

2(0) + 3y = 12

Substitute x = 0 into the equation

3y = 12

Simplify

y=4

Divide both sides by 3

7 6 5 4 3 2 1

−7−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7

7 6 5 4 3 2 1 −7−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7

Plot the coordinate (0, 4).

1 2 3 4 5 6 7

Sketch the line.

x 1 2 3 4 5 6 7

Idea summary The x-intercept occurs at the point where y = 0. The y-intercept occurs at the point where x = 0. We can use the x and y-intercepts to sketch a line.

3.01 Linear graphs mathspace.co

3.01 Practice questions Understanding 1

1–8

Explorer

Adventurer

1–8

Trailblazer

Complete the table: Gradient

y-intercept

y = 3x − 4 y = −5x + 2 y=4 x=3 2

To sketch a linear graph using its intercepts, you need to: A

Find the maximum and minimum points on the graph

Locate where the line crosses the x-axis and y-axis

Determine the gradient of the line

Calculate the midpoint of the line

What is the equation of the line parallel to the y-axis and passing through the point (6, 2)?

What is the equation of the line parallel to the x-axis and passing through the point (−2, −1)?

What are the equations of the x and y-axes?

Identify the x-intercept and y-intercept: a b

(2, 3), (0, −2), (5, −1), (6, 0), (3, 3) x y

−7 0

3 10

Find the gradient, m, and y-intercept, c, for each line: a

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

0 7

−2 5

x 1

2 3 4 5

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

x 1

2 3 4 5

y = 2x + 9

4x + 2y = 3

y=5

9d–h, 11, 13b–d, 14, 15b–d, 16e–h, 17b–d, 19, 20ab, 21

For each equation:

2 3 4 5

x y

x=2

9d–h, 10, 12, 13c–e, 14cd, 15d–f, 16e–h, 17cd, 19–21

−1

Complete the table of values:

Sketch the line on a Cartesian plane.

y=x−3

y=x+2

y = 2x − 7

y = 3x + 2

y = −x + 4

y = 2x − 1

y=3−x

y = −3x + 5

For

Complete the table of values:

Sketch the line.

x y

−2

For y = 2x + 3: a

Complete the table of values. x y

9a–e, 11, 13a–c, 14, 15a–c, 16a–d, 17a–c, 18, 19, 21

Fluency 9

Is the point (−2, 5) on the line? a

Ex 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

2 3 4 5

5 4 3 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

−1

Sketch the line on a Cartesian plane.

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the y-value when x = 3.

Find the x-value when y = −4.

Sketch the line on a Cartesian plane.

For 2x − 3y − 6 = 0: a

Complete the table of values. x y

−1

Find the coordinates of the y-intercept.

Find the coordinates of the x-intercept.

Find the gradient.

Find the y-value when x = 2. 3.01 Linear graphs mathspace.co

Ex 2

Ex 3

Ex 4

Sketch the lines on a Cartesian plane: a

A line that has an x-intercept of 4 and whose gradient is 3.

A line that has a y-intercept of 7 and whose gradient is −3.

A line that has a y-intercept of 4 and whose gradient is .

A line that has a y-intercept of −3 and whose gradient is 6.

A line that has an x-intercept of −6 and whose gradient is −3.

Sketch each linear relationship: a

A line that passes through the point (2, 3) and has a gradient of 1.

A line that passes through the point (0, −3) and has a gradient of 2.

A line that passes through the point (−1, 4) and has a gradient of −1.

A line that passes through the point (3, 0) and has a gradient of 3.

For each linear equation: i

Find the gradient.

Find the coordinates of the y-intercept.

iii

Use the gradient and y-intercept to sketch the line.

y = 2x + 4

y = −x + 5

3x − 2y + 6 = 0

4x − y + 4 = 0

Find the x-intercept.

Find the y-intercept.

iii

Use the intercepts to sketch the line.

y = 2x − 3

y = 2x + 3

y = −2x + 3

y = −2x + 5

y = 3x − 2

y = −3x + 1

y = −3x−4

y = 4x + 3

2x + y = 4

Sketch the graph by ﬁnding any two points on the line: y = −2x − 1

y = 4x − 2

Sketch the line using the y-intercept and any other point on the line: y = 2x + 7

y = 4x − 3

Find the x-intercept of the line 3x − 7y = 21.

For each linear equation: i

Find the x-intercept.

Find the y-intercept.

iii

Sketch the line.

3x + 2y = 6

4x − 7y = 28

y = −2x − 4

y = −3x + 6

4x − 3y = −24

10x + 9y = −90

Sketch the lines using the gradient and y-intercept: a

y = 5x − 2

For each linear equation:

a 18

y = 6x − 3

y = −6x + 2

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reasoning

22, 23

23, 24

23–25

Explain how the gradient of a line affects the direction in which the line slants. Why does a positive gradient slant upwards from left to right, while a negative gradient slants downwards?

Justify why knowing the x-intercept and y-intercept of a linear equation is suﬃcient to sketch its graph.

Complete the table: Gradient

y-intercept

y = kx + d y = −px + r y=k 25

Consider the linear equation y = mx + c. If we change m to 2m, how will the graph change? Explain your reasoning.

Problem-solving 26, 27 26

27–29

28, 30, 31

A toy manufacturer estimates that the number of toys they sell each week will decrease linearly as the price increases. Their prediction is shown in the table: Price Sales

$10 500

$20 400

$30 300

Sketch a graph using the table of values and estimate how many toys they would sell if the price was $25. 27

A car rental company charges a ﬂat fee of $60 per day plus $0.60 per kilometre driven. If a customer drives 100 kilometres in a day, they will be charged $120. How much would it cost for a customer to rent a car for a day and drive 250 kilometres?

Suppose a company’s profit (P ) in thousands of dollars is linearly related to the amount of goods (x) sold in thousands. The company breaks even (P = 0) when they sell 3000 goods, and they make a profit of $5000 when they sell 6000 goods. Determine the equation of the line that represents the company’s profit.

Sandro has been given the task of designing a straight path from the fountain to the library in a park. He knows the coordinates of the fountain (3, 5) and the library (7, 9). The gradient of the path needs to be 1. Write the equation of the path.

The cost (C) of manufacturing a certain product is a linear equation of the quantity (q) produced. The cost of producing 20 units is $250 and the cost of producing 60 units is $450. Find the equation that describes the cost as a function of quantity.

A runner is planning a straight route for their daily run. They want to start at point A and ﬁnish at point B. They know A has coordinates (2, 3) and B has coordinates (6, 8). They want the route to have a constant gradient. Write the equation of the route. 3.01 Linear graphs mathspace.co

3.02

quations with more than E 3 steps

After this lesson, you will be able to… • solve multi-step equations by using appropriate steps like expanding brackets and combining like terms. • apply the method of clearing fractions by finding the common denominator in equations with fractional terms. • demonstrate the process of isolating variables using inverse operations in equations. • identify errors in solving equations and correct them.

Equations with more than three steps We’ve solved linear equations with up to three steps. For three or more steps, we will have more terms that we need to move or combine. For the equations with more than three steps, we may follow these steps: 1. If necessary, expand any brackets and/or multiply both sides by the common denominator of any fractions. 2. Combine like terms on both sides of the equation. 3. Get all the variable terms on one side and all the constants to the other side using inverse operations. 4. Solve for the variable. We can still check our answer by substituting the solution into the original equation.

Example 1 Solve: 4 (2x + 5) − 3 = 37

Create a strategy Follow the steps for solving equations with more than three steps and solve for x.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Apply the idea Start with the equation

Expand the brackets

Combine like terms

Subtract 17 from both sides

Divide both sides by 8

Simplify the fraction

Reﬂect and check Substitute

into the original equation to check. Start with the equation

Replace x with

Evaluate multiplication inside the brackets

Evaluate addition inside the brackets

Multiply

Evaluate

Both sides are equal, so x =

is the solution.

Example 2 Solve: 9 (x − 108) − 3 (x + 96) = 96 − 6 (x − 48)

Create a strategy Expand the brackets and solve for x.

Apply the idea Start with the equation

Expand the brackets

Simplify

Add 6x to both sides, add 1260 to both sides

Evaluate

Divide both sides by 12

Evaluate

3.02 Equations with more than 3 steps mathspace.co

Reﬂect and check As an alternative strategy, we can divide each term by 3 before expanding brackets to work with smaller numbers: Start with the equation

Divide each term by 3

Evaluate

Expand the brackets

Simplify

Add 2x to both sides, add 420 to both sides

Evaluate

Divide both sides by 4

Evaluate

Both strategies yield the same solution, x = 137.

Example 3 Solve this equation:

Create a strategy We can start by multiplying both sides by 2 to clear the fractions. Then, we can expand the brackets and solve for a.

Apply the idea Start with the equation

Multiply both sides by 2

Expand the outer brackets

Expand the inner brackets

Combine like terms

Add 63 and subtract 6a from both sides

Combine like terms

Divide both sides by 7

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reﬂect and check Let’s check our solution by substituting a = 6 back into the original equation. Write the original equation Substitute a = 6 Evaluate the expressions inside the brackets Evaluate the expressions inside the brackets and multiply Evaluate the fractions Combine like terms Evaluate Both sides are equal, so a = 6 is the correct solution.

Idea summary To solve equations with more than three steps, start by expanding any brackets and clearing any fractions if necessary. Then combine like terms, get all the variable terms to one side, and solve for the variable.

3.02 Practice questions Understanding 1

1, 2

Explorer

1, 2

Adventurer

1–3

Trailblazer

Complete the solution for the given equation. 5 − 2(3x − 4) = 3(2x + 1) − 7x 5 − 6x⬚8 = ⬚x + ⬚ − 7x 13 − 6x = −x + 3

13 = ⬚x + 3

⬚ = 5x ⬚=x

Start with the equation ⬚ the brackets ⬚

Add ⬚ to both sides

Subtract ⬚ from both sides Divide both sides by ⬚

3.02 Equations with more than 3 steps mathspace.co

For each equation, is the given solution correct or incorrect? a

x = 2, for −2x + 13 = −4 (2x + 1) + 3 (5x − 2)

y = 3, for 3 (3y − 8) − 5 (3y − 8) = 4 (y − 2) − 6 (y − 2)

p = 4, for 2 (3p + 7) + 3 (p − 3) = 8 (p + 10)

d = −9, for

Write the equation without any fractions. Do not solve. a

Fluency 4 Ex 1

5 Ex 2

7 Ex 3

4–6, 7a–e

4–7

Solve: a

−4a − 11 = −5(a − 2a)

15 − (3b + 1) = 6 (b + 1) + 2b

−6y − 4 (4y − 8) = 3 (−12y − 1)

4 (3h − 9) − 12 = 12 − 8 (h − 9)

− (−8 − 3d) = −2 (1 − d) + 2d

12h + 4 = 5 (6h + 5) + 60

3q + 4 = 6 (5q − 5) + 88

8c − 6 = −8 (6c − 7) + 42

4 (8y + 5) = 3 (6y + 8) + 108

−5 (−5d + 5) = 8 (4d − 8) + 37

Solve: a

9 (2y − 5) − 3y = 10 − 5 (y + 9)

4 (3a + 4) + 5 (a − 3) = 6 (a + 10)

3 (2b + 1) − 3 (3b − 1) = 2 (b + 1)

8 − 6 (2h + 1) = 4 (3h − 1) + 6 (h + 10)

2 (3c + 5) − (3c − 9) = 12 − 2 (3c + 12)

Solve: a

16 (5z + 8) + 2 (3z − 5) = 7 (z + 11) − (2z − 8)

2 (3t − 8) − 4 (3t − 8) = 12 (t − 2) − 6 (t − 2)

−8 (9d + 8) − 6 (8d + 2) = −12 (d + 2) − 7 (d − 3)

9 (c + 5) − 6 (2c + 8) = −5 (c − 2) − 6 (c + 3)

12 − (z − 12) − 6 (2z + 8) = −5 (−z − 2) + 8 (z + 2)

Solve: a

4, 5

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reasoning 8

8–10, 12

9–12

Two students, Clara and David, attempted to solve the equation 2z + 6 = 3z − 5. Their workings are shown: Clara’s working: 1

2z + 6 = 3z − 5

−z + 6 = −5

z−6=5

z = 11

David’s working:

2z + 6 = 3z − 5

−z + 6 = −5

−z = −11

−11 = z

Which student provided the correct solution and why?

Explain the mistake made by the other student.

Alex attempted to solve the equation 4 (2x + 5) − 3 = 37. His working is shown:

a 10

4(2x + 5) − 3 = 37

Start with the equation

4(2x + 5) = 40

Add 3 to both sides

2x + 5 = 10

Divide both sides by 4

2x = 15

Add 5 to both sides

x = 7.5

Divide both sides by 2

What was his mistake?

Solve the equation correctly.

Two students, Alice and Bob, attempted to solve the equation 2 (x + 4) − 3 (x − 3) = x + 7. Their workings are shown: Alice’s working: 1

2 (x + 4) − 3 (x − 3) = x + 7

2x + 8 − 3x + 9 = x + 7

−x + 17 = x + 7

2x = 10

x=5

Bob’s working: 1

2 (x + 4) − 3 (x − 3) = x + 7

2x + 8 − 3x + 9 = x + 7

−x + 17 = x + 7

−2x = 10

x = −5

Which student provided the correct solution and why?

Explain the mistake made by the other student.

3.02 Equations with more than 3 steps mathspace.co

Solve the equation:

(14p − 28) = −8p + 16 in two different ways.

Method 1: Expand the brackets ﬁrst

Method 2: Clear the fraction ﬁrst

Was one method easier than the other? Explain your answer.

Write an equation that could take at least four steps to solve that meets the criteria. Justify your answer. a

A solution of a = 3

Problem-solving 13, 14, 16

13–17

A solution of b = 14–18

If the perimeter of this triangle is 263 cm, ﬁnd the value of x.

7x − 21

5x + 28

3x − 14

The cost of a cricket ball is 136 cents more than the cost of a rugby ball. a

Write an expression for the price of a cricket ball in dollars.

If 8 cricket balls and 9 rugby balls cost $119, write a linear equation representing the situation.

Find the cost of a rugby ball.

Find the value of a cricket ball.

One side of a parallelogram is 4 cm shorter than an adjacent side ( y). The perimeter of this parallelogram is 12 cm. a

Write an expression for the length of the shorter side.

Write an equation for the perimeter.

Find the dimensions of the parallelogram.

A rectangle with a height of 4x + 8 cm and a width of 4x has the same perimeter as a square with side length 3x + 9 cm. Find the value of x.

This rectangle has a perimeter of 126 + 3y centimetres.

4y + 8

Find the value of y. 4y + 3

Right now, Bianca’s father is 42 years older than Bianca. 4 years ago, her father was 4 times as old as her. Solve for y, Bianca’s current age.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

3.03

Equations involving fractions

After this lesson, you will be able to… • apply the three steps in solving equations with fractions. • use the lowest common multiple to clear fractions from equations. • apply inverse operations to isolate the variable and solve equations. • recognise when an algebraic denominator makes a solution invalid. • solve equations that have algebraic expressions in the denominators.

Equations with algebraic fractions To solve equations involving algebraic fractions, we can solve as we did with equations of three or more steps, but at some point we will clear the fractions. We can clear fractions by ﬁnding the lowest common multiple of the denominators and multiplying both sides by it. Using the lowest common multiple will eliminate the fractions and make it easier to solve the equation.

Exploration Let’s look at part of the process for solving this equation. 1

Start with the equation

Multiply by the lowest common multiple

Simplify the fraction

What do you notice about the coeﬃcients outside the brackets in the last step and the denominators in the ﬁrst step?

Will this pattern always hold? For example, does it work for

What will happen to the constant, 1, when solving

? ?

Here are some things to remember: • We often start with the inside brackets and work our way out. • The distributive law can still be used if there are more than two terms in the grouping symbol. • We must multiply the whole side of an equation by the lowest common multiple, not just the fractional terms. It is essential to simplify the equation and check your solution to ensure accuracy in your ﬁnal answer. 3.03 Equations involving fractions mathspace.co

Example 1 Solve:

Create a strategy We can start by expanding the brackets, and then clear the fraction. Once we have done that we can solve by using inverse operations.

Apply the idea Start with the equation

Expand the brackets

Multiply both sides by 8

Combine like terms on the left-hand side

Subtract 15x from both sides

Subtract 21 from both sides

Divide both sides by −14 and simplify

Reﬂect and check An alternative method for solving this problem involves ﬁrst multiplying through by 8, and then expanding the brackets.

Keep in mind that when using this strategy, we multiply each group, not each individual term.

Example 2 Solve:

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fraction. Then we can solve using inverse operations to isolate the variable.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Apply the idea Start with the equation

Multiply both sides by the lowest common multiple, 10

Distribute the 10

Simplify the fractions

Expand the brackets

Add 2x to both sides

Subtract 50 from both sides

Divide both sides by 27

Reﬂect and check We can substitute the value of x to check the answer. Start with the equation

Substitute

Multiply both sides by the lowest common multiple, 10

Evaluate the brackets then expand

Multiply both sides by 27

Evaluate both sides

Since both sides are equal, we can conﬁrm that our answer is correct.

Example 3 Solve:

Create a strategy First, we need to ﬁnd the lowest common multiple to clear the fractions. Then we can solve using inverse operations to isolate the variable.

3.03 Equations involving fractions mathspace.co

Apply the idea Start with the equation Multiply both sides by the lowest common multiple, 12 Distribute the 12 Simplify the fractions Expand the brackets Combine like terms Subtract 73 from both sides and add x to both sides Divide both sides by −7 and simplify

Reﬂect and check We can substitute the value of x to check the answer. Start with the equation

Substitute

Write each term as a fraction with a denominator of 12 Multiply both sides by 12 to clear them Evaluate the brackets Expand the brackets and simplify Multiply both sides by 7 Evaluate both sides Since both sides are equal, we can conﬁrm that our answer is correct.

Idea summary Solving equations with algebraic fractions involves simplifying the equation, clearing fractions by multiplying both sides by the lowest common multiple, and solving for the variable using inverse operations.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Algebraic denominators If we have an algebraic expression in the denominator, then we can multiply both sides by the algebraic expression. When we simplify, we should get an expression which we can solve using previous strategies. Since we can’t divide by 0, the value which makes the denominator zero cannot be a possible solution. For example,

cannot have a solution of x = 2. We sometimes write x ≠ 2.

Example 4 Solve:

Create a strategy To solve this equation, we can multiply both sides by the lowest common multiple, which is 7(2x + 5), to clear the fractions. Then we can solve for the variable using inverse operations.

Apply the idea Start with the equation

Multiply both sides by 7(2x + 5)

Remove the common factor

Expand the brackets and evaluate

Subtract 30 from both sides

Divide both sides by 12 and simplify

Reﬂect and check We can use cross-multiplication as an alternative method for solving problems with fractions.

Start with the equation

Cross multiply to eliminate the fractions

Expand the brackets and evaluate

Subtract 30 from both sides

Divide both sides by 12 and simplify

3.03 Equations involving fractions mathspace.co

Idea summary If we have an algebraic denominator, the lowest common multiple will be an algebraic expression. We can multiply both sides of the equation by the lowest common multiple and then simplify and solve. Since we can’t divide by 0, the value which makes the denominator zero cannot be a possible solution.

3.03 Practice questions Understanding 1

For the equation a

Ex 1

3x + 2 = 15

Is each statement true or false? a

x = 1 is a solution for the equation

is a solution for the equation

is a solution for the equation 4x − 3 = 3

is a solution for the equation

Fill in the blanks: b

Fluency

5a–dg, 6, 7a–d

1–4

Trailblazer

, in solving for x, would this be a valid next step?

1–4

Adventurer

What is the lowest common denominator of each set of fractions? a

1–3

Explorer

5, 6, 7a–f, 8a–e

Solve: a

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5gh, 6e–h, 7, 8

Ex 2

Ex 3

Solve: a

Ex 4

Solve: a

9, 10, 12

9–12

Reasoning 9

9, 12

Dylan attempted to solve the equation 1

8 × (3x + 6) = 4 − 2 × x

24x + 48 = 4 − 2x

26x = −44

. Here is his work:

Cross multiply Expand the brackets and simplify Add 2x to both sides and subtract 48 from both sides Divide both sides by 26 and simplify

Identify where Dylan has made an error and explain what it is.

Solve the equation correctly, showing the correct steps of work.

Complete the equation so it is true for all values of x.

Give an example of a linear equation with algebraic fractions that has no solutions. Justify your answer.

3.03 Equations involving fractions mathspace.co

Faith looked at the equation

and noticed that the fraction could be simpliﬁed

since it had the same numerator and denominator. She then wrote down 1 = 1 as her next line of working and said that x = 1 was the solution to the equation. a

Does x = 3 satisfy the equation?

Does the equation have any other solutions?

What value is not a possible solution?

Problem-solving 13, 16 13

13–16

14–18

A construction company has spent $15 000 000 to develop new cranes, and wants to limit the cost of development and production of a single crane to $4000. Given that the production cost of each crane is $2000, the cost for development and production of x cranes is given by 2000x + 15 000 000 dollars. a

Write an expression for the cost of one crane.

If the cost of one crane must equal $4000, ﬁnd x, the number of cranes that must be sold.

Fiona, a tennis player, has won 40 out of 55 matches in her career. Find x, the number of matches she must win in a row to raise her win percentage to 80%.

Solve for u:

If x = 2 and

At an annual charity event, the money raised was distributed in several different ways. One-third of the funds came from the silent auction, art sales comprised one-fourth of the total, and direct in-person donations made up one-ﬁfth of the revenue. The food booth raised $250. How much money was raised at the charity event? Round your answer to two decimal places.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

, ﬁnd the value of k.

, ﬁnd the value of r.

3.04

Rearrange equations

After this lesson, you will be able to… • apply correct algebraic operations such as addition, subtraction, multiplication, and division to rearrange equations effectively. • solve real-world problems that require making different variables the subject.

Rearrange equations In mathematics and science, there are many formulas that are written in a specific way.

Exploration Let’s look at the formula for speed:

s d t 1.

speed distance time

Use the formula to ﬁnd each of the following: • The speed at which a person travels if they drive 100 km in 75 min • The distance travelled when walking 5 kph for 30 min • The distance travelled by a car driving 85 kph for 5 hr • The time it will take to travel 1 000 km if you walk 80 m/min • How fast a plane is travelling if it can ﬂy 2 789 km in 6 hr • How long it will take to walk to the store 2 km away if you walk 100 m/min

Which ones took the most effort to solve?

If we always wanted to ﬁnd the distance, given time and speed, how could we write the formula to make these calculations more eﬃcient?

By rearranging a formula for a variable of interest, we can make these calculations more efficient. To rearrange an equation and make a speciﬁc variable the subject, we need to perform various algebraic operations such as addition, subtraction, multiplication, and division. The key is to isolate the desired variable on one side of the equation while maintaining the balance of the equation. Here are some general steps to follow: • Identify the variable that you want to make the subject. • Perform algebraic operations on both sides of the equation to isolate the desired variable, remembering to use the correct order of operations. • Simplify the equation as much as possible, and always double-check your steps to ensure accuracy.

3.04 Rearrange equations mathspace.co

Example 1 Rearrange kx = m + nx to make x the subject.

Create a strategy Notice the like terms. Rearrange so that all the terms in x are on one side of the equation.

Apply the idea Write the equation

Subtract nx from both sides

Take out the common factor of x

Divide both sides by (k − n) to make x the subject

Reﬂect and check When the variable we want to make the subject appears more than once in the equation, we should get them all on the same side and factorise it out.

Example 2 Make M the subject of

Create a strategy We need to have one instance of M in the equation, so we ﬁrst need to clear the fraction and then factorise.

Apply the idea Start with the equation

Multiply both sides by (M + A)

Simplify

Expand the brackets

Subtract M from both sides

Subtract PA from both sides

Factorise each side of the equation

Divide both sides by (P − 1)

Simplify

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reﬂect and check We can substitute values for A and P, to see if we get the same value solving for M in the original equation and using the rearranged equation. However, this does not guarantee we haven’t made any errors. It is worth checking every step for errors.

Example 3 Make R the subject of V = IR − E.

Create a strategy I is multiplied by R, then E is taken away to give V. Reverse these operations.

Apply the idea Write the equation

Add E to both sides

Divide both sides by I

Write R as the subject

Example 4 Find the gradient and use the gradient and intercept method to sketch the graph of 6x + 2y − 12 = 0.

Create a strategy Make y the subject of 6x + 2y − 12 = 0, which will put the equation in the form y = mx + c. From this form, we can identify the gradient (m) and the y-intercept (c).

Apply the idea Make y the subject of 6x + 2y − 12 = 0. 6x + 2y − 12 = 0 2y − 12 = −6x

Subtract 6x from both sides

2y = −6x + 12

Add 12 to both sides

−6 x 12 y= + 2 2

Divide both sides by 2

y = −3x + 6

Simplify

Identify the gradient and y-intercept. Gradient is −3 and the y-intercept is 6.

3.04 Rearrange equations mathspace.co

Sketch the graph. Plot the y-intercept (0, 6) and then apply the gradient of −3. Step across 1 unit and down 3 units. 7 6 5 4 3 2 1

−1

Draw the line. 7 6 5 4 3 2 1

−1

x 1

Idea summary We can rearrange equations and formulas to make different variables the subject. This can make it easier to do repeated calculations and for graphing equations that are not in the form y = mx + c. Carefully using inverse operations, we can rearrange equations in a similar way to solving equations. It is very important to keep the equation balanced by doing operations on both sides. However, instead of operating on numbers, we are operating on variables.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

3.04 Practice questions Understanding

1–4

Explorer

Why do we rearrange equations?

For the formula, Q = m × c × ΔT:

1–4

Adventurer

1–4

Trailblazer

• Q is the amount of heat absorbed in Joules, (J) • m is the mass in grams (g) • c is the heat capacity (J/g °C) • ΔT is the temperature change, (°C) We are given the amount of heat absorbed, the mass, and temperature change for ten different substances. Which variable should we make the subject to ﬁnd the unknown values more eﬃciently? 3

To make p the subject of R = pq − 4pt2, which option could be a ﬁrst step? A B

Divide pq − 4pt2 by p

Multiply both sides of the equation by 4pt2

Factorise p from pq − 4pt2

Add p to one side of the equation

Is each statement true or false? a

When we rearrange a formula, we will get a numerical value.

When rearranging a formula, the formula may involve numerical and algebraic terms.

We use inverse operations to make a variable the subject.

When we rearrange a formula, we will eliminate all of the other variables.

Fluency Ex 1

5a–d, 6a–e, 7a–c, 8ab, 10, 11a–f, 12ab, 14–17

y=x+z

y = 5 (8 + x)

3y = 2x + z

y = 5 (3 − x)

Make m the subject of these equations:

b = −2am + 10

y = 3mx − 5

x = 8k (n + m)

x = −7m (p + q)

y = −5mx − 14

x = −7k (l − m)

8x3 = 4y − 12

Make y the subject of these equations: a

Ex 2

5–9, 11e–j, 12cd, 13–15, 18, 19

Make x the subject of these equations:

5–7, 8a–c, 9, 10, 11d–h, 12bc, 14–16, 18

9x2 + 3y = 12

Make k the subject of these equations: a

3.04 Rearrange equations mathspace.co

Ex 3

Ex 4

Make p the subject of the equation M = pq − 7pr2.

Make v2 the subject of the formula for kinetic energy,

Make y the subject in each equation:

y − x + 1 = 0

−4y − 8x + 12 = 0

7y + 14x − 21 = 0

−6y + 24x − 30 = 0

5y − 15x + 25 = 0

3y + 6x − 9 = 0

−2y + 4x + 8 = 0

10y − 20x − 30 = 0

6x + 4y − 18 = 0

4x − 3y − 12 = 0

Identify the gradient and y-intercept and graph each equation. a

5x + 2y − 10 = 0

−4x + y + 16 = 0

6x + 3y − 12 = 0

−12x + 4y − 24 = 0

The bend allowance for sheet metal is given by the formula

Make T, the thickness of the sheet, the subject of the formula. 14

Make r2 the subject of these formulas: a

The volume of a cone is given by the formula

The area of a circle is given by the formula A = π r2.

The area of a sector of a circle is given by the formula

Make the mentioned variables the subject in the following equations: a

E = mc2, make m the subject

P = I2 R, make R the subject

, make h the subject

A = P (1 + r)t, make P the subject

V = π r2 h, make h the subject

D = b2 − 4ac, make c the subject

, make z the subject

g = hx2 k, make h the subject

, make c the subject

d = uv4 − t, make u the subject

The formula for the force of an object is F = ma, where F is the force, m is the mass of the object, and a is the object’s acceleration. Write an equation that could be used to solve for the mass of the object.

The formula for the perimeter of a rectangle is P = 2l + 2w, where P is the perimeter, l is the length, and w is the width. Write an equation that could be used to ﬁnd the length of the rectangle.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Betsy knows the formula to convert Celsius into Fahrenheit is

. She wants to

convert a temperature from Fahrenheit to Celsius. Find an equation that she could use to do this conversion. 19

The displacement of an object is given by used to ﬁnd v0.

Reasoning

20, 22

. Write an equation that could be

20–23

Describe the similarities and differences of solving 3x + y = z and 3x + 8 = 2. Explain your thinking.

Two students, Alice and Bob, are given the Pythagora’s theorem equation to rearrange: c2 = a2 + b2 Alice rearranges the equation to make a2 the subject, while Bob rearranges the equation to make b2 the subject. Their results are as follows: Alice: a2 = c2 − b2 Bob: b2 = c2 + a2 Determine which student, if any, is correct in their rearrangement of the equation and explain your reasoning.

Pamela attempted to make r the subject of the equation

and followed these steps:

Multiplied both sides by R + r

Divided both sides by L

Subtract R from both sides of the equation

What was Pamela’s mistake? Explain and correct her error. 23

Ruﬁno has submitted the following work to write the equation for the mass of an object, m, given the kinetic energy, KE, and velocity, v:

Was Ruﬁno correct? If so, explain each step of his working. If not, ﬁx his error(s).

3.04 Rearrange equations mathspace.co

Problem-solving 24, 25 24

The area, A, of a triangle is given by the formula and b is the length of its base. a b

24–27

, where h is the height of the triangle

Write an equation that could be used to solve for h.

Find the height of a triangle with a base of 12 cm and an area of 18 cm2.

The volume of a sphere is given by the formula a

24, 25, 27

, where r is the radius of the sphere.

Rearrange the formula to make, r the subject of the formula.

Calculate the radius of a sphere with a volume of 91 cm3 to two decimal places.

When a certain amount of money, P, is invested, its future value, A, is given by the formula , where r is the rate per time period and n is the number of time periods. a

Make P, the principal, the subject of the equation.

How much must be invested now, if it is to grow to $3000 in 10 years with interest of 4%

compounded annually? Round to the nearest whole dollar. 27

The volume of the cylinder that ﬁts exactly into a sphere is given by the formula:

a b

Rearrange the formula so that r is the subject.

What is the radius of the sphere if the volume of the cylinder is 13.75π mm3 and the height of the cylinder is 5 mm? Round your answer to the nearest millimetre.

3.05

arallel and perpendicular P lines

After this lesson, you will be able to… • understand the characteristics of parallel and perpendicular lines. • apply the understanding of parallel and perpendicular lines to find equations of such lines. • interpret and solve problems involving parallel and perpendicular lines.

Parallel lines Lines that never intersect are called parallel lines.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Interactive exploration

Discover this concept in action online

Parallel lines are lines that have the same gradient.

When two lines are parallel, they never cross and have no points in common.

3 2 1 −4 −3 −2 −1

mathspace.co

x 1

−1

−2 −3 −4

Let’s look at how we can identify parallel lines given their equations. Equation form

Characteristic of parallel lines

y = mx + c

Parallel lines have the same m value.

Examples y = 2x – 1 y = 4 + 2x

For every straight line, there exists an inﬁnite number of lines parallel to it. 4

Here is the line y = x.

y=x+3

Two more lines in the same family of parallel lines are also graphed:

3 y=x

y=x+3

1 −4 −3 −2 −1

−1

x 1

−2

y=x–4 Notice that they have the same gradient (m-value) but different x and y-intercepts (c-values).

−3 −4 y = x − 4

The gradient of any vertical line is undefined. But by definition, any two vertical lines are parallel.

3.05 Parallel and perpendicular lines mathspace.co

Example 1 Consider these questions about parallel lines: a Are the lines y = 3x + 2 and y = 3x – 5 parallel?

Create a strategy Lines are parallel when the gradients are the same. Identify the gradients for each equation. The equations are in the form y = mx + c. In this form, m represents the gradient.

Apply the idea y = 3x + 2, will have a gradient of 3. y = 3x − 5, will have a gradient of 3. So these lines are parallel. b Which of these three lines are parallel? y = –3x + 2 −3y – x = 5 y + 3x = 7

Create a strategy Lines are parallel when the gradients are the same. The equations will need to be rearranged into the format of y = mx + c to compare the gradients.

Apply the idea y = −3x + 2 is in the form y = mx + c. It has a gradient of −3. −3y – x = 5 This needs to be rearranged.

the equation y− x =5 −3 Write

−3 y = x + 5

−3 y x + 5 = −3 −3 1 5 y = − x − 3 3

Add x to both sides Divide both sides by −3 Write in gradient-intercept form

y + 3x = 7 This needs to be rearranged. y + 3x = 7 y = −3x + 7

Write the equation Evaluate

The lines that are parallel are y = −3x + 2 and y + 3x = 7 as both have a gradient of −3.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 2 Find the equation of the line that is parallel to

and passes through the point (9, −1).

Create a strategy First, ﬁnd the gradient of the given line. Next, substitute the gradient and the given coordinates into the equation y = mx + c, and solve for c.

Apply the idea The gradient of

. Since the new line is parallel to the given line, the gradient of

the new line is also . Equation of a line in gradient-intercept form Substitute −1 =

1 ×9+c 3

Substitute x = 9 and y = −1 Evaluate the multiplication Subtract 3 from both sides

Substituting the values of m and c into the gradient-intercept form of a line, the equation is:

Reﬂect and check y −4 −3 −2 −1

−1

x 1

We can graph the given line with the line we found to see that they are parallel.

−2 −3 −4 −5 −6 −7 −8

Idea summary If two lines are parallel, they have the same gradient. If two lines have the same gradient, they are parallel. All vertical lines are parallel to each other.

3.05 Parallel and perpendicular lines mathspace.co

Perpendicular lines Lines that meet at right angles (90°) are called perpendicular lines.

Interactive exploration

Discover this concept in action online

mathspace.co

Perpendicular lines have gradients whose product is equal to −1. For two perpendicular lines, y = m1 x + c1 and y = m2 x + c2: m1 × m2 = −1 We say that m1 is the negative reciprocal of m2. This means the two gradients have opposite signs and are reciprocals of each other. Here are some examples of negative reciprocals:

If the gradient of the ﬁrst line is m1, then we can ﬁnd the gradient m2 of a perpendicular line using the formula:

Special case: a vertical line and a horizontal line are perpendicular but do not have gradients which have a product of −1. This is because the gradient of a vertical line is undeﬁned.

Example 3 Find the equation of a line that is perpendicular to y = 6x + 10, and has the same y-intercept. Give your answer in the form y = mx + c.

Create a strategy Find the gradient of the perpendicular line and the y-intercept of the given line, then substitute both values into the equation y = mx + c.

Apply the idea The gradient of y = 6x + 10 is 6, and the negative reciprocal of 6 is perpendicular line is

, so the gradient of the

. The y-intercept is c = 10. So the equation of the line is:

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reﬂect and check y 15 12

y=−

We can check that the product of the gradients is equal to −1, and we can graph both lines to see they are perpendicular.

y = 6x + 10

1 x + 10 6

6 3 x −6

−3

Idea summary Two lines are perpendicular if their gradients are negative reciprocals of each other. To test if lines are perpendicular, multiply the gradients together. If the result is −1, then the lines are perpendicular. Vertical and horizontal lines are perpendicular to each other, even though their gradients do not have a product of −1.

3.05 Practice questions Understanding 1

1–4

Explorer

1–5

Adventurer

The equations y = 2x, y = 2x + 5 and y = 2x − 7 have been graphed on the same Cartesian plane: a

What do all of the equations have in common?

What do all the lines have in common?

Trailblazer 2–6

10 8 6 4 y = 2x + 5 2 −4

−2

−2 −4 −6

y y = 2x

x 2

y = 2x − 7

3.05 Parallel and perpendicular lines mathspace.co

If two non-vertical lines are perpendicular, what is the product of their gradients?

For the given graph: a

Which two lines are parallel?

Find two pairs of perpendicular lines.

y = 3x + 15

10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

Complete the statement:

The x-axis and the y-axis are a pair of ⬚ lines.

1 y=− x 3 1 y = − x−9 3

y = 2x − 1

y=x

x=5

If the equation of a line is y = mx + c, and m > 0, which of the statements is true? A

Any line parallel to this line will have a negative gradient.

Any line perpendicular to this line will have a positive gradient.

Any line perpendicular to this line will have a gradient of zero.

Any line perpendicular to this line will have a negative gradient. 7–9, 10ab, 11, 12, 14a–d, 15, 16, 18a

8–12, 14a–d, 15–18

8–11, 13–18

Which of these lines are parallel to y = 7x + 3? a

y = 7x − 3

y = 6x + 3

y = 7x + 4

y = 7x

y = 6x + 4

y = −7x + 3

y = 3x + 7

y = −3x − 7

Are these pairs of lines parallel? a

y = −3x − 2 and y = −3x + 9

y = −2x − 5 and y = −2x − 8

y = 7x + 8 and y = −5x + 8

y = 4x − 1 and y = 4x − 6

x = 4 and y = 5

y = 7x − 5 and y = −7x + 6

y = 3x − 5 and

y = −8x − 2 and 9x + 7

The line L passes through the point (9, −5) and is parallel to the line y = −5x + 2. a

100

y = −1

Fluency Ex 1

x 2 4 6 8 10

Which of these lines is parallel to the y-axis? A

y = 1.9x

Find the gradient of line L.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Find the equation of line L.

Ex 2

Find the equation of the line described: a

Parallel to the line y = −2x + 9 and passes through the point (−3, 1).

Parallel to the line y = 8x − 3 and crosses the y-axis at 5.

Parallel to the line y = −3x − 8 and crosses the y-axis at −4.

Parallel to the x-axis and passes through (−10, 2).

Parallel to the y-axis and passes through (−7, 2).

Find the gradient of the line perpendicular to: a

A line with gradient 6

c 12

y = −3x

3y − x = 5

x + 3y = −10

y = 9x

y = −9x + 5

9y = −x + 2

y = x + 1 and y = x − 1

y = 3x − 7 and y = −3x + 6

y = −3x + 6 and

and

and y = −(3x − 27)

y = 3x − 5 and

and

The line L is perpendicular to y = 5x − 4 and crosses the y-axis at 1. Find the gradient of line L.

Find the equation of line L.

Given the lines L1: 8x = −7y + 3 and L2: 7y − 8x − 7 = 0 a

b c 17

y=8−x

Are these pairs of lines perpendicular?

a 16

Are these lines perpendicular to y = 9x + 2? a

A line with gradient

Are these lines perpendicular to y = 3x + 2? a

Find the gradient of line L1.

Find the gradient of line L2.

Are the two lines perpendicular?

Given the lines L1 : 5x + 8y − 10 = 0 and L2 : a c

Find the gradient of line L1. Are the two lines parallel?

Find the gradient of line L2.

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Ex 3

Find the equation of the line described: Perpendicular to

Perpendicular to y = −6x + 10 and has the same y-intercept.

Perpendicular to

Perpendicular to the x-axis and passes through (−8, −1).

Perpendicular to the y-axis and passes through (−8, −8).

Reasoning 19

and passes through (−8, 5).

and passes through the point (0, 6).

19, 20

19, 21

19–21

A farmer wants to create an enclosure for his sheep. He puts fencing along these lines:

What is the shape of the enclosure? Justify your answer. 20

If every point on the line y = 2x + 2 is shifted 2 units up, ﬁnd the equation of the new line.

Every point on a particular line is shifted 6 units down. The resulting line has equation y = 2x − 2. Find the equation of the original line.

Problem-solving 22a, 23–25 22

22, 23, 25, 28, 30

22, 25–33

Each pair of lines are parallel. Find the value of b. a

0 = 5x − 9y + 6

y = bx + 4

0 = bx + 3y − 2 23

The line L passes through (−9, 10) and (2, −1). a

Find the gradient of line L.

Find the equation of the line that passes through (10, 8) and is parallel to line L.

The line L passes through (1, 1) and (3, 13). a

Find the gradient of line L.

Find the equation of the line that has a y-intercept of (0, −4) and is perpendicular to line L.

A construction company is building a new road that runs parallel to an existing road. If the equation of the existing road is y = 0.5x + 3, ﬁnd the equation of the new road that passes through the point (2, 6).

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The line L passes through (3, 2) and (−2, 4). Find the equation of the line that has a y-intercept of 1 and is parallel to line L.

Each pair of lines are perpendicular. Find the value of k. a

5x + 4y + 8 = 0

kx + 9y − 8 = 0 28

−6x + 6y − 8 = 0 kx + 10y + 8 = 0

The line L is perpendicular to y = 10x + 8 and passes through the point of intersection of the lines y = x + 5 and 8x − 10y + 60 = 0. a

Find the gradient of line L.

Find the point of intersection of y = x + 5 and 8x − 10y + 60 = 0.

Find the equation of the perpendicular line L.

The line y = 2x + 4 is perpendicular to the line cx − 4y + 7 = 0. Find the value of c.

The 4 vertices of square ABCD have been plotted on a Cartesian plane.

Find the gradient of side AB.

Find the gradient of side BC.

Find the product of the gradients in parts (a) and (b).

Are sides AB and BC perpendicular?

5 4 D

2 C

1 1

The points P(0, −2), Q(−3, −2), R(0, 4) and S(2, 2) are graphed on the Cartesian plane:

x 5

y 4 R

Find the gradient of PQ.

Find the gradient of RS.

Are PQ and RS parallel?

Are QR and PS parallel?

What type of quadrilateral is PQRS?

−4 −3 −2 −1 Q

−1

S x 1

−2 P −3 −4

The vertices of a quadrilateral are A(2, 2), B(6, 4), C(8, 10), and D(4, 8). a

Find the gradient of each side.

Is quadrilateral ABCD a parallelogram?

Is quadrilateral ABCD a rectangle?

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A quadrilateral ABCD has been graphed on the Cartesian plane. a

Find the gradient of each side: i

iii

What type of quadrilateral is ABCD? Explain your answer.

3.06

y x

−1 1 2 3 4 5 6 7 8 9 −1 D −2 −3 −4 A −5 C −6 −7 −8 B −9

Linear models

After this lesson, you will be able to… • understand the roles of independent and dependent variables in a linear model. • interpret linear models in real-world contexts, including understanding the significance of the gradient and vertical intercept. • explore the concept of negative gradients in the context of decreasing quantities. • understand the characteristics of linear graphs in real-world scenarios. • interpret linear relationships from a table of values. • solve real-world problems involving linear models.

Linear models In linear relationships we have two variables, independent and dependent variables which are x and y respectively. These will often represent physical quantities such as time, distance, cost, mass or temperature. Instead of being labelled x and y, they may be labelled with letters or names that better represent those quantities. For example, if we were modelling the rate of fuel being used in a car, the independent variable may represent the distance travelled, in kilometres, and the dependent variable may represent the volume of fuel, in litres, in the car’s fuel tank. Instead of x and y, we might use d and V as our variables. Each of these variables has units of measurement associated with them.

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Fuel consumption of a mid-sized Sports Utility Vehicle V (L) 70 60 50 40 30 20 10

d (km) 100 200 300 400 500 600 700 800 900

As we know already, the two key features of a linear equation, or straight-line graph, are the gradient and the y-intercept. In linear modelling situations, the y-intercept is often referred to as the vertical intercept, because the vertical axis may be labelled with a variable other than y. The vertical intercept represents an initial value. In our example above, the vertical intercept is 63 litres. It represents the volume of fuel in a full tank, before the car began its journey. In a real-world context the gradient represents a rate of change. Using our example above, the gradient would represent the volume of fuel used per distance travelled. In other words, the gradient is a measure of the car’s fuel consumption. The graph above is decreasing as the fuel is being used, so it has a negative gradient. If we divide the ‘rise’ of −63 by the ‘run’ of 900 we get a gradient of −0.07. This means fuel is being consumed at a rate of 0.07 litres per kilometre (or 7 litres per 100 kilometres). Because most physical quantities like distance, volume or time do not contain negative values, the graphs of most linear models tend to exist only in the ﬁrst quadrant of the Cartesian plane (like the example above). This is not always the case though - for example, temperature is a physical quantity that can have negative values.

Idea summary In linear modelling situations: • • • •

The vertical intercept represents the initial value. The gradient represents the rate of change. The independent variable goes on the horizontal axis. The dependent variable goes on the vertical axis.

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Linear relationship from a table of values Real world situations often involve the collection of data, which is commonly displayed in a table of values. The independent variable is always displayed in the ﬁrst row of the table, followed by the dependent variable, in the second row. We may ﬁrst want to decide whether the values in the table represent a linear relationship. To do this, we could plot the values on a Cartesian plane. If we can draw a straight line through all of the points, then we have a linear relationship. Once we know that the relationship between the variables is linear, we can work out the gradient and vertical intercept, and express the relationship as a linear equation.

Example 1 Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

a Graph the linear relationship represented in the table.

Create a strategy Write the values in the table as points, plot the points on a Cartesian plane, and draw the line that passess through the points.

Apply the idea y 70 60 50 40 30 20 10

x 10

20 30 40 50 60

b Write an equation linking the number of litres of petrol pumped (x) and the cost of the petrol ( y).

Create a strategy Find the gradient of the line and the y-value of the y-intercept, then write the equation of the line in the form y = mx + b.

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Apply the idea Find the gradient of the line: Use the coordinates (0, 0) as (x1, y1) and (10, 16.40) as (x2, y2). We can find the gradient of the line: Write the gradient formula

Substitute the values

Divide the change in y by the change in x

Evaluate Since the y-intercept is at (0, 0), b = 0. The equation of the line is y = 1.64x

c How much does petrol cost per litre?

Create a strategy The gradient of the line represents the unit rate of the cost of petrol per litre.

Apply the idea The gradient of the line is

and represents the increase in the price of petrol per 1 litre.

In other words, the cost of petrol per litre is $1.64

Reﬂect and check We can also ﬁnd the cost of petrol per litre by dividing the cost of the petrol ( y) by the number of litres (x). Divide 16.40 by 10

Evaluate

d How much would 47 litres of petrol cost at this unit price?

Create a strategy Using the equation from part (b), substitute x = 47.

Apply the idea y = 1.64x

Write the equation

= 1.64 × 47

Substitute x = 47

= $77.08

Evaluate

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Idea summary The independent variable is often displayed in the first row of the table, followed by the dependent variable. From a table of values we can plot the graph and determine the equation for the relationship. We can use the relationship to determine the value of one of the variables given the value of the other.

3.06 Practice questions Understanding 1

y = 2x + 3

y = x2 + 2x + 1

1–4

y = 3x3 + 2x + 1

f (x) = 2x+1

The rate of change is variable.

The rate of change is constant.

The rate of change is exponential.

The rate of change is never zero.

Complete the sentences with ‘horizontal’ or ‘vertical’: b c d

The independent variable is plotted on the ⬚ axis. The dependent variable is plotted on the ⬚ axis. The line y = 5 is a ⬚ line.

The line x = −2 is a ⬚ line.

The initial value is the starting value or the y-intercept. Find the initial value for these linear models: a

C = 250 + 7.55x

Fluency 5

1–4

Trailblazer

Which of the statements is a characteristic of a linear relationship?

Adventurer

Which of the equations represents a linear model? A

1–4

Explorer

T = 12.5 + 1.5m

5–11, 16, 20

What is the gradient of the line?

What does the gradient represent? A

The initial value

The speed

Cost of petrol

Time travelled

5, 7, 12–15, 17–19

The horizontal axis represents the time in hours and the vertical axis represents the distance travelled in kilometres. a

4x + 5y = 40

F = 0.6k − 5

5, 7, 12–15, 17–19

Distance (km) 575 460 345 230 115 t (hrs) 1

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The following graphs represent the cost and revenue in dollars for x items produced and sold, respectively.

$ 600 500

C = 15x is the cost equation, indicating each item costs $15 to make.

400

R = 0.5x2 + 50 is the revenue equation.

300

Identify each equation on the graph.

Explain the meaning of the intersection points of the graphs.

200 100 Items (x) 10

Ross deposits $100 into a savings account each week.

Tammy’s parents record her height every year for the ﬁrst 12 years of her life.

Julia collected lollies from a pinata at a party and eats two lollies per day.

Janet starts running and doubles the distance she runs each day.

A dam used to supply water to the neighbouring town had this data recorded for its volume over a number of months:

Plot the points on the Cartesian plane.

Is this relationship linear?

1 112

2 106

3 110

4 80

After Sally starts running, her heartbeat increases at a constant rate as shown in the table: Number of minutes passed (x) Heart rate ( y)

Identify whether the situation is represented by a linear or non-linear equation:

Month (M ) Volume in billions of litres (V )

0 75

2 81

4 87

Complete the table.

Graph the linear relationship represented in the table.

By how much is her heartbeat increasing each minute?

6 93

8 99

10 105

A diver starts at the surface of the water and begins to descend below the surface at a constant rate given by y = 0.8x a

Complete the table to show the depth of the diver over 4 minutes: Number of minutes passed (x) Depth of diver in metres ( y)

0 0

Graph the linear relationship represented in the table.

Calculate the increase in depth each minute.

Calculate the depth of the diver after 24 minutes.

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11 Ex 1

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol (C)

30 36

40 48

Graph the linear relationship represented in the table. Write an equation linking the number of litres of petrol pumped, x, and the cost of the petrol C.

Find the cost of petrol per litre.

Calculate the cost of 75 L petrol.

The time in minutes and the temperature in degrees Celsius (°C) are given in the table: 1 8

2 13

3 18

4 23

5 28

By how much is the temperature increasing each minute?

Find the initial temperature.

Form an equation relating x and y.

Graph the linear relationship represented in the table.

The table shows the temperature of a metal plate, in °C, after an amount of time in minutes: 1 10

2 15

3 20

4 25

5 30

Graph the linear relationship represented in the table.

By how much is the temperature increasing each minute?

Find the initial temperature.

Form an equation relating x and T.

Chirping crickets can be used as an indication of how hot or cool it is outside. For a particular species this data was recorded: Number of chirps per minute (x) Temperature (T °C)

110

20 24

Time (x) Temperature (T)

10 12

Time (x) Temperature ( y °C)

0 0

80 16

110 19

140 22

160 24

Graph the linear relationship represented in the table.

Determine the temperature when the crickets make 130 chirps each minute.

How many chirps per minute will the crickets make if the temperature is 27 °C?

How many chirps per minute will the crickets make if the temperature is 20 °C?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Emma is gathering data on the growth of her plant as shown in the table: Day (d) Height of plant (h cm)

6 14

8 18

Graph the linear relationship represented in the table.

How tall was the plant when Emma started tracking the growth?

Determine the height of the plant on day 12.

On which day will the plant be 24 cm tall?

Write a linear equation to represent the height, h, as an equation of the day, d.

When a plane is coming in to land, the pilot requires the plane to lose altitude at a rate of 5 m/s. At the beginning of a particular descent the plane starts at an altitude of 8500 m. The altitude of the plane at various times after starting the descent are shown in the table: 0 8500

200 7500

400 6500

1600 500

Graph the linear relationship represented in the table.

Determine the altitude of the plane 500 seconds after starting the descent.

How many seconds did the plane take to descend to the ground?

Write a linear equation to represent the altitude of the plane, A, as an equation of the number of seconds after commencing descent, t.

Katrina is going to catch a taxi into the city. She knows that the taxi has a ﬂat fee of $0.30 and that the fare rises by $0.10 per kilometre travelled. The fares for particular distances are shown in the table: Distance travelled, d (km) Fare, F ($)

4 10

Descent (t seconds) Altitude (A)

2 6

6 0.90

7 1.00

8 1.10

9 1.20

Graph the linear relationship represented in the table.

Write a linear equation to represent the fare, F, as an equation of the journey’s length, d.

Calculate the fare for a 14 km journey.

Find the distance travelled for a trip that costs $0.60

Justin is looking into the details of his mobile phone plan. He knows the costs for several call lengths as shown in the table: Length of call in minutes (t) Cost (C)

2 $1.00

6 $2.20

10 $3.40

14 $4.60

Graph the data in the table.

How much will it cost to make an 8 minute call?

Find the length of a call that costs $1.60

Determine the rule that connects the cost of a call, C, to the length of the call, t.

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Beth’s income is based solely on the number of hours she works, and she is paid a ﬁxed hourly wage. She earns $750 for working 30 hours. a

Identify the two variables in the question and assign a variable name to each.

Sketch a graph that displays her income against her hours worked.

Form an equation relating the variables.

Calculate Beth’s income when she works 25 hours.

Calculate the number of hours that Beth must work to earn $125.

The cost for a business to operate can be expressed in terms of the total number of hours it has operated for. The cost is $150 an hour. a

Identify the two variables in the question and assign a variable name to each.

Form an equation relating the two variables.

Find the total cost for the business to operate for 22 hours.

Find the number of hours that the business needs to operate to incur a total cost of $5400.

Reasoning 21

22, 24

21, 23, 25

21, 25, 26

If a line were plotted on each of these graphs, what would the gradient represent? a

Petrol used (L)

Amount earned ($)

Hours worked (h)

Distance (km)

Heartbeat (b)

Minutes running (m)

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Altitude in metres (a)

Number of seconds flying (f)

The graph shows the conversion between Country A and Country B’s currency:

b c

Use the graph to convert 8 of currency A to currency B.

Use the graph to convert 2 of currency B to currency A.

Discuss the usefulness and limitations of this model.

12 8 4 2

B 1

A baseball is thrown vertically upward, and the velocity of the baseball, V in metres per second after T seconds is given by V = 120 − 32T. a

Complete the table of values: Time (T ) Velocity (V )

What does the negative value of V, when T = 4, mean in context?

Discuss the usefulness and limitations of this model.

A racing car starts the race with 250 litres of fuel. From there, it uses fuel at a rate of 5 litres per minute. a

Complete the table of values: Number of minutes passed, x Amount of fuel left in tank, y

2 3 4 5 6 7 8 9 10

Determine an algebraic rule linking the number of minutes passed, x, and the amount of fuel left in the tank, y.

Describe how the amount of fuel in the car is changing over time.

Discuss the usefulness and limitations of this model.

A ball is rolled down a hill. The table shows the velocity, in m/s of the ball after a given number of seconds: Time in seconds, t Velocity, V

0 18

1 19.7

2 21.4

3 23.1

4 24.8

5 26.5

Find and interpret the gradient of the line.

Find and interpret the vertical intercept of the line.

Determine the rule that connects the velocity, V, to the time in seconds, t.

Find the velocity of the ball after 17 seconds, rounded to one decimal place.

Discuss the usefulness and limitations of this model.

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The graph shows the temperature of a room after the heater has been turned on:

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Write an equation to represent the temperature of the room, y, in terms of time, t.

Find the temperature of the room after the heater has been turned on for 30 minutes.

Discuss the usefulness and limitations of this model.

d e

Temp (°C)

5 3 1

t (mins) 1

Problem-solving 27, 28, 32, 33 27

29–31, 33

29–31, 34, 35

The graph shows the conversion between miles and kilometres:

Convert 10 miles to kilometres.

How many kilometres is 1 mile equal to?

14 12 10 8 6 4 2

mi 1

2 3 4 5 6 7

8 9 10

The graph shows the conversion between temperatures in Celsius and Fahrenheit: 50

°C

40 30 20 10 −20

−10

°F 10

−20 −30 −40 −50

Use the graph to convert as indicated:

114

40 °C to Fahrenheit

50 °F to Celsius

−20 °F to Celsius

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100

Petrol costs a certain amount per litre. The table shows the cost of various amounts of petrol in dollars: Number of litres (x) Cost of petrol ( y)

0 0

10 16.40

20 32.80

30 49.20

40 65.60

Calculate the cost of 47 L petrol. 30

The graph shows the conversion between temperatures in Celsius and Fahrenheit:

°F

Use the graph to convert 10 °C into Fahrenheit.

0 °C is 32 °F. For every 1 °C increase, by how much does the Fahrenheit temperature increase?

100 80 70 60

Would 80 °F be above or below normal body temperature? Normal body temperature is approximately 37 °C.

50 40

Write the rule for conversion between Celsius (C) and Fahrenheit (F).

Convert 35 °C into Fahrenheit.

30 20 10 5

°C 10 15 20 25 30 35 40

The number of ﬁsh in a river is approximated over a ﬁve year period. The results are shown in the table: Time in years (t)

Number of fish (F )

6000

5700

5400

5100

4800

4500

Find and interpret the gradient of the line.

What is the value of F when the line crosses the vertical axis?

Write an algebraic equation for the line relating t and F.

How many ﬁsh remain in the river after 9 years?

Determine the number of years it takes for there to be 1800 ﬁsh remaining in the river.

The table shows Peter’s earnings from ironing shirts: Shirts ironed Earnings ($)

0 4

2 8

4 12

6 16

8 20

Find and interpret the gradient of the line.

Find and interpret the y-intercept of the line.

What will be his total earnings if he irons 14 shirts?

How many shirts will he have to iron in order to earn $28?

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The graph shows the conversion between miles and kilometres: a

If a car is driving 32 km/hr in a school zone, at what speed is it travelling in miles per hour?

A road sign says the speed limit is 128 km/hr. Calculate this speed limit in miles per hour.

22 20 18 16 14 12 10 8 6 4 2

km 4

The graph shows the amount of water remaining in a bucket that was initially full before a hole was made in its side. a

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Write an equation to represent the amount of water remaining in the bucket, y, in terms of time, x.

Find the amount of water remaining in the bucket after 54 minutes.

32 Water remaining (L) 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 5

Time (mins)

20 25 30 35

The table shows the linear relationship between the number of plastic chairs manufactured and the total manufacturing cost: Number of plastic chairs Cost (dollars)

116

8 12 16 20 24 28 32 36

5 135

10 185

15 235

Find and interpret the gradient of the line.

Find and interpret the y-intercept.

Form an equation relating the number of plastic chairs and the cost.

Find the total cost of manufacturing 25 plastic chairs.

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3.07

Graph linear inequalities After this lesson, you will be able to… • graph linear inequalities on the Cartesian plane. • apply test points to verify if they satisfy the given linear inequality, and shade appropriate regions.

Graph linear inequalities We can represent a linear relationship using a line on the Cartesian plane, for example, y = −2x − 5.

2 1 −5 −4 −3 −2 −1

−1

x 1

−2 −3 −4

The line indicates the set of all points on the Cartesian plane that make the equation true.

Interactive exploration

Discover this concept in action online

−5 −6

mathspace.co

There are two key considerations when plotting inequalities. 1. Is the region shaded above or below the line? 2. Is the line included (solid) or excluded (dotted)?

The inequality y > 2x + 3 is satisfied for y-values greater than the expression 2x + 3. This corresponds to y-values ABOVE the line, which is the shaded region.

5 4 3 2 1 −4 − 4 − −3 3 − −2 2 −1

−1

−2

x 1

The line corresponds to the equation y = 2x + 3. Since this is not included in the strict inequality y > 2x + 3, the points on the line are not included in the solution. We indicate this by using a dotted or dashed line.

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5 4 3 2 1 −4 −3 −2 2 − −11

−11

x 1

Compare this to the inequality y ≤ 2x + 3. This represents all of the y-values less than or equal to 2x + 3. This includes all of the points BELOW the line, but also the points ON the line itself. We show that the line is included in the inequality by using a solid line.

−2 2

Example 1 Determine whether or not each of the following points satisfies the inequality y < 2x − 5: a (1, −4)

Create a strategy Substitute x = 1 into the right side of the inequality, and compare the result to the value of y, which is −4.

Apply the idea y < 2x − 5

Start with the inequality

y<2×1−5

Substitute the value of x

y < −3

Evaluate

−4 < −3

Compare to −4

This result matches the inequality, and so this point satisﬁes the inequality. b (−3, −9)

Create a strategy Substitute x = −3 into the right side of the inequality, and compare the result to the value of y, which is −9.

Apply the idea y < 2x − 5

Start with the inequality

y < 2 × (−3) − 5

Substitute the value of x

y < −11

Evaluate

−9 > −11

Compare to −9

This result does not match the inequality, and so this point does not satisfy the inequality.

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c (3, 1)

Create a strategy Substitute x = 3 into the right side of the inequality, and compare the result to the value of y, which is 1.

Apply the idea y < 2x − 5

Start with the inequality

y<2×3−5

Substitute the value of x

y<1

Evaluate

1=1

Compare to 1

This result does not match the inequality, and so this point does not satisfy the inequality. The fact that the two values are the same tells us that this point lies on the boundary line y = 2x − 5, but due to the inequality symbol < the boundary is not included in the inequality.

Example 2 Consider the inequality 3x − 4y ≥ 24. a Find the x-intercept of the boundary line 3x − 4y = 24

Create a strategy The x-intercept occurs when y = 0.

Apply the idea 3x − 4y = 24 3x − 4 × 0 = 24

Start with the equation Substitute y = 0 into the equation

3x = 24

Simplify

x=8

Solve for x

The x-intercept for the boundary line is (8, 0).

b Find the y-intercept of the boundary line 3x − 4y = 24.

Apply the idea 3x − 4y = 24 3 × 0 − 4y = 24

Start with the equation Substitute x = 0 into the equation

−4y = 24

Simplify

y = −6

Solve for y

The y-intercept for the boundary line is (0, −6).

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c Does the origin (0, 0) satisfy the inequality?

Create a strategy Substitute x = 0 and y = 0 into the inequality and see if it is true.

Apply the idea 3x − 4y ≥ 24 Start with the inequality 3 × 0 − 4 × 0 ≥ 24 Substitute the values into inequality 0 ≥ 24 Evaluate the left-hand side The statement is false, so the origin does not satisfy the inequality.

d Graph the inequality and shade the appropriate region.

Create a strategy Draw the boundary line using the intercepts found in parts (a) and (b). Decide which side to shade based on the answer to part (c) which states that the origin is not in the region.

Apply the idea 1 −1 −1 −2

y x 1

2 3 4 5 6 7 8 9

−3 −4

Since the inequality is ≥ we know the boundary line is included so it is solid. The boundary line goes through (8, 0) and (0, −6). We know that (0, 0) is not in the required region, so we shade the opposite side.

−5 −6 −7 − 7 −8 − 8 −9 − 9

e Do points on the line 3x − 4y = 24 satisfy the inequality?

Create a strategy Is the boundary included or excluded?

Apply the idea The inequality is 3x − 4y ≥ 24. The inequality symbol ≥ means “greater than or equal to”, so points that satisfy 3x − 4y = 24 are included. Points on the boundary line satisfy the inequality. 120

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Example 3 A linear inequality has points (1, 4) and (−1, 10) on its boundary line. The inequality is satisﬁed by (5, 5). Points on the boundary line itself do not satisfy the inequality. a Find the gradient of the boundary line.

Create a strategy

Apply the idea

Use m =

Let (1, 4) = (x1, y1) and (−1, 10) = (x2, y2)

Substitute the values into the gradient formula Simplify numerator and denominator Calculate

The gradient of the boundary line is −3 b Find the equation of the boundary line in the form y = mx + c.

Create a strategy Use the gradient found in part (a) and one of the points that lie on the boundary and substitute into y = mx + c.

Apply the idea y = mx + c

Start with y = mx + c

4 = −3 × 1 + c

Substitute m = −3 and (1, 4)

c=7

Evaluate c

The equation of the boundary line is y = −3x + 7, which may be expressed as y = 7 − 3x c State the inequality that satisﬁes the given conditions.

Create a strategy We are told that the point (5, 5) satisﬁes the inequality. Substitute x = 5 into the expression for the line and then compare this to the value y = 5 to see which way the inequality should go. We are also told that points on the boundary line do not satisfy the inequality, so the relationship will either be < or > and not ≤ or ≥.

Apply the idea 7 − 3x = 7 − 3 × 5

Substitute 5 for x

= −8

Evaluate

5 > −8

Compare to the y-value of 5

So the inequality is y > 7 − 3x. 3.07 Graph linear inequalities mathspace.co

121

d Graph the inequality.

Create a strategy Draw the boundary line. The given points (1, 4) and (−1, 10) can be used to draw the boundary. The boundary line should be dashed. Shade the side that contains (5, 5).

Apply the idea 10 9 8 7 6 5 4 3 2 1 −1 −1

Since the inequality is > we know the boundary line will be dashed.

The boundary line goes through (−1, 10) and (1, 4). We know that (5, 5) is in the required region, so we shade that side.

x 1

2 3 4 5 6 7 8 9

Example 4 Bernie sells his homemade pizza sauce in two sizes of containers: 200 mL and 500 mL. One week he has 4 L of sauce to sell. a Let x be the number of 200 mL bottles he ﬁlls, and y be the number of 500 mL bottles he ﬁlls. Write an inequality, with y as the subject, to represent this situation.

Create a strategy The amount of sauce in 200 mL bottles, added to the amount of sauce in 500 mL bottles, must be less than, or equal to, the amount of sauce available, 4000 mL.

Apply the idea x of the 200 mL bottles contains 200x mL. y of the 500 mL bottles contains 500y mL. The amount in bottles must be less than or equal to the amount available Divide each term by the common factor of 100 Subtract 2x from each side of the inequality Divide each term of the inequality by 5

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b Graph the solutions.

Create a strategy The boundary line may be expressed as 2x + 5y = 40. We can find the intercepts for this line by setting x = 0 to find the y-intercept and y = 0 to find the x-intercept. We can then decide which side of the boundary line to shade.

Apply the idea To ﬁnd the y-intercept set x = 0 2 × 0 + 5y = 40

Substitute x = 0

5y = 40

Simplify

y=8

Divide both sides by 5

To ﬁnd the x-intercept set y = 0 2x + 5 × 0 = 40

Substitute y = 0

2x = 40

Simplify

x = 20

Divide both sides by 2 The boundary line is drawn between the two intercepts.

The boundary is solid as the relationship is ≤.

A possible solution to the inequality is (0, 0) (which equates to ﬁlling no bottles), so we need to shade the side of the line that contains the origin.

6 5 4 3 2 1

x 2 4 6 8 10 12 14 16 18 20

c Which combinations of bottles of sauce could Bernie ﬁll? Select all that apply. A

No 200 mL bottles and eight 500 mL bottles

Four 200 mL bottles and ten 500 mL bottles

Ten 200 mL bottles and four 500 mL bottles

Twenty 200 mL bottles and no 500 mL bottles

Create a strategy As we have already graphed the line and shaded the region we can plot the points on the graph to determine if the points are in or out of the region.

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Apply the idea The four points we need to test are: y

A: (0,8)

B: (4, 10)

C: (10, 4)

8 A

D: (20, 0)

7 6 5 C

4 3 2 1

D x 2 4 6 8 10 12 14 16 18 20

Points A, C and D are in the acceptable region.

Reflect and check If the region had not already been plotted we could use the equation to test each set of coordinates. We can express the inequality as 2x + 5y ≤ 40 A: x = 0, y = 8. 2 × 0 + 5 × 8 = 40

Substitute x = 0, y = 8

This satisﬁes the inequality, so A is a possible solution. B: x = 4, y = 10. 2 × 4 + 5 × 10 = 58

Substitute x = 4, y = 10

This does not satisfy the inequality, so B is not a possible solution. C: x = 10, y = 4. 2 × 10 + 5 × 4 = 40

Substitute x = 10, y = 4

This satisﬁes the inequality, so C is a possible solution. D: x = 20, y = 0. 2 × 20 + 5 × 0 = 40

Substitute x = 20, y = 0

This satisﬁes the inequality, so D is a possible solution.

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Idea summary The following are features of two variable inequalities: • • • •

A line is solid if the symbols are ≥ or ≤. A line is dotted if the symbols are > or <. If the symbols are ≥ and >, then the shaded region must be above the line. If the symbols are ≤ and <, then the shaded region must be below the line.

3.07 Practice questions Understanding 1

Explorer 1, 3–5

1–5

Adventurer

1–5

Trailblazer

The two entertainment options for the school holidays are movies and ice-skating. • Movies, m, cost $12 each • Ice-skating, s, costs $21 per session The inequality 12m + 21s ≤ 150 represents the total amount of money you can spend. What does the point (0, 0) on this graph signify?

The point (0, 0) signiﬁes that there is no money spent on movies or ice-skating sessions.

The point (0, 0) signiﬁes that you spent all your budget on movies and ice-skating sessions.

The point (0, 0) signiﬁes that you attended only movies.

The point (0, 0) signiﬁes that you attended only ice-skating sessions.

Are these statements true or false? a

If a point lies in the shaded region of the graph of an inequality, it is not a solution to the inequality.

The point (0, 0) lies in the solution region of the inequality x + y > 0.

If an inequality has < or > symbol, the graph of the inequality will have a dashed line.

In the context of budgeting, the inequality 10m + 20s ≤ 100 could represent spending on movies (m) and snacks (s) with a total budget not exceeding $100.

The inequalities 3x − 2y > 6 and 3x − 2y ≥ 6 will have exactly the same graph.

Changing 2x + 3y ≤ 20 to 2x + 3y ≤ 30 will shift the line upwards on the graph.

The inequality x ≥ 0 represents all points to the right of the y-axis, including the y-axis itself.

When graphing an inequality, we shade the region that contains the solutions to the inequality.

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What does this shaded region mean in the context of a problem? A

The shaded region represents the line of the graph.

The shaded region represents all possible combinations of x and y-values that make the inequality false.

The shaded region represents all possible combinations of x and y-values that make the inequality true.

The shaded region looks nice and has no mathematical significance.

Is the inequality represented using a solid or dashed line on the graph? a

y ≤ 6x − 3

y ≤ 5x − 3

y ≥ 3x − 2

y>8

y ≥ 6x − 2

y < −9

6bc, 7b, 8, 10–13, 14c–h, 15, 16a, 17bc, 18

6, 7c, 9, 11, 12, 14c–l, 16d, 17cd

Is the coordinate pair a solution to the inequality? a

(6, 20); y ≥ 4x + 5 b

(3, 5); 2y ≤ x

(3, 2); 3x + 2y ≥ 12

(−1, 7); y ≥ 2x2 − 4x

For each inequality, identify if the points satisfy it or not. a

x − y > 10 i

b c

(−2, −9)

(15, 20)

iii

(16, 7)

(10, −6)

(−2, −3)

iii

(0, 0)

(2, 7)

(−2, 4)

iii

(−4, 29)

(2, −13)

y > 3x i

(12, 20)

y ≥ −8x − 3 i

126

y < −2x − 7

6ab, 7a, 8, 10–13, 14a–f, 15, 16ab, 17ab, 19

Fluency 6

Is the shaded region above or below the line? a

Ex 1

y < 5x − 7

(6, −50)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

For each graph: i

Find the equation of the boundary line.

Write the inequality that represents the shaded region.

−4 − 4 − −3 3 − −2 2 − −11

−11 − − 2 −2 − 3 −3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

7 y 6 5 4 3 2 (0, 2) 1

4 3 (−3, 2)

3 − −4 −3 −11 −2 2 −

−3 − 3 (−3, −4) − 4 −4

10 8 6

x 3

−4 −3 −2 2 − −11

− 2 −2 − 4 −4 − 6 −6 − −8 8 − −10 10

x 1

x 3

(0, 7)

y 4 (0, 4)

2 (0, 0) 1 2

4 2

(−1, 3)

−2

−11 −

−1

− 2 −2

1 −4 −3 −2 −1

7 y 6 5 4 3 2 (−2, 1) 1

(−2, 4)

(3, 2)

−4 − 4 − −3 3 − −2 2 −1 −1 1 2 −2 −3 − −4 − 4 −5 − 5 (0, −5) −6 − 6 −7 − 7

−4 − 4 − −3 3 − −2 2 − −11

−11

(3, 0) x 1

−2 2

−3

−3 3

−4

−4 4

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Write the inequality that describes the shaded region: a

6 5 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 3 −4 − 4 −5 − 5 −6 − 6

6 5 4 3 2 1

−6 − 6 − −5 5 − −4 4 − −3 3 − −2 2 − −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6

6 5 4 3 2 1

x 2 3

The line y = x − 5 is graphed on the Cartesian plane. a

Graph the solution to y > x − 5.

Graph the solution to y ≤ x − 5.

−6 − 6 − −5 5 − −4 4 − −3 3 −2 −1 −1 −2 −3 −4 −5 − 5 − −6 6

7 y 6 5 4 3 2 1 −3 −2 −1−1

x 1

2 3 4 5 6 7

−2 −3 −4 −5 −6 −7

5 4

3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

Ex 2

128

x 1

2 3 4 5

For 3x + 4y = 12: a

Find the x-intercept.

Find the y-intercept.

Graph the line on a Cartesian plane.

Write a related inequality that describes the region where (−18, 15) is located.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

For y = −3x + 4: a

Find the x-intercept.

Find the y-intercept.

Graph the inequality y ≤ −3x + 4.

Do these points satisfy the inequality y ≤ −3x + 4? i

e 13

(3, 4)

Graph the boundary line.

Does the point (0, 0) satisfy the inequality?

iii

Graph the inequality.

y≥x+5

Ex 4

(2, 2)

y > 2x − 4

2x + 3y < 12

6x + 3y > 12

y > 3x − 5

y < 4x − 7

y ≥ −3x + 5

Graph the inequality:

(5, −5)

Do the points on the line satisfy the inequality y ≤ −3x + 4?

y ≤ 4x − 3

iii

For each inequality:

Ex 3

(4, −8)

f −4x + y < 5

−3x + y < 7

h −2x + y < 9

3x + y < 11

For each set of conditions: i

Find the gradient of the boundary line.

Write the equation of the boundary line in the form y = mx + c.

iii

Write the inequality that satisﬁes the given conditions.

A linear inequality has solutions of (0, −3) and (−2, −7) on its boundary line and is not satisﬁed by (1, 4).

A linear inequality has solutions of (0, −4) and (−4, −20) on its boundary line and is satisﬁed by (−2, −8).

By using a graph of the inequality, find the maximum value of y with the given conditions. a

2x + y ≤ 10, when x = 3

3x + y ≤ 9, when x = 2

, when x = 3

, when x = 4

By using a graph of the inequality, find the minimum value of y with the given conditions. a

x − 2y ≤ 6, when x = 10

2x − 3y ≤ 12, when x = 8

, when x = 5

, when x = 4

Roxanne has set aside $14.00 in her shopping budget for fruit this week. Currently, oranges are on sale for $1.40 each, while apples are on sale for $1.75 each. a

Write an inequality relating the number of oranges, x, and the number of apples, y, that Roxanne buys.

Graph the solution to the inequality.

Can Roxanne buy these combinations of fruit? i

6 oranges and 7 apples

3 oranges and 4 apples

iii

7 oranges and 4 apples

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A book seller makes a proﬁt of $8 on every book sold online and $5 on every book sold in-store. The company wants to make a proﬁt of at least $240 a day selling books through online and in-store sales. a

Write an inequality relating the number of books sold online, x, the number of books sold in-store, y, and the target proﬁt.

Graph the solution to the inequality.

Will the combination of books sold in-store and online be enough for the book seller to meet the target proﬁt?

15.1 online and 29.1 in-store

25 online and 5 in-store

iii

13 online and 23 in-store

20 online and 20 in-store

Using the graph, ﬁnd the minimum number of total books ( from both online and in-store) that need to be sold to reach the target proﬁt.

Reasoning 20

21, 22

Explain the differences between the graphs of: a

4x + 5y > 20 and 4x + 5y ≥ 20

4x + 5y ≤ 20 and 4x + 5y ≥ 20

Can the inequality x + y > 10 have integer solutions where both x and y are less than 3? Explain your reasoning.

Can the inequality x − y > 10 have integer solutions where both x and y are less than 3? Explain your reasoning.

130

A gardening club wants to invest in new tools and plants. They can buy a set of tools for $12 and a bundle of plants for $21. They have a budget of $150. a

Given the inequality 12t + 21p ≤ 150 (where t represents the number of tool sets and p the number of plant bundles), explain why the club can’t buy more than 12 tool sets or more than 7 plant bundles.

Suppose the club decides to buy 5 sets of tools. Explain why the maximum number of plant bundles they can afford is 4.

Suppose the club decides to buy an equal number of tool sets and plant bundles. What is the maximum number of each they can afford? Explain your reasoning.

Suppose the budget for the gardening club was raised to $180. They decide to buy an equal number of tool sets and plant bundles. What is the maximum number of each they can afford?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Problem-solving 23, 24, 26, 27 23

23, 25–28

23, 25–30

Applicants for a particular job are asked to sit a numeracy test and verbal reasoning test. Successful applicants must obtain a minimum combined score of 43 for both tests. Write an inequality relating the applicant’s score on the numeracy test, x, and the verbal reasoning test, y.

Athletes applying for a speciﬁc sports scholarship must pass both a physical and a written test. The minimum combined score for both tests is 50. Write and graph an inequality relating the athlete’s score on the physical test, x, and the written test, y.

Rochelle is being careful with her spending so she can later purchase a house. She has allocated no more than $855 each month for both travel expenses and eating out. On average, every time she eats out it costs $19, and every travel journey costs $9. Write an inequality relating the number of times she eats out, x, and the number of her travel journeys each month, y.

Edward is setting up a monthly budget to save for a new car. He plans to spend no more than $720 each month on groceries and utility bills. On average, each grocery bill is $30, and each utility bill is $60. Write and graph an inequality relating the number of utility bills, x, and the number of grocery bills each month, y.

In the new basketball season, William is looking to beat his personal best of scoring 102 points in total for the whole season (this does not include points from fouls). a

If he is to score x ‘two pointers’ and y ‘three pointers’ throughout the new season, write an inequality relating x, y and the number of points he hopes to score.

Graph the solution to the inequality.

In which given scenarios will William beat his personal best for the season? A

He scores 17.6 ‘two pointers’ and 27.9 ‘three pointers’

He scores 23 ‘two pointers’ and 24 ‘three pointers’

He scores 4 ‘two pointers’ and 27 ‘three pointers’

He scores 31 ‘two pointers’ and 8 ‘three pointers’

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Sean is thinking about how to use his remaining spending money for snacks. A pack of dried fruit costs $6 and a bag of mixed nuts costs $4. He has $48 remaining. a

Write an inequality relating the number of packs of dried fruit, x and the number of bags of mixed nuts, y that Sean buys.

Sketch the region containing the points corresponding to all the different ways Sean could spend his remaining spending money on these snacks.

Supposing that Sean does not need to spend all of it, how many different ways can Sean spend his remaining money on some combination of dried fruit and mixed nuts?

Oprah is thinking about how to use her last arcade tokens. A game of Table Tennis costs 2 tokens and a game of Frog Days costs 5 tokens. She has a total of 30 tokens left. a

Write an inequality relating the number of times Oprah plays Table Tennis, x, and the number of times she plays Frog Days, y.

Construct a graph of the region containing the points corresponding to all the different ways Oprah could spend her remaining tokens across the two games.

Interpret the value of the x-intercept of the boundary line in the given context.

Interpret the value of the y-intercept of the boundary line in the given context.

Monica works part-time as a tutor and also as a sales associate. The number of hours she works in each role varies from week to week, and she needs to earn at least a certain amount weekly to cover her expenses. The graph of the inequality relating the number of hours she works as a tutor, x, and as a sales associate, y, is shown. a

Write the rule for the inequality.

Find Monica’s hourly rate of pay for each job.

Find the amount of money she needs weekly to cover her expenses.

40 35 30 25 20 15 10 5

x 5 10 15 20 25 30 35 40 45

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3 Chapter review 1

For the equation y = 4x + 7: a

Complete the table of values: x y

b 2

−1

Sketch the line on a Cartesian plane.

For the linear equation

Find the gradient.

Find the y-intercept.

Use the gradient and y-intercept to sketch the line.

Sketch a line that passes through the point (−2, 7) and has a gradient of −2.

Sketch the graph of y = −3x + 8 by ﬁnding any two points on the line.

Javier tracked the number of apps on his phone over a number of months. Month (M ) Apps (A)

1 118

3 104

4 96

Plot the points on a Cartesian plane.

Is this relationship linear?

The time in minutes and the temperature in degrees Celsius (°C) are given in the table: Time (x) Temperature (y)

2 106

1 8

2 11

3 14

4 17

5 20

By how much is the temperature increasing each minute?

Find the initial temperature.

Form an equation relating x and y.

Graph the linear relationship represented in the table.

Rice costs a certain amount per kilogram. The table shows the cost of various amounts of rice in dollars: Number of kilograms (x) Cost of rice (C)

0 0

10 32

20 64

30 96

Complete the table.

Graph the linear relationship represented in the table.

Find the cost of rice per kilogram.

Calculate the cost of 24 kg of rice.

Find the amount of rice that costs $44.80.

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133

Solve: a

g 10

W = Fd, make F the subject

Q = n2L, make n the subject

, make H the subject

Rearrange the formula to make, r the subject of the formula.

Calculate the radius of a cylinder with a volume of 200 cm3 and a height of 10 cm. Round your answer to two decimal places.

Are these lines parallel to y = 3x + 5? a

y = −3x + 5

y = 3x + 2

3x + y = −5

y = 3x

x−2

4y = x

Are these lines perpendicular to y = −4x + 1? a

S = I (1 + rt), make I the subject

The volume of a cylinder is given by the formula V = π r2h, where r is the radius of the base and h is the height of the cylinder. b

Make the mentioned variables the subject in the following equations:

x+5

y = −4x + 5

The graph shows the conversion between metres and feet:

If a plank of wood is 10 ft in length, how long is it in metres?

In softball, the ﬁrst base is 60 ft from home plate. Calculate this length in metres.

6 5 4 3 2 1

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ft 2 4 6 8 10 12 14 16 18 20 22

Is the coordinate pair a solution to the inequality? a

(2, 2) ; x + 4y > 11

(−1, 6) ; y ≤ 3x2 − 2x + 1

For each graph, write the inequality that represents the shaded region. a

−4 − 4 −3 −3 −2 2 −1 −1

−4 − 4 −3 −2 −1 −1

−2

−3

−3 − 3

−4

−4 − 4

Find the x-intercept.

Find the y-intercept.

Graph the line on a Cartesian plane.

Write a related inequality that describes the region where (−12, 9) is located.

For y = −2x + 2: a

Graph the inequality y ≤ −2x + 2.

Do these points satisfy the inequality y ≤ −2x + 2? (2, 3)

(3, −6)

iii

(4, −2)

(1, 2)

Do the points on the line satisfy the inequality y ≤ −2x + 2?

Graph the inequality: a

For 3x + 5y = 15:

2x − 3y < 15

A linear inequality has solutions of (0, −3) and (7, −31) on its boundary line and is satisﬁed by (3, −11). a

Find the gradient of the boundary line.

Write the equation of the boundary line in the form y = mx + b.

Write the inequality that satisﬁes the given conditions.

Throughout university, Ben works as a waiter, getting paid $14 per hour, and as a valet, getting paid $18 per hour. The number of hours he works in each job can vary from week to week, and he needs to be able to at least cover his weekly expenses of $210. Write an inequality relating the number of hours he works as a waiter, x, and valet, y.

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Big ideas Quadratic equations can be represented graphically and solved through various methods, and their properties can be determined using the discriminant.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10A13 solve simple quadratic equations using a range of strategies, including null factor law VC2M10AA02 devise and use algorithms and simulations to solve mathematical problems VC2M10AA07 factorise monic and non-monic quadratic expressions and solve a wide range of quadratic equations derived from a variety of contexts VC2M10AA10 experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns

Quadratics Chapter outline 4.01 Quadratic equations 4.02 Quadratic graphs 4.03 Solve by factorisation 4.04 Solve by completing the square Investigation: Use bisection to solve quadratics Investigation: Derive the quadratic formula 4.05 The quadratic formula 4.06 The discriminant 4.07 Optional: Further quadratic equations Chapter 4 review

138 146 170 181

188

Developers use quadratics to control the speed and physics of race cars in video games!

4.01 Quadratic equations After this lesson, you will be able to… • identify the key features of quadratic graphs and how they relate to the solutions of the equation. • recognise that quadratic equations can have zero, one, or two solutions. • apply the null factor law to solve quadratic equations.

Quadratic graphs The standard form of a quadratic equation is y = ax2 + bx + c, where a ≠ 0. For example, the quadratic equation y = 2x2 − 3x + 5 has the values a = 2, b = −3, c = 5. The simplest quadratic equation is y = x2 and has the values of a = 1, b = 0, c = 0. The graph of a quadratic equation is called a parabola. This parabola is concave up as it opens upwards. A parabola will be concave up when the value of a is positive, such as y = x2.

y 7 6 5 y = x2

The point (0, 0) is the minimum point on the curve and is called the turning point or vertex.

4 3 2 1

−2

−1

x 1

−1

y 1 −2

−1

x 1

If the value of a is negative, such as y = −x2, the graph opens downwards. It is called concave down and the turning point is a maximum.

−2 y = −x2 −3 −4 −5 −6 −7

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There are many other key features including the intercepts and axis of symmetry. These key features will depend on the values of a, b, and c in the quadratic equation. axis of symmetry

y x x-intercepts Maximum x

y-intercept

y-intercept Minimum

Concave up with two x-intercepts

axis of symmetry

Concave down with no x-intercepts

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 For the parabola deﬁned by the equation y = −x2 + 3: a Is the parabola concave up or concave down?

Create a strategy For a parabola in the form y = ax2 + bx + c, concavity depends on the x2 coeﬃcient a: positive for concave up, negative for concave down.

Apply the idea Looking at the equation, the coeﬃcient of x2 is −1. So the parabola is concave down.

b What is the y-value of the y-intercept of the parabola?

Create a strategy To ﬁnd the value of the y-intercept, set the x-value equal to 0.

Apply the idea y = −x2 + 3 =3

Rewrite the equation Substitute x = 0

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139

Idea summary A quadratic equation is of the form y = ax2 + bx + c, where a ≠ 0. The graph is called a parabola. If a > 0, the graph is concave up as it opens upwards. If a < 0, the graph is concave down as it opens downwards. The minimum or maximum point on the curve is called the turning point or vertex. The turning point and whether it is concave up or down can tell us about the number of solutions.

Quadratic equations Quadratic equations can be written in the form ax2 + bx + c = 0. Quadratic equations can come in many forms such as: x2 = 16, (x − 5)2 = 25, 2x2 + 5x = 17 Quadratic equations can have zero, one or two solutions. This is related to the number of x-intercepts on the graph. For example: x2 = −4 has no solutions as a squared number is always positive. The related quadratic graph will have no x-intercepts.

x2 = 0 has one solution only, x = 0. The related quadratic graph will have one x-intercept.

x2 = 9, has two solutions, x = 3 and x = −3. We can also write this as x = ±3. The related quadratic graph will have two x-intercepts.

The symbol ± means “plus or minus”. We can use this as a shorthand for both the positive and negative of a number.

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Example 2 Solve for x: a x2 = 16

Create a strategy To ﬁnd x, we need to reverse the square operation.

Apply the idea x = ±4

Take the square root of both sides

Reﬂect and check Remember to always use the plus or minus sign (±) when dealing with perfect square answers.

b 81x2 − 64 = 0

Create a strategy Isolate the term with x and reverse the square operation.

Apply the idea Add 64 to both sides

Divide both sides by 81

Take the square root of both sides

Create a strategy Isolate x using inverse operations.

Apply the idea

Add 3 to both sides

Multiply both sides by 16

Take the square root of both sides

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Example 3 Solve (7 − x)2 = 81.

Create a strategy We can directly reverse the square operation on both sides, without expanding the binomial square on the right-hand side.

Apply the idea 7 − x = ±9

Take the square root of both sides

We need to further simplify both equations 7 − x = 9 and 7 − x = −9. 7−x=9

Write the first equation

−x = 2

Subtract 7 from both sides

x = −2

Divide both sides by −1

7 − x = −9

Write the second equation

−x = −16

Subtract 7 from both sides

x = 16

Divide both sides by −1

Idea summary Equations where the highest power is two are quadratic equations. They can be written in the form ax2 + bx + c = 0 and can have zero, one or two solutions. If we can rearrange a quadratic equation into the form x2 = k, then we can solve the equation by taking the positive and negative square roots. That is, x = ± . The symbol ± means “plus or minus”.

The null factor law The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. For example, if ab = 0 then either a = 0 or b = 0 We can use this to solve quadratic equations that are not in the form x2 = k. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

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Example 4 Solve: a (x − 6) (x + 7) = 0

Create a strategy Using the null factor law, set both factors to 0 and solve the resulting linear equations.

Apply the idea x−6=0 x=6

Set the ﬁrst factor to 0 Add 6 to both sides

x+7=0

Set the second factor to 0

x = −7

Subtract 7 from both sides

So the solutions are x = 6 and x = −7.

b 4m (m + 5) = 0

Create a strategy Using the null factor law, set both factors to 0 and solve the resulting linear equations.

Apply the idea 4m = 0

Set the ﬁrst factor to 0

m=0

Divide both sides by 4

m+5=0

Set the second factor to 0

m = −5

Subtract 5 from both sides

So the solutions are m = 0 and m = −5.

Idea summary The null factor law states that if a product of two or more factors is equal to 0, then at least one of those factors must be equal to 0. If we can rearrange a quadratic equation into the form (x − p) (x − q) = 0, then we know that either x − p = 0 or x − q = 0. We can solve the quadratic equation by solving each of these linear equations.

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4.01 Practice questions Understanding

How many solutions can a quadratic equation have?

A quadratic equation can be written in the form ax2 + bx + c = 0. Write the values of a, b and c for these quadratic equations:

2x2 − 7x + 5 = 0

−3x2 + 8x = 7

x2 = 16

−x2 = 4x − 3

A quadratic equation can be written in its factorised form (x − a) (x − b) = 0. Write the values for a and b for these quadratic equations. (x − 3) (x − 5)

Ex 4

144

(x − 6) (x + 3)

x(x − 4)

The equation x2 − 144 = 0 has a positive integer solution of x = 12. Find its other solution. 6, 7a–f, 8a–c, 9a–i

6, 7a–g, 8c–f, 9a–i

6, 7e–l, 8c–h, 9c–l

Are the graphs of these quadratic equations concave up or concave down? a

Fluency

Ex 3

Trailblazer 2–5

Given that we know 5 × 5 = 25. What is

Ex 2

1–5

Adventurer 2–5

Ex 1

Explorer

y = −x2 + 4x − 2

y = 3x2 + 8x

y = −x (x + 2) + 5

Solve: a

a2 = 2

k2 = 25

x2 = 121

t2 = 294

x2 − 121 = 0

4y2 = 100

15 = x2 − 10

36 = 25y2

−3k2 = −12

81k2 + 8 = 24

−16 = −81x2

Solve: a

(z + 3)2 = 49

(m − 7)2 = 81

(x − 6)2 = 2

(−3 + p)2 = 64

81 = (2 − n)2

(7 − x)2 = 81

(8l + 9)2 = 256

10 ( p2 − 7) = 930

x (2x − 9) = 0

Solve: a

j (j + 7) = 0

−5m (m + 5) = 0

(10x − 9) = 0

(4k − 9)2 = 0

(−3 + 7x)2 = 0

(e − 4) (e − 2) = 0

(s + 6) (s + 7) = 0

(8x − 5) (3x − 7) = 0

(3w + 8) (5w − 7) = 0

(3x − 17) (2x − d) = 0

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Reasoning

10, 11, 13

12–14

If a quadratic equation is given in the form x2 = k, how can we ﬁnd the solutions for x?

Explain why the quadratic equation x2 − 9 = 0 has two solutions.

A quadratic equation has two solutions, x = 4 and x = −3. Explain how to write the equation in the form (x − a) (x − b) = 0.

Jade used the null factor law to solve the equation (x − 4) (x + 6) = 0 and found the solutions x = −4 and x = 6. Is Jade’s answer correct? Explain why.

Explain and justify why the equation x2 + 4 = 0 has no solutions.

Problem-solving 15–17 15

15, 17, 18

17–20

The Widget and Trinket Emporium has released the forecast of its revenue over the next year. The revenue R (in dollars) at any point in time t (in months) is described by the equation: R = −(t − 12)2 + 4 a

Is the graph of R concave up or concave down?

Does the revenue reach a maximum or minimum point?

When will the revenue be zero?

The equation 4x2 + kx + 16 = 0 has one solution: x = 2. Find the value of the coeﬃcient k.

One solution to the equation ax2 − 32x − 80 = 0 is x = 4. Find the value of the coeﬃcient a.

Neville needs a sheet of paper x cm by 13 cm for an origami giraffe. The local origami supply store only sells square sheets of paper. The lower portion of the image below shows the excess area, A, of paper that will remain after Neville cuts out the x cm by 13 cm piece. The excess area, in square centimetres, is given by the equation: A = x (x − 13) a

At what lengths of x will the excess area be zero?

Does the graph of A have a minimum or maximum turning point?

For what value of x will Neville be able to make an origami giraffe with the least amount of excess paper?

13 x A

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The equation px2 + qx − 20 = 0 has solutions coeﬃcients p and q.

Given ax2 − c = 0: a b

and x = 4. Find the value of the

Solve for x, in terms of a and c. For what values of a and c, would there be: i

No solutions

One solution

iii

Two solutions

4.02 Quadratic graphs After this lesson, you will be able to… • understand the different forms of quadratic equations and how they relate to the graph of the equation. • identify the turning point and interpret its significance in the graph of a quadratic. • understand the transformations of a parabola, including translations, dilations and reflections. • calculate the y-intercept and x-intercepts of a parabola and understand their significance in sketching the graph. • apply knowledge of quadratic graphs to solve problems.

Transformations of parabolas We can graph a quadratic using transformations that can be identified from the equation. Let’s explore the possible transformations and how they are represented in the equation and graph.

Interactive exploration

Discover this concept in action online

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mathspace.co

A parabola can be vertically translated by increasing or decreasing the y-values by a constant number. A translation of y = x2 up by k units gives the equation y = x2 + k. y 4 y = x2 + 2

3 2

y = x2

x −4 −3 −2 −1

This graph shows y = x2 translated vertically up by 2 to get y = x2 + 2, and down by 2 to get y = x2 − 2.

y = x2 − 2

−1 −2

Similarly, a parabola can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value and the translation must be squared together. That is, to translate y = x2 to the left by h units we get y = (x + h)2. y 6 5 y = x2

4 3 y = (x + 2)2 2

y = (x − 2)2

1 −4 −3 −2 −1

x 1

−1

This graph shows y = x2 translated horizontally left by 2 to get y = (x + 2)2 and right by 2 to get y = (x − 2)2.

−2

A parabola can be dilated by multiplying every y-value by a constant number greater than 1. So to enlarge the parabola y = x2 by a scale factor of a, we get y = ax2. We can compress a parabola by using a scale factor between 0 and 1. y 6 5 y = 2x2 4

y = x2

This graph shows y = x2 vertically enlarged by a scale factor of 2 to get y = 2x2 and compressed by

3 2

1 −4 −3 −2 −1

−1

x2 2

x 1

a scale factor of 2 to get

−2

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y 4 3 2 y = x2

1 −4 −3 −2 −1

−1

x 3

−2

Finally, we can reflect a parabola across the x-axis by multiplying by −1. So to reflect y = x2 across the x-axis, we get y = −x2. Notice that reflecting will change the concavity (in this case from concave up to concave down).

−3 −4

y = −x2

All of these transformations can be performed on a quadratic and are represented by the equation y = a (x − h)2 + k.

Example 1 For the parabola y = −(x + 2)2 − 5: a Which successive transformations turn y = x2 into the equation y = −(x + 2)2 − 5? A

Reﬂection across the x-axis, horizontal shift 2 units to the left and vertical shift 5 units down.

Horizontal shift 2 units to the right, vertical shift 5 units down, reﬂection across the x-axis.

Reﬂection across the x-axis, vertical shift 2 units down and horizontal shift 5 units to the left.

Vertical shift 2 units up, horizontal shift 5 units to the left, reﬂection across the x-axis.

Create a strategy Identify the values of a, h, and k in turning point form where: • a represents a reflection and/or dilation • h represents a horizontal translation • k represents a vertical translation

Apply the idea y = (x + 2)2 shifts y = x2, 2 units to the left. y = −(x + 2)2 reﬂects y = (x + 2)2 about the x-axis. y = −(x + 2)2 − 5 shifts y = −(x + 2)2 down 5 units. So the answer is option A.

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b State the coordinates of the turning point of the curve.

Create a strategy Apply the transformations from part (a) to the turning point.

Apply the idea y 2 1 −4 −3 −2 −1

x 1

−1

−2 −3

The turning point of y = −(x + 2)2 − 5 is the turning point of y = x2 reﬂected across the x-axis, horizontally shifted 2 units to the left and shifted vertically 5 units down. As we can see in the graph, the coordinates of the turning point of the curve are: (0 − 2, 0 − 5) = (−2, −5).

−4 −5 −6

Reﬂect and check Because the graph has been reflected across the x-axis, the parabola is concave down. This means the turning point will be a maximum.

c Plot the graph of the equation.

Create a strategy Use the transformations and turning point we found in the previous parts.

Apply the idea 3 2 1

−4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 −5 −6 −7 −8 y = −(x + 2)2 − 5

Starting with the top dashed curve, y = x2, we need to reﬂect the curve across the x-axis, shift the curve 2 units to the left, then shift this curve down 5 units to get the ﬁnal parabola.

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Idea summary The parabola y = x2 can be transformed by: • Vertical translation by k units: y = x2 + k • If k > 0, like y = x2 + 3, then it is translated upwards. • If k < 0, like y = x2 − 1, then it is translated downwards. • Horizontal translation by h units: y = (x − h)2 • If h > 0, like y = (x − 2)2, then it is translated to the right. • If h < 0, like y = (x + 3), then it is translated to the left. • Dilated by a scale factor of a: y = ax2 • If a > 1, or a < −1, like y = 2x2, then it is vertically enlarged (stretched). • If −1 < a < 1 (with a ≠ 0), like y = 0.5x2, then it is compressed. • Vertically reﬂected across the x-axis: y = −x2 All of these transformations can be performed on a quadratic and are represented by the equation y = a (x − h)2 + k.

Quadratics in turning point form A quadratic equation in turning point form is written as y = a(x − h)2 + k, where a, h, k are numbers. We can find the key features and then use these to graph the parabola. The turning point is (h, k). The axis of symmetry is x = h. To find the y-intercept, substitute x = 0 and evaluate. To find the x-intercepts, if required, substitute y = 0 and solve the equation using inverse operations. Going the other way, we can also find the equation of a quadratic from the graph using these steps. 1. Identify the turning point from the graph and fill in the values of h and k in y = a(x − h)2 + k. 2. Substitute in any other point on the curve in for x and y to solve for the value of a. 3. Write the equation.

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Example 2 For the quadratic equation y = −2 (x − 3)2 + 8: a Is the parabola concave up or concave down?

Create a strategy Concavity depends on the sign of a: positive for concave up, negative for concave down.

Apply the idea Since the quadratic equation is y = −2 (x − 3)2 + 8 and a = −2 < 0, the parabola is concave down.

b Is the parabola more or less steep than y = x2?

Create a strategy Look at the value of a. Use the fact that as the value of a gets further from 0, the parabola becomes more steep and narrow.

Apply the idea Since we have that a = −2, and −2 is further away from zero than 1 is, the parabola is more steep than y = x2.

c Find the coordinates of the turning point.

Create a strategy A quadratic of the form y = a(x − h)2 + k has a turning point at (h, k).

Apply the idea The turning point of the quadratic equation y = −2 (x − 3)2 + 8 is at (3, 8).

d Find the y-intercept.

Create a strategy Substitute x = 0 into the equation and solve for y.

Apply the idea y = −2(x − 3)2 + 8 Write the equation = −2(0 − 3)2 + 8 Substitute x = 0 = −2(9) + 8

Evaluate the power

= −10

Evaluate

The y-intercept is at (0, −10).

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e Sketch the graph of this quadratic. Include the axis of symmetry.

Create a strategy Plot the y-intercept and turning point and draw a curve through the points.

Apply the idea Since the axis of symmetry occurs at x = h, then the axis of symmetry for y = −2(x − 3)2 + 8 is x = 3. 10 8 6 4 2

y (3, 8)

1 2 3 4 5 6 7 8 9 −2 −1 −2 −4 −6 −8 −10 (0, −10)

Reﬂect and check Graphing using as many key features as possible helps us to identify any errors we may have made when ﬁnding the key features.

Example 3 A parabola has the equation y = x2 + 4x − 1: a Express the equation of the parabola in the form y = (x − h)2 + k by completing the square.

Create a strategy Add and subtract

into the expression and factorise the perfect square part.

Apply the idea For x2 + 4x – 1 we need to add and subtract x2 + 4x − 1 = x2 + 4x + 4 − 4 – 1

Factorise the perfect square part

Simplify the constant

= (x + 2) − 4 – 1 = (x + 2) – 5

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b Find the y-intercept of the curve.

Create a strategy Substitute x = 0 into the equation of the parabola.

Apply the idea y = (0)2 + 4(0) – 1

Substitute 0 to x

y = −1

Evaluate

Reﬂect and check We can check that we have correctly completed the square by ﬁnding the y-intercept using that form as well. y = (x + 2)2 – 5

Write the equation

y = (0 + 2) – 5

Substitute x = 0

y = 4 – 5

Evaluate the brackets and indices

y = −1

Evaluate the subtraction

This is the same value we found when using the standard form, which tells us we correctly completed the square.

c Find the turning point of the parabola.

Create a strategy We can find the turning point using the equation found in part (a).

Apply the idea The equation in turning point form is: y = (x + 2)2 − 5. We can write this as y = (x − (−2))2 − 5. The coordinates of the turning point are (−2, −5).

Reﬂect and check If we hadn’t found the equation in turning point form, we could have used the idea that the axis of symmetry is at

From the equation, y = x2 + 4x − 1, a = 1, b = 4.

Substitute a and b Evaluate

The equation of the axis of symmetry is x = −2, which is also the x-coordinate of the turning point. y = x2 + 4x – 1

Write the equation

y = (−2) + 4(−2) – 1

Substitute x = −2

y = −5

Evaluate

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d Is the parabola concave up or down? A

Concave up

Concave down

Create a strategy Look at the coefficient of x2. A positive coefficient of x2 means the parabola is concave up while a negative coefficient of x2 means the parabola is concave down.

Apply the idea Since there is a positive coeﬃcient of x2 in the equation of the parabola y = x2 + 4x − 1, the parabola is concave up. So, the correct answer is Option A.

e Plot the curve y = x2 + 4x − 1.

Create a strategy Plot the intercepts and turning point and draw a curve through the points.

Apply the idea 4 3 2 1 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

x 1 2 3 4

Idea summary If an equation of a parabola is in the form y = a (x − h)2 + k, then: • The turning point form will be (h, k). • The axis of symmetry will be x = h. • The y-intercept, can be found by setting x = 0 and evaluating to get (0, a(h)2 + k). These key features can be used to plot the parabola.

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Quadratics in factorised form A parabola will always have one y-intercept and can have zero, one or two x-intercepts. A quadratic equation in factorised form is written as y = a (x − p) (x − q) where p and q are the x-values of the x-intercepts. y 7

−4 −3 −2 −1

2 1 −1

−1

x 1

−1

−2

x 1

−3

−4

y = (x − 4)2 We can graph a quadratic in factorised form using these steps: 1. Find the y-intercept by substituting x = 0, this gives us y = apq. 2. Find the x-intercepts by setting y = 0 and using the null factor law. 0 = a (x − p) (x − q), so that x = p or x = q, and the two x-intercepts are ( p, 0) and (q, 0). 3. Find the axis of symmetry by ﬁnding the average of the two intercepts, 4. The x-value of the turning point is the axis of symmetry, so to ﬁnd the y-value, we can substitute the x-value into the original equation and evaluate for y. If the equation of a parabola factorises into the form y = a (x − p) (x − q), then it is possible to use these key features to plot the parabola.

Example 4 Consider the graph shown: a Find the coordinates of the turning point.

Create a strategy Find the point on the parabola where the direction changes from downward to upward.

Apply the idea Based on the given graph, the coordinates of the turning point of the parabola are (−3,−6).

10 8 6 4 2 −7 −6 −5 −4 −3 −2 −1 −2

x 1

−4 −6

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b Find the equation in the form y = a(x − h)2 + k.

Create a strategy Use the coordinates of the turning point found in part (a) to fill in h and k in the quadratic equation of the form y = a(x − h)2 + k. We can then use another point on the curve to find the value of a.

Apply the idea Since the coordinates of the turning point of the parabola are (−3, −6), then the equation is y = a(x − (−3))2 − 6 which simplifies to y = a(x + 3)2 − 6 To determine the value of a, we can use one of the points that passes through the curve of the parabola, such as (0, 12).

Write the equation

Substitute the values of x and y

Add 6 to both sides

Evaluate the addition

Evaluate the power

Divide both sides by 9

Evaluate So, the quadratic equation of the given graph is: y = 2(x + 3)2 − 6

Reﬂect and check We could have used any point we can identify from the graph, it does not need to be the y-intercept at (0, 12). For example, we could also have used (−5, 2), (−1, 2), or (−6, 12).

Example 5 For the equation y = (2 − x) (x + 4): a State the y-value of the y-intercept.

Create a strategy Substitute x = 0 into the equation and solve for y.

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Apply the idea y = (2 − x) (x + 4) Write the equation y = (2 − 0) (0 + 4) Substitute x = 0 =8

Evaluate the product

b Determine the x-values of the x-intercepts.

Create a strategy Substitute y = 0 into the equation and solve for x using the null factor law.

Apply the idea y = (2 − x) (x + 4)

Write the equation

0 = (2 − x) (x + 4)

Substitute y = 0

2 − x = 0 Equate the ﬁrst factor to 0 x = 2 Add x to both sides x+4=0 x = −4

Equate the second factor to 0 Subtract 4 from both sides

So the x-values of the x-intercepts are x = 2, x = −4.

c Determine the coordinates of the turning point of the parabola.

Create a strategy The x-coordinate of the turning point is halfway between the x-intercepts.

Apply the idea Find the average of the x-intercepts Evaluate Substitute x = −1 into the equation Evaluate The coordinates of the turning point are (−1, 9).

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d Plot the graph of the parabola.

Create a strategy Plot the intercepts and the turning point and draw a curve through the points.

Apply the idea 9 8 7 6 5 4 3 2 1 −5 −4 −3 −2 −1

−1 −2 −3 −4

x 1

Idea summary If a quadratic equation can be factorised into the form y = a (x − p) (x − q), then: • The y-intercept will be (0, apq). • The x-intercepts will be ( p, 0) and (q, 0). • The x-value of the turning point is

. The y-value of the turning point will

be found by substituting in the x-value turning point into the equation. These points can be used to plot the parabola.

Quadratics in standard form A quadratic equation in standard form is written as ax2 + bx + c = 0, where a ≠ 0. If a > 0, the graph is concave up and the turning point is a minimum. If a < 0, the graph is concave down and the turning point is a maximum. To graph the parabola, we can: 1. Find the axis of symmetry using

2. Find the turning point by substituting in the axis of symmetry for x and evaluating for y. 3. Find the y-intercept by setting x = 0 and evaluating to get y = c. If it is able to be factorised, we can use the null factor law to ﬁnd the x-intercepts. The turning point and y-intercept can provide a sketch of the graph. If other points are required, they can be found by substituting x-values into the equation. 158

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Example 6 For the equation y = 4x + x2: a Factorise the expression 4x + x2.

Create a strategy Take out the common factor of the two terms.

Apply the idea The common factor of the two terms is x. 4x + x2 = x(4 + x)

b Solve for the x-values of the x-intercepts of y = 4x + x2.

Create a strategy Substitute y = 0 into the equation and solve for x using the null factor law.

Apply the idea y = 4x + x2 Write the equation y = x (4 + x) Factorise 0 = x (4 + x) Substitute y = 0 x = 0 Equate the ﬁrst factor to 0 4+x=0

Equate the second factor to 0

x = −4 Subtract 4 from both sides So the x-values of the x-intercepts are x = 0, x = −4.

c Determine the y-value of the y-intercept of the graph.

Create a strategy Substitute x = 0 into the equation then solve for y.

Apply the idea y = 4x + x2

Write the equation

= 4 × 0 + 02

Substitute x = 0

Evaluate

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d Find the coordinates of the turning point.

Create a strategy Use the equation

and substitute the value of x into the equation of the parabola.

Apply the idea From the equation, y = 4x + x2, a = 1, b = 4. Substitute a and b

Evaluate The equation of the axis of symmetry is x = −2, which is also the x-coordinate of the turning point. y = 4x + x2 Write the equation = 4(−2) + (−2)2 Substitute x = −2 Evaluate

= −4

The coordinates of the turning point are (−2, −4).

Reﬂect and check We could also have used the average of the x-intercepts to ﬁnd that

e Plot the graph of the parabola.

Create a strategy Plot the intercepts and turning point and draw a curve through the points.

Apply the idea y 4 3 2 1 −4 −3 −2 −1

−1

x 1

−2 −3 −4

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Example 7 For the quadratic y = 3x2 − 6x + 8: a Find the axis of symmetry.

Create a strategy Use the formula

Apply the idea From the equation y = 3x2 − 6x + 8, we have a = 3 and b = −6.

Substitute the values of a and b

Evaluate the numerator and denominator

Simplify the fraction

Reﬂect and check The turning point lies on the axis of symmetry, which means x = 1 is the x-value of the turning point.

b Find the turning point of the parabola. Give your answer in the form (a, b).

Create a strategy Substitute x = 1 into y = 3x2 − 6x + 8 to ﬁnd the y-coordinate of the turning point.

Apply the idea y = 3(1)2 − 6(1) + 8 =5

Substitute 1 to x Evaluate

So, the turning point of the parabola is (1, 5).

c What is the y-intercept of this quadratic curve?

Create a strategy Substitute x = 0 into y = 3x2 − 6x + 8 to ﬁnd the y-intercept.

Apply the idea y = 3(0)2 − 6(0) + 8 =8

Substitute 0 to x Evaluate

So, the y-intercept is (0, 8).

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d Plot the graph of the equation.

Create a strategy Plot the y-intercept and the turning point and draw a curve through the points.

Apply the idea y 11 10 9 8 (0, 8) 7 6 5 (1, 5) 4 3 2 1 −4 −3 −2 −1

−1

x 3

Idea summary A quadratic equation in standard form is written as y = ax2 + bx + c, where a ≠ 0. To graph the parabola, we can: 1. Find the axis of symmetry using

2. Find the turning point by substituting in the axis of symmetry for x and evaluating for y. 3. Find the y-intercept by setting x = 0 and evaluate to get y = c. If a > 0, the graph is concave up and the turning point is a minimum. If a < 0, the graph is concave down and the turning point is a maximum. If it is able to be factorised, we can use the null factor law to ﬁnd the x-intercepts.

4.02 Practice questions Understanding 1

162

Explorer 1–4

Adventurer 1–4

Describe the basic shape of a parabola.

Given that a general quadratic is y = ax2 + bx + c: i

What does the sign of a tell us?

What does changing the value of c do?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Trailblazer 1–4

Complete the following statement and ﬁll in the key features of a parabolic curve. The graph is concave ⬚.

y 10 8 6 4 2 –2

Ex 2

Ex 3

What feature is at the point (h, k) on the parabola deﬁned by y = a (x − h)2 + k?

What features are at the points ( p, 0) and (q, 0) on the parabola defined by y = a (x − p) (x − q)? 5ab, 6ac, 7a, 8a–c, 9, 10ad, 11ab, 12, 13ab, 14a, 15a–e, 16a–c

Fluency

Ex 1

5ac, 6ac, 7ab, 8ace, 9, 10b–d, 11bc, 12, 13bc, 14a, 15c–g, 16c–f

5ad, 6bd, 7cd, 8d–f, 9, 10d–f, 11cd, 12, 13cd, 14, 15h–l, 16d–h

For each equation: i

Describe the transformations from y = x2.

What is the turning point?

iii

Sketch the graph of the parabola.

y = (x − 3)2 − 4

y = −(x + 3)2 − 6

y = 3 (x + 1)2 + 4

y = −0.5 (x − 6)2 + 1

For each quadratic: i

Is the parabola concave up or down?

Is the parabola more or less steep than the parabola y = x2?

iii

Find the coordinates of the turning point.

Find the y-intercept.

Sketch the graph of this quadratic. Include the axis of symmetry.

y = 2 (x − 1)2 − 3

y = 3 (x + 1)2 + 2

y = −3 (x + 5) − 4

For each quadratic equation: i

Write in the form y = a (x − h)2 + k.

Find the coordinates of the turning point.

iii

Sketch the graph of the parabola.

y = x2 + 6x

y = −x2 + 4x − 3

y = −x2 − 8x − 18

y = x2 + x + 3

4.02 Quadratic graphs mathspace.co

163

Ex 4

Each of the graphs has an equation of the form, y = a(x − h)2 + k. Find the values of a, h and k for each graph and write the equation. a

2 −3

−2

−1

x 1

−2

−1

y 12

8 6

x 1

−2

−6

−5 −4 −3 −2 −1 −2

−8

−4

−5 −4 −3 −2 −1

−1

−3 −2 −1

−1

−3

−2

−4

−3

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−2

x 1

164

−2

−4

x 1

2 −3

−1

−4

x 1

The parabola y = x2 has been translated 2 units to the right and 6 units down. a

Ex 5

Find the equation of the parabola.

Sketch the graph of the parabola.

For each quadratic: i

Find the y-intercept.

Find the x-intercepts.

iii

Find the axis of symmetry.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

y = x (x + 4)

y = x (4 − x)

y = (x − 2) (x − 6)

y = (x − 3) (x + 1)

y = (5 − x) (x + 1)

y = (1 − x) (3 − x)

A parabola has x-intercepts at x = 1 and x = −7. State the equation of the parabola in the form y = a (x − p) (x − q) for each of these cases, and make a sketch of the curve. a

The turning point is a minimum turning point and the y-intercept is at (0, −7)

The turning point is a maximum turning point and the y-intercept is at (0, 14)

The turning point is a minimum turning point and the y-intercept is at (0, −21)

The turning point is a maximum turning point and the y-intercept is at (0, 3.5)

For the graph: y 14 12 10 8 6 4 2 −2 −1 −2

x 1

−4

Find the equation of the parabola in factorised form.

What is the turning point?

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These parabolas each have x-intercepts at x = ± . What is the equation of each parabola in the form y = a (x − p) (x − q) and y = a (x − h)2 + k? a

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

Ex 6

166

9 8 7 6 5 4 3 2 1

x 1

−4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

7 6 5 4 3 2 1

−1 −2 −3 −4 −5 −6 −7

7 6 5 4 3 2 1

−4 −3 −2 −1

7 6 5 4 3 2 1

−4 −3 −2 −1−1 −2 −3 −4 −5

For each quadratic: i

Factorise to ﬁnd the x-intercepts.

Express the equation in the form y = a (x − h)2 + k by completing the square.

iii

Find the coordinates of the turning point.

Sketch the graph of the parabola.

y = x2 − 12x + 32

y = −2x2 − 8x + 10

For each quadratic: i

Describe the concavity of the parabola.

Factorise, and ﬁnd the x-intercepts.

iii

Find the y-intercept.

Find the axis of symmetry.

Find the coordinates of the turning vi point.

Sketch the graph and identify these key features on it.

y = x2 − 4x + 3

y = x2 − 36

y = 4 − x2

y = x2 + 4x + 3

y = x + 6x + 5

y = x2 − 8x + 15

y = −x2 + 2x + 24

y = −x2 − 2x + 8

y = x − 2x − 8

y = 8 + 2x − x2

y = x2 + 2x − 3

y = 2x2 + 9x + 9

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Ex 7

For each quadratic: i

What is the axis of symmetry?

What are the coordinates of the turning point?

iii

What is the y-intercept?

Sketch a graph of the parabola.

y = x2 − 4x + 1

y = x2 + 6x + 7

y = −4x2 − 4x + 1

y = −(x2 + 8x + 11)

y = x + 6x − 2

y = −x2 + 10x − 23

y = −x2 + 8x − 9

y = x2 + 2x − 11

Reasoning 17

17, 18, 23

17, 18, 20, 23

19, 21, 22, 24

Three different ways of expressing a quadratic equation are: y = a (x − p) (x − q), y = a (x − h)2 + k, and y = ax2 + bx + c. How can you ﬁnd the y-intercept for each form? Explain which form you would use to identify the y-intercept if you had all three available.

Determine whether each statement is true or false. Justify your answer. a

The turning point of a quadratic graph is always at the origin (0, 0).

If the coeﬃcient of the x2 term in a quadratic equation is negative, the graph will open upwards.

All quadratic graphs have a line of symmetry.

The y-intercept of a quadratic graph is always equal to the constant term in the quadratic equation.

The solutions of a quadratic equation set equal to zero can be of a quadratic equation can be determined by ﬁnding the x-intercepts of the corresponding quadratic graph.

Given a quadratic of the form y = a (x − h)2 + k, what are the conditions on a, h, and k so that: a

The turning point is a minimum turning point and the parabola has two x-intercepts?

The turning point is a maximum turning point and the parabola has one x-intercept?

The turning point is a minimum turning point and the parabola has no x-intercepts?

The turning point of a parabola is located at (2, 5). a

Determine the equation of the parabola if the y-intercept is at (0, 9).

The parabola is translated 2 units to the right and 7 units down. Determine the equation of the transformed parabola.

Explain the change that would be made to the equation in part (a) if the graph is translated a units to the right and b units upwards.

For the equations: A = (x − 2) (x − 4) and B = 2 (x − 2) (x − 4) a

Which key features are the same for A and B?

Which key features are different for A and B?

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Marvin wants to understand how turning point form and standard form of a quadratic function are related. Consider the turning point form: y = a(x − h)2 + k a

Expand a(x − h)2 + k and write the equivalent equation to y = a(x − h)2 + k in standard form.

Determine the value of b in the equation from part (a).

Determine the value of c in the equation from part (a).

Explain how the equations are related.

Describe the relationship between the turning point of a quadratic graph and the coeﬃcients a and b of y = ax2 + bx + c.

Explain what possible differences there can be, if: a

Two different quadratic curves have the same intercepts.

Three different quadratic curves have the same turning point and share one common intercept.

Problem-solving 25–27 25

25–28

26–30

The height h, in metres, reached by a ball thrown in the air after t seconds is given by the equation: h = 10t − t2 a

Complete the table of values for h = 10t − t2: t

Graph the relationship h = 10t − t2.

Find the height of the ball after 5.5 seconds have passed.

Find the maximum height reached by the ball.

When an object is thrown into the air, its height above the ground is given by the equation: h = 163 + 38s − s2 where s is its horizontal distance from where it was thrown.

168

Graph the relationship h = 163 + 38s − s2.

Find s, how far horizontally the object is from where it was thrown at the point when it reaches its greatest height above the ground.

Find the maximum height reached by the object.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Over the summer, Susana and her friends build a bike ramp to launch themselves into the local lake. Susana decides that the shape of the ramp will be parabolic, and thinks that the parabola is given by the equation:

1 ( x + 2)2 + 2 4 x

where y is the height in metres above the ground, and x is −6 the horizontal distance in metres from the edge of the lake.

If the ramp starts 6 m back from the edge of the lake, how high is the start of the ramp?

At what height will the rider leave the ramp?

At what other distance x is the rider also at a height of 3 m?

Graph the shape of the ramp on a Cartesian plane.

The main cable on a suspension bridge is approximately parabolic in shape. A point (x, y) on the parabola represents the height, y m, above the roadway x m from the leftmost supporting pole. The lowest point of the main cable is 10 m above the roadway and occurs 150 m from the leftmost post. The main cable meets the supporting poles 90 m above the roadway. Determine an equation for the parabola formed by the main cable. y

(150, 10) x

Elise is the owner of a restaurant. Elise wants to install new wooden ﬂoors in several rooms. The rooms in the restaurant are square. The wood costs $6.25 per square metre. The cost of the ﬂooring in terms of its side length is shown by the quadratic function C = 6.25x2. a

Determine how much Elise should expect to spend on ﬂooring if the room has side lengths of 12 m.

Determine how much the price would change if the side lengths decreased by 3 m.

Graph the quadratic model from part (a).

Describe what changes and what stays the same about the graph of the quadratic model if the cost per square metre decreases.

A quadratic equation in factorised form is y = a (x − m) (x − 12). If the equation has a turning point at the point (8, 15), show how we can ﬁnd the values of a and m.

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4.03 Solve by factorisation After this lesson, you will be able to… • solve quadratic equations by factorisation. • identify x-intercepts on a parabola and use them to solve a quadratic equation. • find the equation of a graph by identifying the intercepts and using them to write an equation.

Solve by factorisation We have seen how to solve quadratic equations using the null factor law. We can use these steps to solve factorisable quadratic equations: 1. Gather together all terms to one side of the equation and make it equal to 0, in the form ax2 + bx + c = 0. 2. Factorise the quadratic using any appropriate techniques, such as: • Finding the highest common factor • Using patterns like a2 − b2 = (a − b) (a + b) and a2 + 2ab + b2 = (a + b)2 • Splitting the middle term and using grouping • For x2 + bx + c, ﬁnding two numbers whose sum is b and whose product is c • Using the cross product method 3. Use the null factor law to split the quadratic equation into two linear equations and make each linear equation equal to 0. 4. Solve the linear equations to get the solutions to the original quadratic equation. 5. Check the solutions by substituting back into the original equation.

Interactive exploration

Discover this concept in action online

170

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

mathspace.co

Example 1 Solve the quadratic equations: a 3y − 15y2 = 0, which has two solutions. Solve for both values of y.

Create a strategy Factorise 3y − 15y2 and use the null factor law to solve.

Apply the idea 3y − 15y2 = 0 Write the equation 3y(1 − 5y) = 0 Factorise the quadratic expression 3y = 0 Set the ﬁrst factor equal to 0 y = 0 Divide both sides by 3 1 − 5y = 0 Set the second factor equal to 0 5y = 1 Add 1 to both sides Divide both sides by 5 So the solutions are y = 0 and

Reﬂect and check We can substitute y = 0 and

to check that these are the solutions of 3y − 15y2 = 0.

For y = 0: LHS = 3(0) − 15(0)2 =0

Substitute 0 for y Evaluate

= RHS For

Substitute

for y

Evaluate the power 15 Simplify 25

Evaluate the subtraction

4.03 Solve by factorisation mathspace.co

171

b x2 − 22x + 121 = 0

Create a strategy Factorise x2 − 22x + 121 and use the null factor law to solve.

Apply the idea x2 − 22x + 121 = 0

Write the equation

(x − 11) (x − 11) = 0

Factorise the quadratic expression

x − 11 = 0 x = 11

Set one of the factors equal to 0 Add 11 to both sides

Reﬂect and check Notice that x2 − 22x + 121 is in the form a2 + 2ab + b2, where a = x and b = −11. x2 − 22x + 121 = 0 2

Write the equation

Use the pattern

Factorise

(x) + 2 (x) (−11) + (−11) = 0 (x − 11) = 0 x − 11 = 0 x = 11

Set one of the factors equal to 0 Add 11 to both sides

Example 2 Solve x2 = 17x + 60 for x.

Create a strategy Move all terms to one side, then factorise and use the null factor law to solve.

Apply the idea x2 = 17x + 60 2

Write the equation

x − 17x − 60 = 0

Move all the terms to one side

(x − 20) (x + 3) = 0

Factorise the quadratic expression

x − 20 = 0 x = 20 x + 3 = 0 x = −3

Set the ﬁrst factor equal to 0 Add 20 to both sides Set the second factor equal to 0 Subtract 3 from both sides

So the solutions are x = 20 and x = −3.

172

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Idea summary We can solve most quadratic equations using these steps: 1. Gather together all terms to one side of the equation and make it equal to 0, in the form ax2 + bx + c = 0. 2. Factorise the quadratic expression using any appropriate techniques. 3. Use the null factor law to split the quadratic equation into two linear equations and make each linear equation equal to 0. 4. Solve the linear equations to get the solutions to the original quadratic equation. 5. Check the solutions by substituting back into the original equation.

Equations from graphs As well as drawing graphs from quadratic equations, we can also find the equation from the graph of a parabola. We have previously done this using the turning point, but it is not always easy to identify the coordinates of the turning point from the graph. Sometimes we will know other features, such as the x-intercepts, but not the turning point. To find the equation of a quadratic, we can follow these steps: 1. Identify the x-values of the x-intercepts. These occur at the point where y = 0. 2. Substitute the x-values found in Step 1 for p and q in the equation y = a (x − p) (x − q). 3. Find the value of a by using another known point or by using second differences. The first difference is the difference between consecutive y-values. The second difference is the difference between consecutive first differences. The second difference of any quadratic relationship is a non-zero constant. We can identify the first and second differences from a table of values or from a graph. 11 10 9 8 7 6 5 4 3 2 1 −4 −3 −2 −1

For our basic parabola, y = x2, where a = 1, the first differences are:

1−0=1 4−1=3 9−4=5

The second differences are: 3−1=2 3

5−3=2 x

1 1

In general,

4.03 Solve by factorisation mathspace.co

173

Example 3 Write an equation for the quadratic relation shown by the graph.

y 4 3 2 1 x −4 −3 −2 −1

−1 −2

Create a strategy For this graph, we can easily read the coordinates of the x-intercepts, so we can write an equation using the factorised form of a quadratic given by: y = a (x − p) (x − q) where p and q are the x-values of the x-intercepts.

Apply the idea The coordinates of the x-intercepts are (−1, 0) and (2, 0), so p = −1 and q = 2. To ﬁnd the value of a, we can use one of the points that passes through the curve of the parabola, such as the y-intercept, (0, −1). Write the factorised form of a quadratic equation Substitute p = −1 and q = 2 Substitute x = 0 and y = 1

Evaluate the brackets

Divide both sides by −2

Simplify The quadratic equation of the graph is:

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reflect and check Instead of substituting a point to find the value of a, we can also look at the second differences. For this graph, the first differences are 1, 2, 3, …, so the second differences are all 1.

y 4

3 2 2

x −4 −3 −2 −1

−1

−2

Alternatively, we could have use the coordinates of points on the graph to create a table of values, then calculate the ﬁrst and second differences. x

−1

First Difference −1 − (−1) = 0 0 − (−1) = 1 2−0=2 5−2=3

Second Difference 1−0=1 2−1=1 3−2=1

Idea summary To ﬁnd the equation of a quadratic, we can use these steps: 1. Identify the x-intercepts from the graph or otherwise. 2. Substitute the values for p and q found in Step 1 into y = a (x − p) (x − q). 3. Find the value of a by using another known point or by using second differences.

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Did you know?

Quadratic equations are essential in rocket science! Engineers use them to calculate the perfect launch angle and speed, ensuring that rockets follow an optimal trajectory. Quadratics help model the parabolic path of a rocket’s ascent and descent, allowing scientists to predict where it will land and how high it will go. By analysing factors like launch force, gravity, and air resistance, these equations ensure a safe and precise flight path. Without quadratics, achieving precise landings and controlled flight would be almost impossible!

4.03 Practice questions Understanding 1

176

Adventurer 1, 2c–f, 4, 5cd

Trailblazer 2e–h, 4, 5ad

Which two options are solutions to the equation x (x − 7) = −10? A

Explorer 1, 2a–d, 3, 4, 5ab

x=5

x = −5

x=1

x=2

Find possible values of a and b: a

a + b = 5, ab = 6

a + b = −11, ab = 18

a + b = 8, ab = 16

a + b = −6, ab = 9

a + b = −1, ab = −2

a + b = −7, ab = 12

a + b = 0, ab = −9

a + b = 2, ab = −15

To factorise x2 + 11x + 24, we need to find two numbers: a

What should their product be?

What should their sum be?

What are the two numbers?

Factorise x2 + 11x + 24

One expression for the area of the rectangle below is m2 + 14m + 45. The rectangle is made up of four smaller rectangles. Use the diagram to express the area of the large rectangle in factorised form.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

For the given graphs, ﬁnd the solution(s) to the equation. a

(x + 2) (x − 3) = 0

(x − 1) (x + 4) = 0

2 x

−6

−4

−2

x −6

−4

−2

−4

−6

(x − 3) (x + 3) = 0

−6

−4

−2

x 2

−6

−2

−4

−2

Ex 1

−6

−8

−4

−6

6a–d, 7a–h, 8a–d, 9a–d, 10a–c

2 −2

−4

Fluency

(x + 1) (x − 2) = 0

6ef, 7d–h, 8c–f, 9d–g, 10a–f, 11ab

6e–h, 7g–l, 8e–h, 9g–l, 10d–l, 11cd

Solve: a

x (x − 9) = 0

2m (m − 8) = 0

c (5c − 12) = 0

( y − 6) ( y + 11) = 0

(k − 3) (k − 5) = 0

(3x − 9) (2x − 5) = 0

5 (x + 5) (x − 5) = 0

(7a − 2)2 = 0

Solve: a

x2 − 64 = 0

x2 − 9 = 0

x + 12x = 0

x2 − 8x = 0

x2 + 5x + 4 = 0

x2 + 7x + 10 = 0

x2 − 8x + 15 = 0

x2 − 2x − 24 = 0

x − x − 20 = 0

x2 + 3x − 10 = 0

x2 − 14x + 40 = 0

x2 − 14x + 49 = 0

4.03 Solve by factorisation mathspace.co

177

Solve: a

Ex 2

10x2 − 10x − 120 = 0

4x + 24x + 20 = 0

3x2 − 30x + 75 = 0

3x2 − 21x − 54 = 0

−2x2 + 30x − 112 = 0

−5x2 + 10x + 40 = 0

−3x2 − 3x + 396 = 0

Solve: a

m2 = 14m

m2 = m + 20

x2 = 81

−3m (m + 2) = 0

m − 27m = −182

3 (n2 + 8) = −18n

2y − 6y2 = 0

2y − 10y2 = 0

15 − 11b − 12b2 = 0

6k2 = −7 + 13k

x2 = 24x − 144

Solve: a

x(x + 2) = 48

x2 + 8x + 16 = 36

4x2 = 400

3x2 − 7x − 20 = 0

5x = −320 + 80x

−5x2 = −53x + 72

−10x + x2 = 2 − 7x − x2

−8x + x2 = −6 − x − x2

( y − 3)2 = 2y + 2

( y + 1)2 = 4y + 4

(5x2 − 13x + 6) (2x2 − 13x + 20) = 0

k Ex 3

2x2 − 32x = 0

6x − 27x + 28 = −3x + 3x + 3

For each graph, complete the following: i

What are the solutions of the related quadratic equation?

Determine the y-intercept(s).

iii

Write the equation in factored form.

18 16 14 12 10 8 6 4 2

12 10 8 6 4 2 −2 −1 −2

x 1

−4

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−2

−2 −4

x 2

2 −2

x 2

−4

−6

−12

−4

−14

−6

−16

−8

Reasoning

−8 −6 −4 −2 −2

−8 −10

y 8

12, 15

13–16

For x2 + kx + 24, where k is an integer: a

Show that k = 10 is a possible value of k so that the expression can be factorised.

Find all of the integer values of k that allow the expression to be factorised over the integers.

For each integer value of k in part (b), state the solutions to x2 + kx + 24 = 0.

For x2 + kx + 28: a

Find the largest integer value of k that makes the expression factorisable. Explain.

Find the integer solutions to x2 + kx + 28 = 0 so that the difference between the two solutions is as large as possible.

For x2 + 4x + c: a

Find all positive integer values of c that allow x2 + 4x + c to be factorised as a product of rational linear factors. Explain your process.

Find all possible integer solutions to x2 + 4x + c = 0 for positive integer values of c.

Explain why it is not possible to list all negative integer values of c that allow x2 + 4x + c to be factorised into the product of rational linear factors.

Explain why there are always solutions to x2 + 4x + c = 0 for c < 0.

In general, a quadratic can have 0, 1, or 2 x-intercepts. If its equation can be factorised using rational numbers, how many x-intercepts will it have? Explain.

Explain why we can say that the solutions to (ax + b) (cx + d) = 0 are and , but we cannot say anything about the solutions for (ax + b) (cx + d) = e, where e ≠ 0 without further work.

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Problem-solving 17a–c, 18, 19, 21 17

17d–h, 18, 19, 21

17e–j, 18–22

Solve: a

On the graph of y = x2 − 4, there are two points where y = 12. Find the x-coordinates of these two points.

Find the values of m and n that make both sides of the following equation equal: y2 + my + 60 = ( y + 10) ( y + n)

Software engineers are designing a self-serve checkout system for a supermarket. They notice that the traﬃc through the store during the day is described by the equation: C = −t (t − 12) where C is the number of customers and t is the number of hours after the store opens. To meet the peak demand, the engineers allow for an extra checkout machine to automatically turn on when the number of customers ﬁrst reaches 27 people, and to automatically turn off when it next falls below 27 people. a

Find the times t when the number of customers is equal to 27 people.

After how many hours will the extra checkout machine turn on after opening?

How many hours will the extra machine be on for?

The height of a projectile above the ground is given by y = 10 + 5t − 5t2. Height is measured in metres and time in seconds. Find the time, t, at which the projectile hits the ground.

Write the equation of a quadratic in factorised form that has the following key features and draw a graph of the curve:

180

Two x-intercepts that have the same values with opposite signs

One x-intercept that is a fraction and the other is a prime number

One unique x-intercept that is less than 1 but greater than 0

x-intercepts of x = 7, x = −2 and has a y-intercept at (0, 14)

x-intercepts of x = −1, x = −10 and has a y-intercept at (0, 30)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

4.04 Solve by completing the square After this lesson, you will be able to… • solve quadratic equations by completing the square.

Algebraic method The process for completing the square: 1. Factor out the coeﬃcient of the x2 term, unless it is 1. 2. Halve the coeﬃcient of x and square it, add and subtract this value. 3. Factorise the terms that now form a perfect square. 4. Combine the constant and term left over from completing the square.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Using the completing the square method, rewrite into the form (x + a)2 + b: x2 − 14x + 59

Apply the idea The coefficient of the x term is −14, and half of this is −7. x2 − 14x + 59 = x2 − 14x + 72 − 72 + 59 = (x − 7)2 − 72 + 59 2

= (x − 7) + 10

Add 72 − 72 Factorise the perfect square part Evaluate the constant

4.04 Solve by completing the square mathspace.co

181

Example 2 Solve x2 − 7x + 8 = 0.

Create a strategy Complete the square and then solve.

Apply the idea Complete the square: The coefficient of the x term is −7, and half of this is

. Add

Factorise the perfect square part

Evaluate the exponent

Evaluate the constant

Solve: Collect like terms

Take the square root of both sides

Simplify the square root

Add

to both sides

Evaluate So the solutions are

Simplify using a common denominator and

Example 3 Solve 4x2 + 11x + 7 = 0.

Create a strategy Divide the equation by the coeﬃcient of x2, then complete the square and then solve.

182

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Apply the idea Divide the equation by the coefficient of x2: Write the equation

Divide both sides by 4

Evaluate

Complete the square: The coefficient of the x term is

, and so

Add

Factorise the perfect square part

Evaluate the exponent

Evaluate the constant

Solve: Collect like terms

Take the square roots of both sides

Evaluate

Subtract

Evaluate

Simplify using a common denominator

from both sides

We can see that we can still simplify each x-value.

Write the positive x-value

Evaluate the subtraction

Evaluate Write the negative x-value

Evaluate the subtraction

Simplify the fraction

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Idea summary The following are the steps of completing the square algebraically. 1. Factor out the coeﬃcient of the x2 term, unless it is 1. 2. Halve the coeﬃcient of x and square it, add and subtract this value. 3. Factorise the terms that now form a perfect square. 4. Combine the constant and term left over from completing the square. We can then solve using inverse operations.

4.04 Practice questions Understanding 1

184

Explorer

Adventurer

1–4

Trailblazer 1–4

When solving x2 + 24x = 10, four different students showed their ﬁrst step of working. Identify if there is an error or not. a

Asad: x2 + 24x + 144 = 10

Blanche: x2 + 24x + 144 − 144 = 10

Cordelia: x2 + 24x + 144 = 10 + 144

Dacia: x2 + 24x − 144 = 10

What should be added to x2 + 10x in order to make it a perfect square trinomial?

What should be added to x2 − 8x in order to make it a perfect square trinomial?

Which value would complete the square for x2 + 6x? x

For x2 − 3x − 6 = 0, what should be added and subtracted to the left side of the equation in order to complete the square?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5a–d, 6a–c, 7ab, 8a, 9a–d

Fluency Ex 1

Ex 2

Ex 3

5e–h, 6c–f, 7cd, 8b, 9e–h

Rewrite in the form (x − h)2 + k: a

x2 − 2x + 46

x2 + 2x + 2

x2 − 14x + 56

x2 − 6x + 5

x2 + 3x + 6

x2 − 7x + 14

x2 + 5x − 1

x2 − 2x + 4

Solve: a

(x + 6)2 = 4

(4x − 4)2 = 4

2 (x − 3)2 = 8

(x + 2)2 = 20

(x + 5)2 − 2 = 15

Solve: a

x2 + 18x + 32 = 0

x2 − 6x + 8 = 0

x2 − 2x − 32 = 0

6x2 + 48x + 24 = 0

x2 + 11x + 5 = 0

x − 9x + 8 = 0

Solve: a

5c–f, 6b–e, 7ac, 8b, 9c–f

2x2 − 12x − 32 = 0

Solve, leaving your answers in exact form: a

x2 + 24x + 5 = 0

x2 − 7x + 8 = 0

2x2 − 5x − 4 = 0

2x2 − 8x − 28 = 0

3x2 − 5x + 1 = 0

−4x2 − 20x + 8 = 0

Reasoning

10, 13

10–13

Explain how to ﬁnd the number to add to x2 + bx to make it a perfect square trinomial.

Explain the relationship between solving equations by completing the square and solving equations in the form x2 = k where k is a positive number.

Explain why we might use the method of completing the square to ﬁnd the exact solutions to some quadratic equations rather than by sketching a graph or factoring.

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185

Tyler attempted to solve the quadratic equation x2 − 4x + 3 = 0 using the completing the square method. Tyler’s solution is shown below. Identify the errors in the solution and explain what Tyler should have done instead. Tyler’s solution:

Given

Complete the square

Simplify fractions and exponents

Factorise the perfect square part

Evaluate and simplify

Collect like terms

Take the square root of both sides

Add 4 to both sides

Evaluate 4 + 1 and 4 − 1

Problem-solving 14, 15 14

14, 15

15–17

The revenue y (in millions of dollars) of a company x years after it ﬁrst started is modelled by: y = 12.5x2 − 64x + 135 By completing the square, use this equation to predict the number of years x after which the revenue of the company will reach $1167 million.

A model submarine performs a parabolic dive so that its depth, y, below the surface after covering a horizontal distance x was given by: y = x2 − 20x All distances are in metres. By completing the square:

186

Find the lowest depth of the submarine beneath the surface.

Find the horizontal distance(s) travelled when the submarine is 10 m beneath the surface.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The headlights of a car are shaped like a parabola, to direct all light from the bulb in the same direction. For a particular model of car, the bottom of the headlight can be modelled by the equation:

where x and y are measured in centimetres, and the y-intercept represents the top left point of the headlight. Determine the width and height of the headlight by ﬁrst rewriting the equation in turning point form. Explain your reasoning. y (0, 54)

Consider the quadratic equation ax2 − c = 0. Given the following conditions, state whether the equation has zero, one, or two solutions. a

a and c are positive

a and c are negative

a and c have opposite signs

c is equal to 0

Investigation: Use bisection to solve quadratics

Investigate online

mathspace.co

Investigation: Derive the quadratic formula

Investigate online

mathspace.co

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4 Chapter review 1

Are the graphs of these quadratic equations concave up or concave down? a

y = −x (x + 2) + 5

y = 9x2 + 4x

55 = x2 − 45

36 = (n − 2)2

Solve: a

y = x2 + 6x − 9

x2 = 81

A company’s revenue, R (in dollars), after t months is described by the equation: R = − (t − 12)2 + 36

188

x (2x − 16) = 0

Does the revenue reach a maximum or minimum point?

When will the revenue be zero?

For each equation: i

Describe the transformations from y = x2.

What are the coordinates of the turning point?

iii

What are the coordinates of the y-intercept?

Sketch the graph of the parabola.

y = 2 (x − 4)2

For the quadratic equation y = (x − 1) (x + 3): a

Find the y-intercept.

Find the x-intercepts.

Write the equation of the axis of symmetry.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

For the quadratic equation y = x2 + 4x − 1: a

Write in the form y = a (x−h)2 + k.

Find the coordinates of the turning point.

Sketch the graph of the parabola.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

y = − (x + 1)2 + 7

Solve by factorisation: a c

x2 − 12x = 0 2

−x + 24x − 144 = 0

3x2 − 34x + 75 = 0

What are the solutions of the related quadratic equation?

Write the equation in factored form.

18 16 14 12 10 8 6 4 2

4 2 −6 −4 −2 −2

x 2 4 6 8 10 12

−4 −6 x 1

−8

−10 −12

Solve for y = 0 by completing the square: a

x2 + 9x − 10 = 0

For each graph:

−4 −3 −2 −1 −2 −4

y = x2 − 14x + 40

y = −x2 + 6x − 9

y = 2x2 + 9x + 10

A rocket is launched into the sky and its altitude in metres above the ground after travelling a horizontal distance x metres is given by y = −x2 + 24x By completing the square: a

Find the maximum altitude of the rocket above the ground.

Find the horizontal distance(s) travelled when the rocket is 85 m above the ground.

Chapter 4 review mathspace.co

189

Big ideas Solutions to simultaneous equations and inequalities can be graphically and algebraically determined through various methods. There is often a more efficient method that can be used based on the structure of the equation. Curriculum Outcomes VC2M10A08 solve linear inequalities and graph their solutions on a number line VC2M10A09 solve simultaneous linear equations, using algebraic and graphical techniques including using digital tools VC2M10A16 solve equations graphically or using systematic numerical guess-check-andrefine with digital tools, with consideration of whether all solutions have been found VC2M10AA09 solve linear and non-linear simultaneous equations using graphing or systematic guess-check-and-refine with digital tools VC2M10AA10 experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns

5 Simultaneous equations Chapter outline 5.01 5.02 5.03 5.04 5.05 5.06

Solve graphically Simultaneous inequalities Solve using substitution Solve using elimination Applications of simultaneous equations Investigation: Intersections of curves Optional: Intersections of curves Chapter 5 review

192 200 210 218 225

238

Businesses use optimisation to improve efficiencies. In 1987, a large airline removed a single olive from each of its inflight salads, reducing costs by $ 55,000 a year!

5.01

Solve graphically After this lesson, you will be able to… • plot pairs of linear equations on a Cartesian plane. • identify the point of intersection of two graphs as the solution to the simultaneous equations. • recognise the significance of parallel lines and their impact on solutions. • apply graphical methods to real-life equations to find their solutions. • draw conclusions about the presence or absence of simultaneous solutions based on the properties of the graphed lines.

Solve graphically To ﬁnd the solution to a pair of linear equations, which we call simultaneous equations, we can graph both lines on a single Cartesian plane and find any points of intersection.

Interactive exploration Discover this concept in action online

mathspace.co

If two lines are parallel and are not the same line, they will never intersect.

Example 1 Consider the following linear equations: y = 5x − 7 y = −x + 5 a Sketch the two lines representing these equations on a Cartesian plane.

Create a strategy Use the gradient and y-intercept from the equations to graph the lines.

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Apply the idea 8

For the line y = 5x − 7, m = 5 and c = −7. Plot the y-intercept of (0, −7).

6 4 2

−5 −4 −3 −2 −1 −2

Use the gradient of 5 to find another point, by moving 1 unit to the right and 5 units upwards, which is the point (1, −2). Draw a line through these two points.

−4 −6 −8

y = 5x − 7

For the line y = −x + 5, m = −1 and c = 5. Plot the y-intercept of (0, 5).

8 6

y = −x +5

4 2

−5 −4 −3 −2 −1 −2

1 2 3 4 5

−4

Use the gradient of −1 to find another point, by moving 1 unit to the right and 1 unit downwards, which is the point (1, 4). Draw a line through these two points.

y = 5x −7

−6 −8

b What is the point of intersection for the two graphs?

Create a strategy Use the graphs to identify the point of intersection.

Apply the idea y

The point where the two lines meet is at (2, 3). This is the point of intersection.

8 y = -x +5

The intersection point (2, 3) satisfies both equations, y = 5x – 7 and y = –x + 5.

4 2 -5 -4 -3 -2 -1 -2 -4

x 1 2 3 4 5

When x = 2: • The first equation is y = 5 x 2 – 7 = 3. • The second equation is y = –2 + 5 = 3.

y = 5x -7

-6 -8

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Idea summary We can solve any pair of simultaneous equations by plotting their graphs and ﬁnding the point of intersection. If two different lines have the same gradient, then they will be parallel and will have no point of intersection. So their equations will have no simultaneous solution.

5.01 Practice questions Explorer

Understanding

Adventurer

1–3

Trailblazer

1–3

What does the point of intersection of two straight lines on a graph represent?

If a pair of linear equations has no solutions, what would this look like graphically?

State the number of solutions for the simultaneous equations: a

6 5 4 3 2 1 −6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

5 4 3 2 1 −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

194

x 1 2 3 4 5 6

x 1 2 3 4 5

6 5 4 3 2 1 −6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

x 1 2 3 4 5 6

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

6 5 4 3 2 1

x 1 2 3 4 5 6

−6 −5 −4 −3 −2−1 −1 −2 −3 −4 −5 −6

3 2 x 1 2 3 4 5 6

1 −4 −3 −2 −1

−1

−2 −3 −4

x 1

4b–d, 5ab, 6a–d, 8, 9, 11, 12ab, 13ab

Fluency 4

4ace, 5bc, 6e–h, 7, 9, 11, 12ab, 13ab

4aef, 5bc, 6i–l, 7, 10, 11, 12cd, 13cd

Identify the solution to the simultaneous equations: a

y 10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

10 8 6 4 2

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

5 4 3 2 1

4 3 2 1 −3 −2 −1−1

x 2 4 6 8 10

x 1 2 3 4 5 6 7

−2 −3 −4 −5 −6 −7 −8 −9

2 4 6 8 10

−1−1 −2 −3 −4 −5 −6 −7

10 8 6 4 2

x 2 4 6 8 10

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

f 4

3 2 x 1 2 3 4 5 6 7 8 9

1 −9−8−7−6−5−4−3−2−1 −1

x 1 2 3 4

−2 −3 −4

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Use the graph to solve the simultaneous equations: y

5 4 3 2 1 −2

−1

x − 2y = 4

y=x−2 1

−1 −2 −3 −4 −5

x 8

y = −8 + 2x

y=x−2

y = −8 + 2x

y = −8 + 2x x − 2y = 4 x − 2y = 4 Ex 1

For each pair of linear equations: i

Sketch the lines on the same Cartesian plane.

Find the solution.

y = 4x − 4

y=x+9

y = 8 − 2x y = −x – 9 e

y = 3x + 6

y = −x + 2 i

y = 1.5x

y = −x − 3

y = −1 j

y = 2.5x − 4

y=x+3 y = 4x − 3

2x − 4y = −8

y = 3x + 2

y = 4 − 3x y = 7 − 2x k

y = 4 − 0.5x y = 12 − 1.5x 7

y = −x + 3 −4x + 2y = −8

A band plans to record a demo at a local studio. The cost of renting studio A is $250 plus $50 per hour. The cost of renting studio B is $50 plus $100 per hour. The cost, y, in dollars of renting the studios for x hours can be modelled by the simultaneous equations: • Studio A: y = 50x + 250 • Studio B: y = 100x + 50

Sketch the two lines on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (b) mean in this context?

Michael plans to start taking an aerobics class. Non-members pay $4 per class. Members pay a $10 one-time fee, but only have to pay $2 per class. The cost, y, of taking x classes can be modelled by the simultaneous equations: • Non-members: y = 4x • Members: y = 2x + 10

196

Sketch the two lines on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (b) mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

A company’s cost is represented by C = 0.4x + 2015, and its revenue is represented by R = 3x, where x represents the number of items produced.

y 2790

What coordinates satisfy both equations?

2325

What do the coordinates from part (a) mean in this context?

1860

1395 930 465

x 155 310 465 620 775 930 1085 1240

The cost of manufacturing toys, C, is related to the number of toys produced, n, by the formula C = 400 + 2n. The revenue, R, made from selling n toys is given by R = 4n. a

Sketch the graphs of cost and revenue on the same Cartesian plane.

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

Laura is going on maternity leave. Her living expenses are represented by y = 1200x. Her bank account balance is represented by y = 10 000 + 772.60x. For both equations, x is time in weeks. a

Sketch both equations on the same Cartesian plane using technology.

Estimate the solution to one decimal place.

What does the point of intersection of the two lines mean in this context?

A pair of linear equations has no solutions. One of the equations is y = −3x − 2. Determine if the following can be the other equation: a

y = −3x − 3

y = −3x + 2

A pair of simultaneous linear equations has inﬁnite solutions. One of the equations is y = 4x − 5. Determine if the following can be the other equation: a

y = 8x − 10

Reasoning

b 14, 15, 17, 18

y = 4x + 5

c 15–18

3y = 12x − 15

2y = 8x − 10

16–19

Is it possible to have a pair of linear equations with exactly two solutions? Explain why or why not.

In a pair of linear equations, one equation is a scalar multiple of the other. For example, 2x + 4y = 6 and x + 2y = 3. What can you say about the solutions of the simultaneous equations?

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197

The graph of y = 3x + 2 is shown.

10 8 6 4 2

If a second line intersects this line at the point (0, 2), can we determine the equation of the second line? Explain your answer.

−10 −8 −6 −4 −2 −2 −4 −6 −8 −10

The graph of y = 5x + 3 is shown.

2 4 6 8 10

A second line, y = mx + c, intersects this line at the one point (0, 3).

What is the value of c? Explain your answer.

What value of m is not possible? Explain your answer.

1 −2

−1

x 1

−2

Consider the simultaneous equations: y = −2x − 5 y = −2x + 4 Without graphing, is there a solution? Explain your answer.

A company needs to ﬁnd the best way to distribute goods from multiple suppliers to multiple customers. The distribution amounts must satisfy the simultaneous linear equations: y = 3x + 15 y = 9x − 10 Explain the implications of having non-integer solutions in this context and discuss how the company might handle such solutions.

Problem-solving 20, 21 20

198

20, 21, 23

21–24

A rectangle is to be 3 m longer than it is wide, with a total perimeter of 18 m. a

Let y represent the length of the rectangle and x represent the width. Create two equations that represent this information.

Sketch the two lines on the same Cartesian plane.

Find the length and width of the rectangle.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Rochelle and Mohamad are siblings. Rochelle’s age is 11 more than 4 times the age of Mohamad. The sum of their ages is 21. a

Write a pair of linear equations to represent the situation.

Graph the simultaneous equations.

How old are Rochelle and Mohamad?

Kang and Elvia are working on an assignment together. They have broken down the work into 9 parts of the same size. Kang works at 2 times the speed of Elvia but has 6 pieces of work to do for another subject. a

Write a pair of linear equations to represent the situation.

Graph the simultaneous equations.

Solve the simultaneous equations and interpret the result in terms of this context.

Write a scenario to represent the simultaneous equations and its solution. Explain what the solution means in terms of the scenario. 20 18 16 14 12 10 8 6 4 2

x 2 4 6 8 10 12 14 16 18 20 22 24

Two equations, y1 and y2 represent the growth of two different house plants over time. Use the graph of y1 and y2 to support the claim that the two plants will never reach the same height on the same day. 25 20 15 10 5 −40 −30 −20 −10 −5 −10 y1 −15 −20 y2 −25

x 10 20 30

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5.02

Simultaneous inequalities

After this lesson, you will be able to… • define simultaneous inequalities. • find solutions to simultaneous inequalities and represent them graphically. • interpret and solve problems involving simultaneous inequalities in given contexts. • apply the concepts of simultaneous inequalities to real-world scenarios.

Simultaneous inequalities Simultaneous inequalities is a set of inequalities which have the same variables. The solution to simultaneous inequalities is the set containing any ordered pair that makes all the inequalities true. A solution can also be represented graphically as the region of the plane that satisﬁes all inequalities. 4

1 −4 −3 −2 −1

−11

x 1

−2 2

−4 −3 −2 −1

−1

x 1

−2

−3 3

−3

−4 4

−4

The solutions to simultaneous inequalities in a given context are viable if the solutions make sense in the context, and are non-viable if they do not make sense.

200

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 1 y

Select the inequalities that describe the shaded region. 12

A x + y ≤ 9, x ≥ 3 B x + y > 9, x < 3

C x + y > 9, x > 3 D x + y ≤ 9, x ≤ 3

6 3 x −3

−3

Create a strategy If the lines are dotted and the shaded region is above or on the right of each line, then the inequality symbol is >.

Apply the idea The answer is option C.

Example 2 Consider the lines y = 4x − 2 and y = 3x − 3. a Find the x-coordinate of the point at which the two lines intersect.

Create a strategy

Apply the idea 4x − 2 = 3x – 3 Form an equation by substitution of y

Form an equation by substitution of y and solve.

x − 2 = −3

Subtract −3x from both sides

x = −1

Add 2 to both sides

b Find the y-coordinate of the point of intersection.

Create a strategy

Apply the idea

Substitute x = −1 into one of the equations.

We can use y = 4x − 2. y = 4x – 2 = 4 × (−1) – 2

Substitute x = −1

= −6

Evaluate

5.02 Simultaneous inequalities mathspace.co

201

c Graph the region that satisfies both y < 4x − 2 and y < 3x − 3.

Create a strategy • Plot the intersection point and the y-intercepts of each inequality. • The inequality symbols are both < so the lines must be dotted. • Since it is an “and” inequality, shade the region which satisfies both inequalities.

Apply the idea We found from parts (a) and (b) that the intersection point is at (−1, −6).

y 1 x −2

−1

1 −1

y < 4x − 2 has a y-intercept at (0, −2), while y < 3x − 3 has a y-intercept at (0, −3). The shaded region must be below both lines.

−2 (0, −2) −3 (0, −3) −4 4 −5 − 5 −6 − 6 (−1, −6) −7 − 7

Reﬂect and check We can perform a region test. This is where we take a point such as (0,0) and test if it satisfies the inequality or not by plotting it on the Cartesian plane and checking to see if it falls in the shaded region. We can see that the origin, point (0, 0), does not fall in the region, so (0,0) does not satisfy the inequality.

Example 3 Applicants for a particular university are asked to sit a quantitative reasoning test and verbal reasoning test. Successful applicants must obtain a minimum score of 14 on a quantitative reasoning test and a minimum combined score of 29 for both tests. a Write simultaneous inequalities for this scenario, where x represents the quantitative reasoning test score and y represents the verbal reasoning test score.

Create a strategy

Apply the idea

The quantitative reasoning will show an inequality with 14 as the lowest possible solution. The total of the two scores must sum to 29 or more. 202

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b Sketch the graph of the simultaneous inequalities.

Create a strategy

Apply the idea

Graph each inequality on the same Cartesian plane and shade the region that satisfies both.

y 30 25 20 15 10 5 x 5

c Does the solution

make sense in terms of the context? Explain your answer.

Apply the idea No. This point satisfies both inequalities, but a test score is likely a positive integer or a simple fraction such as

or .

Idea summary A set of simultaneous inequalities is a set of two or more inequalities with the same variables. The solutions of simultaneous inequalities are the values of the variables that make all the inequalities true. It is shown as a shaded region on the Cartesian plane.

5.02 Practice questions Understanding 1

Explorer 1–3

Adventurer 1–3

The graph shows the region for the simultaneous inequalities: y > − 3x + 6 y>4 Determine whether each point is located in the region. a

(−3, 2)

(2, 5)

(−1, −6)

(3, 2)

(4, 4)

Trailblazer 1–4 7 6 5 4 3 2 1 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

x 1 2 3 4 5 6 7

5.02 Simultaneous inequalities mathspace.co

203

The graph shows the region for the simultaneous inequalities: x+y≤4

7 6 5 4 3 2 1

x≥1 y≥0 What are the coordinates of the three vertices of the region?

Which ordered pair satisﬁes the simultaneous inequalities?

−3 −2 −1−1 −2 −3 −4 −5 −6 −7

1 2 3 4 5 6 7

2x − y > −6 2x + 2y < 0 A 4

(2, 6)

(1, −3)

(−6, 1)

The region can only include individual points and cannot include lines.

The solution set to a linear inequality is a region of the graph and can include an infinite number of points.

All the points on the boundary line of a linear inequality are included in the solution if the inequality is “less than or equal to” or “greater than or equal to”. 5ab, 6, 7ab, 8, 10, 11, 13, 14 5, 7ac, 8, 10, 11, 13, 15

5, 7ad, 9, 10, 12, 15

Write the simultaneous inequalities that describe the shaded region: a

5 4 3 2 1 −5 5 −4 4 −3 3 −2 2− −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5

3 2 1

x −4 − 4 − −3 3 −2 2 −1 −1

1 2 3 4 5

−11 −

−2 − 2 −3 −3 −4 − 4

3 2 1 −7 −6 −5 5− −4 4− −3 3− −2 2− −11 −11 −2 − 2 −3 3 −4 4 −5 5 −6 −7

204

Is the statement true or false?

Fluency Ex 1

(−3, 6)

4 3

1 2 3

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

1 −4 −3 −2 −1

−1

−2 −3 −4

Ex 2

Ex 3

For 2x − y < 4 and 6x + 3y > 0: a

Sketch the boundary lines, 2x − y = 4 and 6x + 3y = 0, on the same set of axes.

Does the point (2, 2) satisfy the simultaneous inequalities?

Graph the solution set to the simultaneous inequalities 2x − y < 4 and 6x + 3y > 0.

For each pair of simultaneous inequalities: i

Find the x-coordinate of the point at which the two lines intersect.

Find the y-coordinate of the point of intersection.

iii

Graph the region.

x ≤ 5 and y < 3

y ≤ 3x − 4 and y > −4x + 1

3x + y > 5 and 3x + y < 7

y < 2x − 4 and y < −3x + 6

For y > x − 5, y > −x − 5, and y < 0: a

Graph the lines y = x − 5, y = −x − 5, and y = 0 on the same set of axes.

Does the point (1, −2) satisfy the simultaneous inequalities?

Shade the region described by the inequalities y > x − 5, y > −x − 5, and y < 0.

Calculate the area of the region.

For x > −2, y > 1 and y ≤ −0.75x + 2.5: a

Graph the region which satisfies the inequalities.

Calculate the area of the region.

Applicants for a particular university are asked to sit a numeracy test and verbal reasoning test. Successful applicants must obtain a minimum score of 17 on the numeracy test and a minimum combined score of 37 for both tests. Let x and y represent an applicant’s score on the numeracy and verbal reasoning test respectively. a

It’s not possible to get a negative score therefore y ≥ 0 is one inequality. Write two more inequalities from the information in terms of x and y.

Graph the solution to the simultaneous inequalities.

Determine whether each point represents a score that would make the applicant successful. i

(14, 16)

(33, 18)

iii

(20, 13)

(17, 22)

If an applicant obtains a score of 24 in the numeracy test, ﬁnd the minimum integer score they need to obtain in the verbal reasoning test to be successful.

An airline is checking passengers into two flights, A and B, simultaneously. Due to passenger numbers, there must be at least 10 staff at check-in for flight A and at least 7 staff at check-in for flight B. Since there must be staff on hand for other services, the airline can only allocate at most 23 staff for check-in of both flights. Let x and y represent the number of staff attending check-in of flights A and B respectively. a

Write three inequalities representing the possible values of x and y.

Graph the solution to the simultaneous inequalities.

If 14 staff are allocated to checking in passengers of ﬂight A, what is the maximum number of staff that can be allocated to checking in passengers of ﬂight B?

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An old Air Force plane used for skydiving can ﬁt no more than 32 divers at a time, and has a maximum carrying capacity of 3900 kilograms. When carrying a parachute and equipment an average female’s weight is 100 kilograms and an average male’s weight is 150 kilograms. Let x represent the number of women. Let y represent the number of men. a

Write four inequalities in x and y that represent the number of men and women who can be on the plane at one time.

Graph the inequalities on the same set of axes.

Which combinations of men and women can be on the plane at the same time? A

25 women and 11 men

19 women and 12 men

25 women and 9 men

7 women and 10 men

A bicycle manufacturer employs a mechanic and a painter to construct two types of bikes: racing bikes and mountain bikes. The time each worker spends on each bike is shown (assume both workers can work on the same bike at the same time). Let x represent the number of racing bikes built and y represent the number of mountain bikes built. Each worker can work at most 40 hours in a week.

Racing Bike Mountain Bike

Mechanic 5 hours 8 hours

Painter 10 hours 4 hours

Write an inequality relating x and y to the total time spent working by the mechanic.

Write an inequality relating x and y to the total time spent working by the painter.

Lamborgotti Motors produces two types of car: a sedan and a coupe. Each type of car requires assembly in two different factories. The time that each car is required to be in each factory is given in the table. Factory A Factory B Each factory can operate for 18 hours a day. Sedan 3 hours 4.5 hours Let x represent the number of sedans Coupe 2 hours 1.5 hours produced each day, and y represent the number of coupes produced each day.

Write an inequality for Factory A.

Write inequalities for x and y given the number of cars cannot be a negative number.

Graph the region deﬁned by the inequalities for Factory A and Factory B

Write an inequality for Factory B.

Mahindra produces two types of tractors: an All Purpose and a Harvester. Each type of tractor requires production in two different factories. The time that each tractor is required to be in each factory is given in the table. Each factory can operate for 18 hours a day. Let x represent the number of All Purpose tractors produced each day, and y represent the number of Harvesters produced each day.

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All Purpose Harvesters

Factory A 4 hours 2 hours

Factory B 6 hours 1.5 hours

Write an inequality for Factory A.

Write inequalities for x and y given the number of tractors cannot be a negative number.

Graph the region deﬁned by the inequalities for Factory A and Factory B.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Write an inequality for Factory B.

Reasoning 16

16, 18ab, 20

16, 18ac, 20

17, 18cd, 19

Without graphing, determine whether the simultaneous inequalities have no solutions or infinitely many solutions. Explain your answer. a

y > 4 and y < 4

x ≤ −2 and x ≥ −2

9x − y < 7 and 9x − y > 7

2x − y ≤ 7 and 2x − y ≥ 7

Tracy has experimented with growing two different types of banana trees on her land in different proportions. Let x and y represent the decimal proportion of area Type A and Type B banana trees respectively. a

She has found that bananas grow best if Type A takes up at most 35% of the plantation area. Write two inequalities representing the possible values of x.

She has found that bananas grow best if Type B takes up at least 20% of the plantation area. Write one inequality representing the possible values of y.

The whole area must be planted with a combination of Type A and Type B. Write one inequality representing the possible values of x and y.

Graph the solution to the simultaneous inequalties.

Do these points correspond to appropriate areas for Type A and Type B banana trees? i

(0.28, 0.68)

(0.58, 0.68)

iii

(0.13, 0.57)

(0.09, 0.11)

Is it possible for Type B trees to grow optimally if they take up 100% of the plantation area? Explain your answer.

A company produces cans of cat food and dog food. Each product passes through two machines in their production, and the amount of time each machine takes to process one pallet of pet food is shown in the table. Each machine can operate for 12 hours in a day.

Let x represent the number of pallets of cat food processed each day, and y represent the number of pallets of dog food processed each day.

Cat Food Dog Food

Write an inequality for Machine A.

Can the values of x and y be negative? Explain your answer. Write any additional inequalities if necessary.

Graph the region deﬁned by the inequalities for Machine A and Machine B on a Cartesian plane.

Machine A 2 hours 4 hours

Machine B 3 hours 1.5 hours

Write an inequality for Machine B.

Is it possible for a coordinate point to solve one part of the linear inequality pair but not the other? Explain with a diagram what this would look like.

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A light bulb manufacturer produces two types of light bulbs: incandescent and ﬂuorescent. Let x represent the number of incandescent bulbs produced in a day, and y represent the number of ﬂuorescent bulbs produced in a day. a

If it’s not possible for a number of lightbulbs to be less than zero, write an inequality for x and y to represent this fact.

Up to 200 light bulbs can be produced in one day. Write an inequality to represent this.

The ﬁnal inequality representing the limits on the number of light bulbs that can be produced each day is:

Explain the meaning of this ﬁnal inequality in the context of the question.

Problem-solving 22, 23, 25 21

21, 23–26

There are x surfers competing for wildcard entry into a pro-surﬁng competition, of which Joanne is one. She needs her total score, y, from all judges to be at least 35 in order to be chosen. The total score possible from all judges is at most 50. a

x > 1 is one of the inequalities that can be formed for the situation. Write down the two other inequalities related to y to complete the system.

Graph the solution to the simultaneous inequalities.

If Joanne receives scores of 8.4, 7.3, 7.3, 7.8, 8.9 from the judges, will she be chosen as a wildcard entry?

Whether Joanne is chosen as a wildcard entry is dependent on the number of surfers competing for the spot. True or false?

Roald has 24 centimetres of leftover wood that he is trying to make a rectangular photo frame out of. Let x represent the length of the photo frame and let y represent the width. a

Write three inequalities to represent the situation.

Graph the solution to the simultaneous inequalities.

Are these possible dimensions of the photo frame?

208

21, 23, 25

A length of 6 cm and a width of 4 cm

A length of 7 cm and a width of 6 cm

iii

A length of 9 cm and a width of 8 cm

A length of 7 cm and a width of 4 cm

If Roald wanted to create a square frame, what are the dimensions of the largest possible frame?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Throughout university, Tom works as a mentor, earning $15 per hour, and as a ﬁtness instructor earning $18 per hour. The number of hours he works in each job can vary from week to week, but he never works more than 29 hours in total each week, and he needs to be able to at least cover his weekly expenses of $270. Let x represent the number of hours he works as a mentor and y represent the number of hours as ﬁtness instructor.

Write four inequalities representing the possible values of x and y.

Graph the solution to the simultaneous inequalities.

If Tom works 6 hours as a mentor in one week, ﬁnd the minimum number of hours y he can work as a ﬁtness instructor so that he can cover his expenses.

Is it possible for him to work the same number of hours in both jobs and still be able to cover his expenses?

A warehouse is stocked with two types of storage containers: square boxes and rectangular boxes. The square boxes cost $55 each and have a volume of 2 m3 and the rectangular boxes cost $65 each and have a volume of 7 m3. The warehouse has a total storage space of 200 m3 and there is a budget of $800 to purchase the containers. Let x represent the number of square boxes purchased, and y represent the number of rectangular boxes purchased.

If it’s not possible for a number of boxes to be less than zero, write two inequalities for x and y to represent this fact.

Write an inequality relating x and y for the total budget for the warehouse.

Write an inequality relating x and y for the total storage space of the warehouse.

Eileen is spending the week at a sports camp. She is able to attend coaching sessions for her two favourite sports: basketball and football. She can attend a maximum of 26 sessions and a minimum of 20 sessions per week. Eileen’s basketball coach has told her that she must attend at least 5 basketball sessions. Her football coach has told her that she must go to at least 8 more football sessions than basketball sessions. Let x represent the number of basketball sessions she attends, and y represent the number of football sessions she attends. a

From this information, write four inequalities.

Graph the region deﬁned by the four inequalities.

List the four vertices of the region.

The vertices of the region represent the maximum and minimum number of session Eileen can attend. What is the maximum number of basketball sessions that she can attend?

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Edward is going to a music camp for one week. Edward plays drums and guitar and can attend lessons for both instruments. He can attend a maximum of 18 lessons and a minimum of 13 lessons per week. Edward’s guitar teacher has told him that he must not go to more than two times as many guitar lessons as drum lessons. His drum teacher has told him he must attend at least ﬁve drum lessons. Let x represent the number of drum lessons he attends, and y represent the number of guitar lessons he attends. a

From this information, construct a set of ﬁve inequalities.

Graph the region deﬁned by the ﬁve inequalities.

List the ﬁve vertices of the region.

Using the ﬁve vertices determine the maximum number of guitar lessons that Edward can attend.

5.03 Solve using substitution After this lesson, you will be able to… • utilise the substitution method to solve for both variables in two linear simultaneous equations. • demonstrate how to rearrange equations to isolate a variable when equations are not ready for direct substitution and then apply the substitution method.

Substitute variables When solving simultaneous equations, we are looking for values for the variables which make both equations true. This will be the point of intersection of the graphs of the equations. This section focuses on solving for the point of intersection of two linear equations, but the same process can be applied to any two equations with two unknowns. We can rearrange the equations to suit our purpose as long as we always maintain the equivalence. Let’s look at some methods in the following examples.

Example 1 We want to solve these simultaneous equations using the substitution method.

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y = x + 11

y = 3x + 19

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

a First solve for x.

Create a strategy Substitute the expression for y from equation 1, into the variable y in equation 2 and then solve for x.

Apply the idea x + 11 = 3x + 19

Substitute 1 into 2

11 = 2x + 19

Subtract x from both sides

−8 = 2x

Subtract 19 from both sides

x = −4

Divide both sides by 2

Using substitution, we have found that x = −4.

b Solve for y.

Create a strategy Substitute the value of x into one of the equations and solve for y.

Apply the idea y = x + 11

y = −4 + 11

Substitute x = −4 into 1

y=7

Evaluate

Reﬂect and check To check that x = −4, y = 7 is also a solution to 2 , substitute the value of x into 2 and see if we get the same value for y: y = 3x + 19

y = 3(−4) + 19

Substitute x = −4 into 2

y=7

Evaluate

y = 7 is the same value, this confirms that x = −4, y = 7 is the solution. We can also write this as (−4, 7), the intersection point of the two lines.

Idea summary To solve simultaneous equations using the substitution method: • Substitute the expression for one variable, usually y, into the other equation. • Solve the new equation for the other variable, usually x. • Substitute the result back into one of the equations to find the value of the other variable.

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Rearrange and substitute In the previous example, both equations were already written with y as the subject. As such, it was easy to substitute the expression for y from one equation into the other. When the equations are not written with one of the variables as the subject, we can choose any of the variables to isolate in one equation in order to substitute it into the other.

Example 2 Solve these simultaneous equations using the substitution method. p + 7q = –4

–6p + 2q = –20

Create a strategy Equation 1 looks to be the easiest to rearrange as the coefficient of p is already 1. Use 1 to get an expression for p in terms of q. Then we will substitute that expression into 2 and solve.

Apply the idea p + 7q = –4

p = –4 – 7q Subtract 7q from both sides

We now have an equation for p, we will call this equation 3 . p = –4 – 7q

Substitute the value of p into 2 and solve for q. –6p + 2q = –20

–6(–4 – 7q) + 2q = –20

24 + 42q + 2q = – 20

24 + 44q = –20

2 Substitute p Expand the brackets Collect like terms

44q = –44

Subtract 24 from both sides

q = –1

Divide both sides by –44

Finally, we now find the value for p. p = –4 – 7q

p = –4 – 7 × –1

Substitute q = –1

p = –4 + 7

Evaluate the multiplication

p = 3

Evaluate

So the solution is p = 3 and q = –1.

Reﬂect and check If p = 3 and q = –1 then these values should satisfy both equations. We can do a quick check at 1 p + 7q = –4. If p = 3 and q = –1, then we have 3 – 7 = –4 so this is true. At 2 –6p + 2q = –20, if p = 3 and q = –1 then we have –18 – 2 = –20 which is also true.

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Example 3 A man is ﬁve times as old as his son. Four years ago, the man was nine times as old as his son. Find their present ages. a Use the statement “a man is five times as old as his son” to set up equation 1.

Create a strategy Let x represent the present age of the man, and y represent the present age of the son.

Apply the idea The man is ﬁve times as old as his son, so x is equal to 5 times y. x = 5y

b Use the statement “4 years ago the man was 9 times as old as his son” to set up equation 2. Write this with y the subject.

Create a strategy To ﬁnd their ages 4 years ago, we can subtract 4 from x and y.

Apply the idea We can express the age of the man from four years ago as x − 4 and we can express the age of the son as y − 4. At this time, the age of the man is 9 times the age of the son, this will give us equation 2. 4 years ago the man is 9 times as old as the son Expand the brackets Add 36 to both sides Divide both sides by 9 2 2 is

c Solve for y to ﬁnd the age of the son.

Create a strategy Substitute equation 1 into equation 2.

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Apply the idea 2 Substitute 1 to 2 Multiply both sides by 9 Subtract 5y from both sides Divide both sides by 4 The son is 8 years old.

d Solve for x to ﬁnd the age of the man.

Create a strategy Substitute the value of y into equation 1.

Apply the idea x = 5y

x=5×8

Substitute y = 8

x = 40

Evaluate

The man is 40 years old.

Reflect and check Check the ages for the son (8) and man (40) against the initial statement. A man is 5 times as old as his son, this is true as 5 × 8 = 40. Four years ago, the man was 9 times as old as his son, this is true as 4 years ago the man would be 36, and the son would be 4 and 4 × 9 = 36.

Idea summary To solve simultaneous equations using the substitution method, we may need to rearrange one of the equations so that one variable is the subject before we can perform the substitution.

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5.03 Practice questions Understanding 1

Explorer

Adventurer

1, 2

Trailblazer 1, 3, 4

1–3

A group of students were solving the equations: x+y=6 y = 2x − 3 a

Tomi substituted x from the ﬁrst equation into the second.

Carla substituted y from the second equation into the ﬁrst.

Regie substituted y from the ﬁrst equation into the second.

Junard substituted x from the second equation into the ﬁrst.

Write the equation that each student would then have to solve and comment on the eﬃciency of each. 2

Is (5, 2) a solution of the simultaneous equations? x+y=7 x−y=3

State whether these statements are true or false. a

The method of substitution involves isolating one variable in one equation and substituting this expression into the other equation.

The substitution method can be used to solve any simultaneous linear equations.

Once you solve one of the equations for one variable, you must substitute it back into the ﬁrst equation.

Solving simultaneous equations by substitution will always result in the same solution, regardless of which variable you solve for ﬁrst.

Simultaneous linear equations can have exactly two solutions, no solutions, or inﬁnitely many solutions.

Is (4, 17) a solution of the simultaneous equations? y = 6x – 7 4x + 2y = 67

Fluency Ex 1

5a–e, 6, 7a–d, 8, 10

5c–g, 6, 7c–f, 8, 10

5g–l, 6, 7e–h, 8, 9, 11

Solve the simultaneous linear equations using substitution: a

y = 6x + 56

y = 2x + 24 e

y = 3x + 14

x+y=2 i

y = −3x − 23

y = −5x − 22

y = 5x + 37

y = −4x – 27

y = 7x + 38 5y = 35x + 265 2y = 10x + 54 y = −3x + 22

y = −3x + 9

x + y = 10 6x − y = 0 j

y = 2x – 15

x + 4y = −37 3x − 4y = 20

y = 4x – 28 3x + 5y = 44

5x − 2y = −10

2x + 3y = 48

y = −5x + 3 4x − 5y = 8

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For these simultaneous equations: 3x − 7y = 4 −12x + 28y = −16

Ex 2

Rearrange 3x − 7y = 4 to make x the subject in terms of y.

Substitute your expression for x from part (a) into −12x + 28y = −16 and solve for y.

Solve the simultaneous equations using the substitution method: a

2x + 3y = 26

2x + 7y = 23

5x + 4y = 10

4x + 3y = 17

8x − 5y = 36 5x + 6y = 0 3x + 7y = −17 9x − 4y = 6 e

−7f + 2g = −0.4

−4a + 7b = 11.5

−2.1f + g = 0.36 −2a + 2.1b = 2.25 Ex 3

A mother is currently 10 times older than her son. In two years time, she will be 7 times older than her son. Let x and y be the present ages of the son and mother respectively. This can be modelled by the simultaneous equations: y = 10x y + 2 = 7(x + 2) What are the current ages of the son and the mother?

Farzad bought some fresh produce. He picked up 2 apples and 3 kiwi fruit. The cost of Farzad’s shopping was $10.08. Eileen went to the same shop and bought 5 apples and 7 kiwi fruit. The cost of Eileen’s shopping was $24.08. Let a and k be the price of apples and kiwi fruit respectively. This can be modelled by the simultaneous equations: 2a + 3k = 10.08 5a + 7k = 24.08

Solve the simultaneous equations using the substitution method.

What is the cost of one apple and one kiwi fruit?

The number of new jobs created in Venhurst can be modelled by the simultaneous equations y = 4x − 190 000 y = x + 560 000 where x is the number of jobs created in 2008 and y is the number of jobs created in 2013.

Solve the simultaneous equations using the substitution method.

What does the solution to the simultaneous equations mean in this context?

The members in a group can be modelled by the simultaneous equations y=x+6 x + y = 22 where x and y are the number of adults and children in the group respectively.

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Solve for the values of x and y.

What does the solution to the simultaneous equations mean in this context?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reasoning

12–14

12, 13, 15

12, 13, 15, 16

If you solve simultaneous equations by substitution and ﬁnd that the result is a false statement such as 0 = 5, what does that tell you about the equations?

What does it mean if you solve simultaneous equations by substitution and get a true statement, such as 0 = 0 or 5 = 5?

For these simultaneous equations: 3x + y = 7 2x − y = 4 Explain the steps you would take to solve the simultaneous equations and why you would do it in that way.

Do the equations y = 3x + 5 and y = 3(x + 5) have inﬁnitely many solutions? Explain your answer.

For these simultaneous linear equations: −6x − 2y = −28 2x + 16y = 40 4x − 2y = 12 a

Find the values of x and y that satisfy the ﬁrst two equations.

Does this solution satisfy the third equation?

Is this the only solution to the simultaneous equations? Justify your answer.

Conﬁrm the results by graphing the three linear equations on the same Cartesian plane.

Problem-solving 17, 18, 20a–d 17

17, 19, 20b–e

19, 20c–h, 21

A taxi driver charges a fee for the initial pick up and an additional amount per km travelled. A journey of 10 km costs $105, while a journey of 15 km costs $155. a

Find the amount of the ﬁxed charge and the charge per km.

How much does a person have to pay for travelling a distance of 25 km?

The coach of a tennis team buys 7 rackets and 6 balls for $380. Later, he buys 3 rackets and 5 balls for $175. Find the price of rackets and balls respectively.

Mikhail sells sausages and soft drinks. Each sausage costs $1.50 and each soft drink costs $0.50. At the end of the day, he made a total of $78.50 for a total of 87 sausages and soft drinks combined. How many sausages and soft drinks were sold?

Use the substitution method to ﬁnd the solution(s) for each pair of simultaneous equations. a

y = x2

y = −5x d

(x + 3) + ( y − 3) = 4

(x + 1)2 + ( y + 4)2 = 25 y = −3x − 2

y = (x − 5) (x − 2)

y = 5x + 4 y = x2 + x + 8

y = 24 2

x = −5 g

y = x2 − 5x

x2 + ( y + 7)2 = 7

y = −x + 1 y = −7 h

y = (x − 1)2 x − 3y = 1

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The plan for a proposed tunnel through a mountain side is shown in the diagram: y nnel

m the tu

Exit fro

ed tunn

Entrance to the tunnel

Propos

Sea level

The height of the mountain above sea level is given by the equation height of the tunnel above sea level is given by the equation the height above sea level in kilometres.

, and the , where y represents

Use the substitution method to solve for y, the heights above sea level at which cars will enter and exit the tunnel.

5.04

Solve using elimination

After this lesson, you will be able to… • apply the elimination method to solve equations simultaneously.

Elimination method As long as equivalence is maintained in both the left and right-hand sides of any given equation, we can combine them in specific ways to make other equations that are still true. For example, if we know that:

a=b

c=d

Then we can combine these two equations to create others. We label these new equations in a similar way, so that we can reference them in our working later.

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1 + 2

a+c=b+d

1 − 2

a−c=b−d

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

We can also take multiples of equations before combining them with addition or subtraction, for example: 3× 1 −4× 2

3a − 4c = 3b − 4d

× 1 +3× 2

5 6

Equations 3, 4, 5 and 6 are all different, but they all maintain the equivalence that was present in equations 1 and 2. To solve equations simultaneously, we can use specific combinations of equations to create situations where we can eliminate one of the variables. This is the elimination method.

Example 1 Use the elimination method to solve the pair of simultaneous equations: 3x + y = 10

−4x + y = −11

Create a strategy We can remove the variable y, by subtracting the equations.

Apply the idea 3x + y = −

(−4x + y = −11)

7x =

x =

Now that we have a value for x, substitute this into equation 1 and solve for y. 3x + y = 10

3(3) + y = 10

Substitute x = 3

9 + y = 10 y=1

Evaluate Subtract 9 from both sides

The solution to simultaneous equations is (3, 1).

Reflect and check Substitute the values for x and y into equation 2 to confirm. −4x + y = −11

−4(3) + 1 = −11

Substitute

−12 + 1 = −11

Evaluate

−11 = −11

True

Confirmed that (3, 1) also solves equation 2.

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Example 2 Use the elimination method to solve the pair of simultaneous equations: 2x + y = 4

x = 4y − 34 2

Create a strategy Rearrange 2 to have the same form as 1 , then identify a multiple to eliminate one of the variables. We can remove the variable x, by doing 1 − 2 × 2 .

Apply the idea Start with rearranging 2 by subtracting 4y from both sides, then call it 3 . x = 4y – 34 2

x − 4y = −34

To eliminate x, multiply 3 by 2.

x − 4y = −34

2x − 8y = −68

3 ×2

Then, subtract 3 × 2 from 1 . 2x + y = 4 −(2x – 8y = −68) 9y = 72 y=8 Now that we have a value for y, substitute this into equation 1 and solve for x. 2x + y = 4

2x + 8 = 4

Substitute y = 8

2x = −4

Subtract 8 from both sides

x = −2

Divide both sides by 2

The solution to simultaneous equations is (−2, 8).

Reflect and check Substitute the values for x and y into equation 2 to confirm. x − 4y = −34

−2 − 4(8) = −34

Substitute

−2 − 32 = −34

Evaluate

−34 = −34

True

Confirmed that (−2, 8) also solves equation 2.

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Idea summary To solve equations simultaneously, using the elimination method we want to combine equations so the new equation has only one variable. If the coeﬃcient of the variable we want to eliminate in both equations has the same value but different sign, we can add the equations to eliminate the variable. If the coeﬃcient of the variable we want to eliminate in both equations has the same value and the same sign, we can subtract the equations to eliminate the variable.

5.04 Practice questions Understanding 1

Explorer 1, 2

Adventurer

Trailblazer

1–3

The method of elimination involves: A

Substituting one variable into another equation

Graphing the equations

Adding or subtracting equations to eliminate a variable

Isolating a variable

Explain how you would begin the process to eliminate x from the following set of equations. 5x − 9y = 2 7x − 8y = 1

Comment on whether each statement is true of false. a

The elimination method is best used when the coeﬃcients of one variable are the same in both equations.

When using the elimination method, the solution represents where the lines of the equations intersect.

There are situations when the elimination method will not work.

Fluency 4

4, 5, 6ab, 7a–c, 8a–d, 9

4, 5, 6bc, 7b–d, 8b–f, 10

4, 6b–d, 7d–f, 8d–h, 11

For the simultaneous linear equations:

What number should the second equation be multiplied by to eliminate y?

Eliminate y and ﬁnd the value of x that satisﬁes both equations.

Substitute the x-value to ﬁnd the value of y that satisﬁes both equations.

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For the simultaneous equations: 3x + 7y = −24 4x − 5y = 11

What value can we multiply each equation by to eliminate the variable y?

Solve for x and y.

For each pair of simultaneous equations: i

Rewrite the equations as equivalent equations with lowest common multiple integer coefficients.

Solve for x and y.

0.2x + 0.3y = 0.5

0.5x + 0.4y = 0.2

Ex 1

Use the elimination method to solve each pair of linear equations: a

3x + 6y = 2

2x + 5y = 44

3x + 3y = −7 6x − 5y = −28 c

8x + 3y = −11

2x − 5y = 1

−8x − 5y = 29 −3x − 5y = −39

Ex 2

Use the elimination method to solve each pair of linear equations: a

3x + 2y = 22

x = 4y − 16

2 2x = 3y + 9

2y = 2x − 4

6x = y + 15

4y − x = 1

3x − 2y = −15

7x + 13 = 3y − 11

2.5x + 3y = 11

x − 2y = −5

2 1.5x = 2.5y − 2

4.2x − 1.8y = 13.8 3.4x = 2.2y + 8.2

3 1 13 x+ y = 4 2 2 2 2 2 1 x− y= − 3 3 3 1

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4x − y = 13

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

1 2 1 2

The train in an amusement park has 15 cabs that can hold a total of 56 people. Some cabs hold 4 people while some hold 6 people. Let x be the number of four-passenger cabs and y be the number of six-passenger cabs. This situation can be represented by the simultaneous equations: x + y = 15 4x + 6y = 56

Find the number of four-passenger cabs and six-passenger cabs using the elimination method.

In terms of this context, is the solution valid?

Nine pens and 2 rulers cost $26, while 3 pens and 4 rulers cost $22. Let x and y be the price of the pen and ruler respectively. This can be modelled by the simultaneous equations: 9x + 2y = 26 3x + 4y = 22

Solve the simultaneous equations using elimination.

Explain what the solution means in this context.

Marjan spent $55.25 to purchase 9 ﬂowers. She bought rhododendrons which cost $6.45 each and chrysanthemums which cost $5.75 each. This can be modelled by the simultaneous equations R+C=9 6.45R + 5.75C = 55.25 where R is the number of rhododendrons and C is the number of chrysanthemums that Marjan bought. a

Solve the simultaneous equations using elimination.

Explain what the solution means in this context.

Reasoning 12

12, 14

13, 14

13, 15

Explain how you could use the elimination method to solve the simultaneous equations: 2x − 3y = 8 4x + 3y = 12

Without solving, how can you determine the number of solutions to these simultaneous equations using the elimination method? 3x − 2y = 6 −3x + 2y = −6

Describe a situation where it would be more eﬃcient to use the elimination method rather than the substitution method to solve simultaneous equations and explain why.

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y = 3x2 − 6

Elias solved nonlinear simultaneous equations using elimination. work is shown. Aria chose to solve by graphing, and she found two solutions. a

Explain the mistake Elias made that caused him to only ﬁnd one solution.

Find the second solution.

Original equations

y = −x – 2 y = 3x2 – 6 3y = −3x2 – 6 Multiply equation 2 by 3 4y = −12

Add the equations, solve for y

y = −3 −3 = 3x2 − 6

Substitute y, solve for x

3 = 3x 1 = x2 1=x

The solution is (1, −3).

Problem-solving 17, 20ab 16

6, 18, 20b–d

16, 19, 20d–f

In a game show, Nigel has the opportunity to answer 30 questions. For every question he answers correctly, he earns 6 points, but for every question he answers incorrectly, he loses 3 points. Let x represent the number of correct answers and y represent the number of incorrect answers. a

If Nigel’s total score was 90 points, write a pair of simultaneous equations that represent this problem.

Solve these equations to ﬁnd out the number of correct and incorrect answers Nigel provided.

For two numbers, x and y: • The sum of y and two times x is 27. • The difference between ﬁve times x and y is 43. Find the two numbers.

For two numbers, x and y: • The sum of the x and y is 10. • The difference between six times x and ﬁve times y is 5. Find the two numbers.

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The proportion of the workforce of a particular country that are teenagers is modelled by 3.3x + y = 70.8, where x is the number of years since 2015 and y is the proportion as a percentage. The proportion of the workforce that are pensioners is modelled by 3.2x − y = −64.5. a

Estimate the year in which the proportion of the workforce that are teenagers and the proportion of the workforce that are pensioners are the same.

Estimate the percentage of the workforce that are teenagers or pensioners during the estimated year in part (a).

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Use the elimination method to solve the simultaneous equations. a

x2 − y = −2

−x + y = 4 c

y = 3x2 − 2

x2 + y = 5 −x + y = −1

x2 + y2 = 100

y = x2 2x2 + 10y = 68 e

x2 + y2 = 25

(x + 3)2 + ( y + 1)2 = 25

(x + 3)2 − y = 6

5.05 Applications of simultaneous equations After this lesson, you will be able to… • define variables that represent unknown quantities to translate information into at least two equations. • apply simultaneous equations to real-life scenarios. • define and understand the concept of the break-even point. • perform break-even analysis using graphical representations.

Applications of simultaneous equations Our ﬁrst step is to deﬁne the variables that represent the quantities in the problem. Then translate the pieces of information about the quantities into equations. Lastly, by using either the substitution method, elimination method, or a graphical method, we solve the equations simultaneously.

Example 1 When comparing some test results, Christa noticed that the sum of her geography and science test score was 172, and that their difference was 18. Given that her geography score is x and her science score is y, and that she scored higher in the geography test: a Use the sum of the test scores to form an equation. We will refer to this as equation 1.

Create a strategy Write an equation in terms of x and y.

Apply the idea The sum of her two test scores is 172, where x and y are the scores. Equation 1: x + y = 172

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b Use the difference of the test scores to form an equation. We will refer to this as equation 2.

Create a strategy

Apply the idea

Write an equation in terms of x and y.

The difference between the scores is 18, where x and y are the scores. Since x > y we can subtract y from x. Equation 2: x − y = 18

c Use these two equations to ﬁnd her geography score.

Create a strategy We can use the elimination method to solve for x.

Apply the idea Notice that the coeﬃcient of y in both equations is the same, but with opposite sign. So we can eliminate the variable y by adding the two equations together. +

x + y = 172 x - y = 18 = 190 2x x = 95

So her geography score is 95.

d Find her science score.

Create a strategy Substitute the x-value into one of the equations to solve for y.

Apply the idea x + y = 172

Write equation 1

95 + y = 172

Substitute the value of x

95 + y − 95 = 172 – 95 Subtract 95 from both sides y = 77

Simplify

So her science score is 77.

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Example 2 Consider the following phone plans: • GO SMALL plan: This plan has a $20 monthly base charge and charges 90c per minute for all calls. • GO MEDIUM plan: This plan has a $26 monthly base charge and then charges 70c per minute for all calls. a Complete the table of values for various total monthly call times for the two plans: Call time (in minutes) 20 30 40 50

Total cost for GO SMALL plan (S)

Total cost for GO MEDIUM plan (M )

Create a strategy Add the monthly base charge to the number of minutes times the rate per minute.

Apply the idea For the call time of 20 minutes we can ﬁnd the cost for the GO SMALL plan by: GO SMALL cost = 20 + 0.9 × 20 = $38.00

Add the base charge to the rate times the minutes Evaluate

We can ﬁnd the cost for the GO MEDIUM plan by: GO MEDIUM cost = 26 + 0.7 × 20 = $40.00

Add the base charge to the rate times the minutes Evaluate

Continuing in this way for the rest of the call times gives us this table: Call time (in minutes) 20 30 40 50

Total cost for GO SMALL plan (S) 38.00 47.00 56.00 65.00

Total cost for GO MEDIUM plan (M ) 40.00 47.00 54.00 61.00

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b Sketch a graph for the two plans.

Create a strategy Use the points from the table to graph both plans.

Apply the idea Cost 65 60 55

To include the points from the table, the vertical or cost axis needs to go over 60 and the horizontal or call time axis needs to go over 50. We will go a bit further than this along both axes, in order to see more detail in the graph. We will also label the lines S and M.

50 45 40 35 30 M 25

20 15

To graph the lines we can use two points for each plan from the table. For the GO SMALL plan we can use (20, 38) and (40, 56). For the GO MEDIUM plan we can use (30, 47) and (50, 61).

10 5

Call time 5 10 15 20 25 30 35 40 45 50

c Using the graph, determine how many minutes of calls would need to be made so that the monthly bill costs the same on both plans.

Create a strategy Use the point of intersection.

Apply the idea Cost 65 60

The point of intersection is at (30, 47). This means that for a call time of 30 minutes, both plans will cost 47.

55 50 45 40 35 30 25 20

M S

15 10 5

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Call time 5 10 15 20 25 30 35 40 45 50

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Idea summary To solve a problem with two unknown quantities: • Deﬁne variables to represent the quantities. • Translate the information into two equations. • Use the substitution method, elimination method or graphical method to solve the equations simultaneously.

Break-even point It’s important for businesses to make a profit. The point at which a business makes a profit is known as the break even point. To find this point, we use break-even analysis. Break-even analysis looks at the graphs of cost and revenue together to determine where they intersect. This point of intersection is the break-even point where revenue equals expenses. Here is an example of a break-even analysis for a single day of operation for the company Lovely Lemonade. Lovely Lemonade’s break even analysis 1200 Profit

Money (dollars)

1000

Break even point

800 600 400

Loss

200 1000

2000

3000

4000

5000

Number of drinks (units) Revenue

Expenses

The expenses line (red) starts off higher than the revenue line (green). For a low number of units sold, Lovely Lemonade will lose money. For example, if they only sell 1000 units, the green line tells us they earn $200, the red line tells us they spend $600, and so overall they make $200 − $600 = −$400. The negative sign means overall they lose $400. If they sell 5000 units the graph tells us that they will earn $1100 but only spend $1000. Overall they make $1100 − $1000 = $100, and the positive sign means a profit of $100. The important point is when the two lines intersect - this is the break-even point. In this example we can see that selling 4000 units means they don’t make any profit, but they don’t lose any money either - both their revenue and their expenses are $900. We can also use break-even analysis to compare two different pricing plans for the same product or service to see which one offers the best deal for a particular situation.

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Here’s an example comparison between two energy plans provided by rival companies, Thorgate and Callisto: Thorgate vs Callisto pricing comparison 18 Callisto cheaper

Price (dollars)

15 12

Break even point

9 6 Thorgate cheaper

3 10

Energy per day (kWh) Thorgate

Callisto

In this scenario, the break-even point at (40, 10) tells us that both plans charge $10 for 40kWh. For energy amounts less than the break-even point, we can see that Thorgate is cheaper, and for energy amounts more than the break-even point, Callisto is cheaper. Knowing how much energy someone plans on using can then determine the best plan for them.

Example 3 The cost for a furniture manufacturer to make an armchair is $500 per armchair plus a ﬁxed setup cost of $3500. The armchairs will sell for $650 each. a Write an expression to represent the cost of manufacturing x armchairs.

Create a strategy Add the ﬁxed cost and the variable cost.

Apply the idea The ﬁxed cost is $3500. The variable cost is $500 per armchair, which is $500 × x. Cost = 3500 + 500x

Add the ﬁxed and variable costs

b Write an expression to represent the revenue generated from the sale of x armchairs.

Create a strategy The amount of revenue generated is equal to the number of pieces sold multiplied by the price at which they sell.

Apply the idea Revenue = 650x 230

Multiply the selling price and the number of armchairs sold

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

c Find the break-even point.

Create a strategy Equate the expressions for the cost and the revenue and solve for x.

Apply the idea At the break-even point, the cost is equal to the revenue, so we equate their expressions. Equate the revenue and cost

Subtract 500x from both sides

Simplify

Divide both sides by 150

Evaluate

So the break-even point is when x = 24 since we can only sell whole numbers of armchairs.

Example 4 David decides to start his own yoga class. The cost and revenue from running the class have been graphed: 54

Money (dollars, $)

48 Cost

42 36 30 24 18 12

Revenue

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

a Find the revenue that David makes for each student.

Create a strategy Use the revenue line on the graph.

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Apply the idea 54

For 1 student, x = 1. We move from x = 1 on the x-axis up to the revenue line, then across to the y-axis to get y = 6.

Money (dollars, $)

48 42

So the revenue per student is $6.

36 30 24 18 Revenue

12 6

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

b How many students must attend his class so that David can cover the costs of running the class?

Create a strategy

Apply the idea

The number of students is given by the intersection point of the revenue and cost curves.

The point of intersection is (6, 36). So if David get 6 students, he can cover the cost of running the class.

c How much proﬁt does David make if there are 9 students in his class?

Create a strategy Find the difference between the revenue and the cost.

Apply the idea 54 48 Money (dollars, $)

To ﬁnd the revenue and cost for 9 students, we ﬁnd x = 9 on the x-axis and move up to the two lines and then across to the y-axis from each.

Cost

We can see that cost is $45 and the revenue is $54.

36 30 24 18

Revenue

12 6

x 1 2 3 4 5 6 7 8 9 10 11 Number of students

Profit = 54 – 45 = $9

232

Subtract the cost from the revenue Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Idea summary The break-even point is where the cost and revenue are equal. To ﬁnd the break-even point, we can either: • Look at the graphs of cost and revenue together to determine where they intersect. • Equate the expressions for revenue and cost and solve for the independent variable.

5.05 Practice questions Understanding 1

Explorer 1–3

y = 3x – 1

2x + y = 7

y = 4x + 2

y = −x + 5

3x − 2y = 6

2x + 3y = 5 x−y=1

State whether the substitution method or the elimination method would be more eﬃcient for solving.

Explain why you chose the method in part (i). y = 6x + 56 y = 2x + 24

7x − 4y = 15

7x + 5y = 60

−5x + 16y = 82 25x − 4y = 122

y = −3x − 23 x + 4y = −37

Identify whether each statement is true or false. a

All simultaneous equations can be solved using any method.

Only some simultaneous equations can be solved by graphing.

Graphing is only useful to be used if the exact point of intersection can be determined from the graph.

Fluency 4

y = 2x + 3

For each of the simultaneous equations:

Trailblazer 1–3

For the simultaneous equations, explain which strategy you would use and why: a

Adventurer 1–3

4–9, 14, 16ac

5–8, 10–13, 16bd

5–8, 11–13, 15, 16ab

A family owns two businesses that made a combined profit of $6 million in the previous financial year, with Business B making 2 times as much profit as Business A. Let x and y be the profits (in millions) of Business A and Business B respectively. a

Write a pair of linear equations to represent this situation.

Find Business A’s proﬁt.

Find Business B’s proﬁt.

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Ex 1

When comparing some test results, Akira noticed that the sum of his Geography score and Science score was 184, and that their difference was 22. He also noticed that he scored higher for the Geography test. Let x and y be his Geography score and his Science score respectively.

Ex 2

Write two equations in terms of x and y from the information given.

Find his Geography score.

Find his Science score.

For a phone company, customers can choose between two phone plans: • GO SMALL plan: $24 monthly base charge, and 80c per minute for all calls • GO MEDIUM plan: $32 monthly base charge, and 60c per minute for all calls a

Complete the table of values for various total monthly call times for the two plans: Call time (in minutes) 25 30 35 40

b 7

Ex 3

Total cost for GO SMALL plan

Total cost for GO MEDIUM plan

Find how many minutes of calls would need to be made so that the monthly bill costs the same on both plans.

Toby’s piggy bank contains only 5c and 10c coins. It contains 70 coins with a total value of $3.85. a

Set up two equations by letting x and y be the number of 5c and 10c coins respectively.

Solve the simultaneous equations.

Kerry currently pays $50 a month for her internet service. She is planning to switch to a ﬁbre optic cable service. a

Write an equation for the total cost T of Kerry’s current internet service over a period of n months.

For the ﬁbre optic cable service, Kerry pays a one-off amount of $1200 for the installation costs and then a monthly fee of $25. Write an equation of the total cost T of Kerry’s new internet service over n months.

Determine how many months it will take for Kerry to break-even on her new internet service.

There are 36 employees in a company, and the full-time employees outnumber the part-time employees by 16. Find the number of part-time employees and full-time employees in the company.

Two twin soccer players, Arden and Benton, are having a contest over who will score the most goals in the season. The twins have scored a combined total of 34 goals so far over the season and Arden has scored 6 more goals than Benton. Let x and y be the number of goals scored by Arden and Benton respectively.

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Write a pair of linear equations to model this situation.

Find the number of goals Arden has scored this season.

Find the number of goals Benton has scored this season.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Christa has $4000 to invest, and wants to split it up between two accounts: • Account A earns 7% annual interest. • Account B earns 8% annual interest. Her target is to earn $303 total interest from the two accounts in one year. Let x and y be the amounts, in dollars, that she invests in accounts A and B respectively.

Ex 4

Write a pair of linear equations to represent this situation.

Solve for y, the amount Christa invests in account B.

Solve for x, the amount Christa invests in account A.

The length of a rectangle is 12 units more than the width, and the perimeter of the rectangle is 56 units. Let y be the width and x be the length of the rectangle. a

Write a pair of linear equations to represent this situation.

Solve the simultaneous equations.

Consider the diagram of the rectangle: a

Solve for x and y.

Find the length of the rectangle.

Find the width of the rectangle.

3y + 384

8x + 60 D

B 15y + 48

64x + 12

Maria and Buzz both walk from their houses to the bus stop every morning. Maria walks 1.5 kilometres further, and together they walk 3.3 kilometres. Find the distance Buzz walks.

Qin invested $16 000 in total in two stocks, A and B. In one year, the investment in stock A made a 14% return, while the investment in stock B fell by 6%. The total annual interest from both stocks was $700.

Set up two equations by letting x and y be the amounts that she invested in stocks A and B respectively.

Solve the simultaneous equations.

Explain what the solution means in this context.

A clothing manufacturer is deciding whether to employ people or to purchase machinery to manufacture their line of t-shirts. The cost of employing people to make the clothing is modelled by y = 800 + 60x and the cost of using machinery is modelled by y = 3200 + 20x, where y is the cost and x is the number of t-shirts to be made. a

Sketch a graph that depicts the two cost functions.

For what value of x will the cost of making t-shirts be the same, whether they are made by people or by machines?

For what values of x will it be more cost eﬃcient to use machines to manufacture the t-shirts?

For what values of x will it be more cost eﬃcient to employ people to manufacture the t-shirts?

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Reasoning 17

17, 18

17, 19

Barry is considering opening a restaurant. He has calculated his ﬁxed costs at $10 000 per month. He expects to generate a revenue of $12 per meal, but the variable cost for each meal is $15. a

Write equations for the total cost C and total revenue R for selling x meals.

Can Barry reach the break-even point? Explain.

How does this impact Barry’s restaurant idea and what changes can he make for it to succeed?

Pensacola high school hosts an annual badminton tournament. Student tickets cost $3.00 and general admission tickets cost $5.50. a

If 350 tickets were sold for a total of $1450, how many of each type of ticket were sold?

Explain which method you used to solve and why.

At Soul Food Express you can buy two orders of oxtail stew and a slice of sweet potato pie for $38.49, or you can get ﬁve orders of oxtail stew and three slices of sweet potato pie for $99.22. a

How much is the cost of an order of oxtail stew and a slice of sweet potato pie?

Explain which method you used to solve and why.

Problem-solving 20

20, 21

22–24

A bacteria population doubles every day, following the equation y = 2x. A scientist predicts that the population will increase by 3 times the number of days, following the linear equation y = 3x. At what point will the scientist’s prediction first match the actual population?

A circular garden has a radius of 1 unit and its centre is at the point (2, 1). A straight path is built through the garden, following the equation y = x − 1. Find the points where the path intersects the garden’s boundary.

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For the simultaneous equations: y = −4x2 + 8x y = 2x + c

Determine the value of c that will result in one unique solution to the simultaneous equations.

Determine the values of c that will result in two real solutions to the simultaneous equations.

A chef is experimenting with a new recipe that requires a speciﬁc ratio of ingredients. If 2 is added to the numerator and denominator it becomes

and if 3 is subtracted from the

numerator and denominator it becomes . Find the original ratio of the ingredients.

An engineer is designing a playground that will be placed at the centre of a hyperbolicshaped park. The engineer wants to place a fence along a straight line that intersects the park at two points. The situation can be modelled by the simultaneous equations: • Park: • Fence: Find the points where the fence intersects the park. Round your answers to two decimal places.

Investigation: Intersections of curves Investigate online

mathspace.co

5.05 Applications of simultaneous equations mathspace.co

237

5 Chapter review 1

For each pair of linear equations: i

Sketch the lines on the same Cartesian plane.

Find the solution.

x = −4

2y = 5x − 12

y = −3x − 9 y = x − 3 2

Use the graph to solve the simultaneous equations: y = −3x −13

y = −x − 3

−8

−7

5 4 3 2 1

−6

−5

−4

−3

−2

−1

x + 2y = −6

y = −x − 3

y = −3x − 13 x + 2y = −6 3

−1 −2 −3 −4 −5

x 1

y = −3x − 13 x + 2y = −6

A boutique’s expenses are represented by E = x + 1220, and its sales revenue by S = 1.95x, where x is the number of dresses sold. a

What coordinates satisfy both equations?

What do the coordinates from part (a) mean in this context?

y 3500 3000 2500 2000 1500

E S

1000 500 x 256 512 768 1024 1280 1536

A pair of linear equations has inﬁnitely many solutions. One of the equations is y = 2x + 5. Which of these is the other equation? A

y = 2x + 7

Consider the simultaneous equations:

y = −3x − 4

y = −3x + 7

Without graphing, is there a solution? Explain your answer.

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−y = −2x − 5

A triangular garden has a base and two equal-length sides. One of the equal sides is 2 m longer than the base. The garden has a total perimeter of 22 m. a

Let y represent the length of one of the equal sides and x represent the base. Create two equations that represent this information.

Sketch the two lines on the same coordinate plane.

Find the length of one of the equal sides and the base of the triangle.

For the simultaneous inequalities y ≤ 2x − 1 and y > x − 4: a

Find the x-coordinate of the point at which the two lines intersect.

Find the y-coordinate of the point of intersection.

Graph the solution region.

Write the simultaneous inequalities that describe the shaded region. 8 7 6 5 4 3 2 1

−4 − 4− −3 3− −2 2 − −11−11 −2 2 −3 3 −4 4 −5 5 −6 6

x 1

2 3 4 5

For y > x − 4, y ≥ −x − 4, and y < 1: a

Graph the lines y = x − 4, y = −x − 4, and y = 1 on the same set of axes.

Does the point (2, 0) satisfy the simultaneous inequalities?

Shade the region described by the inequalities y > x − 4, y ≥ −x − 4, and y < 1.

Calculate the area of the shaded region.

ChocDelight produces two types of chocolates: milk chocolate and dark chocolate. Each type of chocolate is made in different machines. The time that each chocolate requires in each machine is detailed in the table. Each machine can operate for 24 hours a day. Let x represent the number of milk chocolates produced each day, and y represent the number of dark chocolates produced each day. a

Write an inequality for Machine A.

Write an inequality for Machine B.

Write inequalities for x and y, given the number of chocolates cannot be a negative number.

Graph the region deﬁned by the inequalities for Machine A and Machine B.

Milk Chocolate Dark Chocolate

Machine A 5 hours 3 hours

Machine B 4 hours 6 hours

Chapter 5 review mathspace.co

239

Mirela has 16 centimetres of leftover cardstock paper that she wants to use to create a rectangular bookmark. Let x and y represent the length and width of the bookmark respectively. a

Write three inequalities to represent the situation.

Graph the solution to the simultaneous inequalities.

Are these possible dimensions of the bookmark?

d 12

A length of 4 cm and a width of 3 cm

A length of 5 cm and a width of 4 cm

iii

A length of 6 cm and a width of 5 cm

A length of 5 cm and a width of 2 cm

If Mirela wanted to create a square bookmark, what are the dimensions of the largest possible bookmark?

Solve the simultaneous linear equations using substitution: a

y = 4x + 40 y = 3x + 30

y = −2x – 11

c 2a + b = 23

4x + 2y = −22

d −3f + 2g = 6

b = −2a + 35

2f + 5g = 53

Use the substitution method to ﬁnd the solution(s) for each pair of simultaneous equations. a

(x + 2)2 + (y − 4)2 = 10

y = 4x + 2

x = −1 y = −x2 + x + 6 14

Solve the simultaneous linear equations using elimination: a

2x + 3y = 8

3x − 6y = 48

x − 3y = −14

x + 2y = 10

0.1x + 0.2y = 0.3

x − y = −22.5

Use the elimination method to ﬁnd the solution(s) for each pair of simultaneous equations. a

x2 − y = 9

x2 + y2 = 80

−x + y = 3 16

A couple runs two restaurants that made a combined profit of $10 000 in the previous financial year, with Fine Diners making 3 times as much profit as Casual Eats. Let x and y be the profits of Casual Eats and Fine Diners respectively.

Write a pair of linear equations to represent this situation.

Find Casual Eats’ proﬁt.

Find Fine Diners’ proﬁt.

Alfred invested $20 000 in total in two bonds, C and D. In one year, the investment in bond C had a 10% return, while the investment in bond D rose by only 4%. The total annual interest from both bonds was $1340. a

Set up two equations by letting x and y be the amounts that he invested in bonds C and D respectively.

Solve the simultaneous equations.

Explain what the solution means in this context.

Juma and Jian both cycle from their homes to the gym every evening. Juma cycles 2.2 kilometres further, and together they cycle 5.8 kilometres. Find the distance Jian cycles.

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Everything connects to everything else. –Leonardo da Vinci

Big ideas • Right-angled triangles have special properties that allow their side lengths to be determined using the Pythagoras’ theorem or ratios. • Trigonometry connects the angle measures and side lengths of a triangle and provides a set of functions that can be used to find unknown angle measures and side lengths.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10M03 solve practical problems by applying Pythagoras’ theorem and trigonometry to right-angled triangles, including problems involving direction and angles of elevation and depression VC2M10ASP02

establish the sine, cosine and area rules for any triangle and solve related problems

VC2M10ASP05

apply Pythagoras’ theorem and trigonometry to solving three-dimensional problems in right-angled triangles

Trigonometry Chapter outline 6.01 Pythagoras’ theorem 6.02 Trigonometric ratios 6.03 Angles of elevation and depression 6.04 Bearings 6.05 Optional: Pythagoras in 3D 6.06 Optional: Trigonometry in 3D 6.07 Optional: Sine and cosine rules Chapter 6 review

244 253 266 276

292

Sundials, such as the famous Intihuatana of Machu Picchu, demonstrate the use of trigonometry to position the hour lines and determine the time of day!

6.01

Pythagoras’ theorem After this lesson, you will be able to… • apply Pythagoras’ theorem to identify right-angled triangles. • use Pythagoras’ theorem to find the length of the hypotenuse and other sides in a right-angled triangle. • apply knowledge of Pythagoras’ theorem to solve real-world problems.

Pythagoras’ theorem Hypotenuse

In a right-angled triangle, the largest angle in the triangle is 90°. The side opposite the right-angle will be the longest side. We call this side the hypotenuse.

Right angle

All three sides of a right-angled triangle are related by the equation shown. The two shorter sides will be called a and b, and the hypotenuse (the longest side) will be c.

a2 + b2 = c2 a, b, c

Sides of a right-angled triangle

a, b

The lengths of the shorter sides

The length of the hypotenuse

This relationship between sides in a right-angled triangle is called Pythagoras’ theorem. We can use this theorem to: • Check whether a triangle is a right-angled triangle • Find the length of the hypotenuse • Find the length of one of the shorter sides in a right-angled triangle

Interactive exploration

Discover this concept in action online

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mathspace.co

Example 1 Use Pythagoras’ theorem to determine whether this is a right-angled triangle.

16 9

a Let a and b represent the two shorter side lengths. First ﬁnd the value of a2 + b2.

Create a strategy Substitute the lengths of the two shorter sides and add them.

Apply the idea a2 + b2 = (9)2 + (16)2

Substitute the values of a and b

= 81 + 256

Evaluate the squares

= 337

Evaluate the sum

b Let c represent the length of the longest side. Find the value of c2.

Create a strategy

Apply the idea c2 = (18)2

Substitute the length of the longest side.

= 324

Substitute the value of c Evaluate

c Is the triangle a right-angled triangle?

Create a strategy

Apply the idea

Compare the results from parts (a) and (b) to determine if a2 + b2 = c2.

337 ≠ 324 The triangle is not a right-angled triangle because a2 + b2 ≠ c2.

Example 2 Which of the following three lengths represent the sides of a right-angled triangle? Select the two correct answers. A (6, 8, 13)

(5, 12, 13)

(2, 4, 6)

(3, 4, 5)

Create a strategy Add the squares of the two smaller numbers. If the result is the same as the square of the biggest number, then it’s a Pythagorean triple.

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Apply the idea The triple (a, b, c) is a Pythagorean triple if c2 = a2 + b2. 132 = 62 + 82

Substitute the sides from option A

169 ≠ 100

Evaluate both sides

132 = 52 + 122

Substitute the sides from option B

169 = 169

Evaluate both sides

62 = 22 + 42

Substitute the sides from option C

36 ≠ 20

Evaluate both sides

52 = 32 + 42

Substitute the sides from option D

25 = 25

Evaluate both sides

The correct answers are options B and D.

Example 3 Find the length of the unknown side as an exact value: a

Create a strategy We can use Pythagoras’ theorem: c2 = a2 + b2.

Apply the idea c 2 = a 2 + b2 2

Write the formula 2

c = 12 + 16

Substitute a and b

c = 144 + 256

Evaluate the squares

Evaluate the sum

c = 400 c=

c = 20

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Take the square root of both sides Evaluate the square root

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Create a strategy Use Pythagoras’ theorem: a2 + b2 = c2.

Apply the idea We can substitute a = x, b = 10, and c = 20 into the formula: a2 + b2 = c2 2

Write the formula 2

x + 10 = 20

Substitute the values

x + 100 = 400

Evaluate the squares

x + 100 − 100 = 400 – 100

Subtract 100 from both sides

x = 300

Evaluate the difference

Take the square root of both sides

Reflect and check When finding a short side, the most important thing to remember is that the two given lengths need to go into different sides of the equation. If we substitute incorrectly, we will end up with the square root of a negative number (calculator error) or a side which is longer than the hypotenuse.

Example 4 A movie director wants to shoot a scene where the hero of the film fires a grappling hook from the roof of one building to the roof of another. If the first building is 37 m tall, the other building is 54 m tall and the street between them is 10 m wide, what is the minimum length, l, of rope needed for the grappling hook? Round your answer to two decimal places.

l 54 m 37 m

10 m

Create a strategy Use Pythagoras’ theorem: c2 = a2 + b2.

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Apply the idea Notice that we have a right-angled triangle, and the length of the grappling hook’s rope will be the hypotenuse. One of the short sides is the width of the street, 10, and the other short side is the difference in heights between the two buildings, 54 − 37. We can substitute a = 10, b = (54 − 37), and c = l into the formula: c2 = a2 + b2

Write the formula

l = 10 + (54 − 37)

Substitute the values

l = 10 + 17

Simplify the brackets

l = 100 + 289

Evaluate the indices

l = 389

Evaluate the addition

Take the square root of both sides

l ≈ 19.72

Evaluate the square root

Idea summary

Pythagoras’ theorem relates the three sides of a right-angled triangle. Sides a and b are the two shorter sides, and the longest side, called the hypotenuse, is c. We can also test to see if a triangle is right-angled by checking to see if its three sides satisfy a2 + b2 = c2.

To ﬁnd the hypotenuse: c2 = a2 + b2 To ﬁnd a shorter side, use: a2 = c2 − b2 or b2 = c2 − a2 We can take the square root of both sides to give us the following formulas:

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6.01 Practice questions Understanding 1

1−3

Explorer

1−3

Adventurer

1−3

Trailblazer

Write Pythagoras’ theorem in a form you could use to ﬁnd: a

The length of z.

The length of x.

Calculate the value of x, rounded to two decimal places. All values of x are positive. a

Ex 1

Ex 2

Ex 3

82 = 52 + x2

Use Pythagoras’ theorem to determine whether this is a right-angled triangle. a

Let a and b represent the two shorter side lengths. First ﬁnd the value of a2 + b2.

Let c represent the length of the longest side. Find the value of c2.

Is the triangle a right-angled triangle?

Fluency 4

x2 = 22 + 32

4a–c, 5a–d, 7ab, 9

4d–f, 5c–f, 6, 7bc, 8a, 9

4f–h, 5e–i, 6, 7cd, 8, 9

Determine whether each of the following sets of values are Pythagorean triples: a

(12, 16, 26)

(4, 8, 12)

(8, 15, 17)

(9, 12, 15)

(12, 16, 20)

(7, 7, 10)

(8, 11, 14)

(7, 24, 25)

Find the length of the unknown side as an exact value: a

d c

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f 5

A right-angled triangle has one side of length 6 and another side of length 4. Find the length of the hypotenuse.

A right-angled triangle has a hypotenuse of length 9 and another side of length 3. Find the length of the third side.

A right-angled triangle has a hypotenuse of length 17 and another side of length 14. Find the length of the third side.

Two poles, one of length 8 m, and the other of length 14 m, stand 8 m apart on level ground. Find the length of wire required to join the tops of the poles. 14 m 8m

Find the length of a diagonal of each shape, rounded to one decimal place where appropriate. a

10 cm

10 mm 5 mm

10 cm

21 m

10 cm 4 cm

20 m 16 cm

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Find the distance between the points P and Q, rounded to one decimal place. a

4 cm 4 cm 6 cm

20 mm

10 mm

4 cm

6 mm

9 Ex 4

18 mm Q

5 mm

Two buildings, one of height 300 m and the other of height 450 m, stand a distance apart on level ground. If a cable 300 m joining the tops of the buildings is 200 m long, ﬁnd, to the nearest metre, the distance between the buildings.

200 m 450 m ?

Reasoning

10, 12a, 12b(i)(ii)

10, 12

11, 12

Can a triangle with sides of length 3 cm, 5 cm and 7 cm be a right-angled triangle? Use Pythagoras’ theorem to justify your answer.

If AB = 10 cm, AD = 8 cm and ∠ADB is a right angle, show that D is the midpoint of BC.

A triangle with sides 3, 4, and 5 units is a right-angled triangle. a

If the triangle is enlarged by a factor of 3, does Pythagoras’ theorem still hold? Justify your answer.

Is a triangle with these sides a right-angled triangle? i

30, 40, 50

Problem-solving 13−16 13

35, 45, 55

iii

3k, 4k, 5k

13−17

Find the length of the diameter AB, of the circle.

Find the exact length x.

21a, 28a, 36a

14−18

24 cm O

A x

7 cm B 9 cm

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Tricia’s house has outer dimensions as shown in the diagram:

A 5.6 m

5.6 m

A drainpipe needs to run from the vertex of the roof, indicated by point A in the diagram, vertically downwards until it is 30 cm above the ground. Find the length of pipe needed in 3.6 m metres, rounded to two decimal places.

3.6 m

10.2 m

Find the value of x, rounded to one decimal place. a

5 cm

Casper produces thin plastic triangles. a

If the triangles need to be edged by a material that costs 3 cents per cm, how much, in dollars, will it cost to edge 100 triangles?

If the triangles need to be painted with a coating that costs 2 cents per cm2, how much, in dollars, will it cost to coat 100 triangles.

20 cm

29 cm

The diagram shows four squares. The outer square has side lengths of 20 cm. The next square is formed by joining the midpoints of the outer square, the third square by joining the midpoints of the second square, and the fourth square by joining the midpoints of the third square. Tamsin is using wire to construct the shape in this diagram. Find, to the nearest centimetre, the length of wire she requires.

Angelia and Beth start their hikes from the same starting point. Angelia walks 900 m south and then 900 m east, before stopping for a lunch break. Beth stops for a break having walked 1200 m south and then 600 m east. If they both walked back to the start point in a straight line, who would walk furthest, and by how far to the nearest metre?

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6.02

Trigonometric ratios

After this lesson, you will be able to… • identify and define sine, cosine, and tangent, and describe the relationship between angles and sides in right-angled triangles. • use trigonometric ratios to find an unknown side or angle in right-angled triangles. • use trigonometric ratios with other geometric principles and formulas to find missing lengths, distances, and angles.

Trigonometric ratios In right-angled triangles, we can use trigonometric ratios to describe the relationship between a given angle and two sides.

Hypotenuse

The trigonometric ratios are:

Opposite

θ A

Adjacent

Exploration We have three right-angled triangles with acute angles of 30° and 60°. 1. Is △ABC ∼ △DEF ∼ △GHI? Explain why or why not. 2. Find the ratio for sin 30° for each of the triangles. 3. What do you notice about each ratio (use a calculator if you need to)? 4. Find the ratio for cos 30° for each of the triangles. 5. What do you notice about each ratio (use a calculator if you need to)? 6. Find the ratio for tan 30° for each of the triangles.

30°

3 30° 2 60°

1 2

30°

2 3

2 60°

60° 1 F

7. What do you notice about each ratio (use a calculator if you need to)? 8. Draw another similar triangle which would have the same ratios.

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The trigonometric ratios will be the same for similar right-angled triangles. This is because the size of the angles in a triangle determines the ratios between the side lengths of a triangle.

Example 1 Evaluate cos (θ ) within △ABC.

A α

Create a strategy Identify which sides we need for the cosine ratio, then ﬁll in the side lengths.

Apply the idea The angle θ is also called ∠B.

Hypotenuse 13

Opposite

so we need the lengths of the adjacent and hypotenuse sides.

θ B

We are using the cosine ratio

Adjacent

With respect to the angle θ, the adjacent side is 5, and the hypotenuse is the longest side of 13. Deﬁntion of cosine Substitute the sides

Idea summary

Hypotenuse

The trigonometric ratios are:

Opposite

θ A

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Adjacent

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Find a side If we are given one side length and one angle, we can use the trigonometric ratios to ﬁnd one of the missing side lengths. We can follow these steps to ﬁnd a missing side in a right-angled triangle: 1. Highlight the reference angle. 2. Identify which sides are the hypotenuse, opposite, and adjacent — label if desired. 3. Choose the ratio which has one of the required sides given and one is the missing side: • Sine: opposite and hypotenuse • Cosine: adjacent and hypotenuse • Tangent: opposite and adjacent 4. Set up the chosen ratio:

5. Solve for the unknown. Remember the following: • For a given angle θ, the value of each trigonometric ratio stays the same no matter the size of the triangle. • Your calculator must be set to degree mode when working with degrees in trigonometric ratios.

Example 2 Find the value of the side length, rounded to two decimal places: a

f 47°

Apply the idea With respect to the angle 47°, the length of the adjacent side is f, and the length of the hypotenuse is 8, so we can use the cosine ratio.

Use the cosine ratio

Substitute the values and f

Multiply both sides by 8

Evaluate and round

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11 g

42°

Apply the idea With respect to the angle 42°, the length of the opposite side is 11, and the length of the hypotenuse is g, so we can use the sine ratio. Use the sine ratio Substitute the values and g Multiply both sides by g Divide both sides by sin 42° Evaluate and round

Reflect and check When the unknown is in the denominator, we need to take care to rearrange properly.

Idea summary We can use the trigonometric ratios to find an unknown side length of a rightangled triangle. Once we set up our equation with a variable representing the unknown side length, we can use inverse operations to make the variable the subject of the equation. Then we can evaluate the expression in our calculators to find the side length.

Find an angle If we are given two sides of a right-angled triangle, we can use the trigonometric ratios to find a missing angle measure. We can follow these steps: 1. Using the unknown angle as the reference angle, label the given sides as opposite, adjacent, or hypotenuse. 2. Fill in the trigonometric ratio with the given information. 3. Use the inverse ratio to solve for the measure of the angle.

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Example 3 Find the value of x to the nearest degree.

A 7

C 25

x B

Apply the idea With respect to the angle with a measure of x, the length of the adjacent side is 7, and the length of the hypotenuse is 25, so we can use the cosine ratio. Use the cosine ratio Substitute the values Apply the inverse ratio Evaluate and round

Reflect and check If we get an error on our calculator, it usually means we haven’t used the right ratio or have substituted incorrectly.

Idea summary To find the measure of an unknown angle of a right-angled triangle, we use a trigonometric ratio to set up an equation relating the angle and two given sides. Then, we can use the inverse ratio to ﬁnd the measure of the angle.

Multi-step problems In some cases, the application of a single trigonometric ratio on a right-angled triangle will not be enough to solve the question.

x cm 43°

In this example, to find the length of h, we must first use our knowledge of trigonometry to find the length of x. Once we find the length of x, we add 70 to x to get the length of h.

h cm 70 m

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13°

16° C 2000 m

Airport

In this example, to find the distance between the two planes, AB, we will need to apply our knowledge of trigonometric ratios to find the distance of the first plane to point C (AC) and then again to find the distance of the second plane to point C (BC). Once we find the distance of both planes to point C, we subtract the distance of the second plane (BC) from the distance of the first plane (AC) to find the distance between the two planes (AB).

Example 4 A girl is ﬂying a kite that is attached to the end of a 23.4 m length of string. The angle between the string and the vertical is 21°. If the girl is holding the string 2.1 m above the ground, ﬁnd the vertical distance from the ground to the place where the string attaches to the kite. Round your answer to two decimal places.

23.4 m 21°

2.1 m

Create a strategy Let’s call the vertical distance from the ground to the kite h. If we could ﬁnd the length of the vertical leg of the triangle, we could calculate h. Let’s call the length of the vertical leg of the triangle x. First, calculate an acute angle in the right-angled triangle to use as a reference angle, then use a trigonometric ratio to ﬁnd the value of x. Then, we can calculate the vertical distance from the kite to the ground.

23.4 m 21°

xm hm

2.1 m

Apply the idea Since the angle between the string and the horizontal is complementary to 21°, the bottom left angle in the triangle must be 69°.

Substitute the values Multiply both sides by 23.4 Evaluate and round

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The vertical distance, h, from the ground to the kite is the sum of the distance from the ground to where the string starts and the vertical distance, x, from the string to the kite. That means h = 2.1 + x, where x = 21.85. So, h = 2.1 + 21.85 = 23.95 The kite is 23.95 m from the ground.

Example 5 Find x and y, rounded to the nearest whole number.

y cm

x cm 64°

Create a strategy Use the tangent ratio to ﬁnd y, then use the cosine ratio to determine x.

Apply the idea With respect to the angle of 64°, the length of the opposite side is 8, and the length of the adjacent side is y, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by y

Divide both sides by tan 64°

Evaluate and round

With respect to the angle of 64°, the length of the adjacent side is 8, and the length of the hypotenuse is x, so we can use the cosine ratio. Use the cosine ratio

Substitute the values

Multiply both sides by x

Divide both sides by cos 64°

Evaluate and round

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Idea summary In multi-step problems, application of a single trigonometric ratio may not be enough to solve the problem. Extra addition, subtraction, or another application of trigonometry may also be needed.

6.02 Practice questions Understanding 1

Explorer 1, 2, 3a–c, 4

Adventurer 3b–d, 4

Trailblazer 3d–f, 4

Name the hypotenuse. a

A B

For the triangle ABC: a

Which side is opposite angle θ?

Which side is adjacent to angle θ?

Which side is opposite angle α?

Which side is adjacent to angle α?

Which angle is opposite the hypotenuse?

A α

θ B

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With reference to the angle θ, ﬁnd the ratio: i

iii b

34 5

25 C

Write down the indicated ratios for each triangle using the side labels: a

tan θ

tan α

sin θ

A α

sin α

A α

θ B

cos θ

cos α

A α

θ B

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5a, 7−8, 10−12, 14, 16, 17a–c

Fluency Ex 1

5b, 7−8, 10−12, 14, 16, 17b–d

5b, 6−7, 9−11, 13, 15−16, 17c–e

Write down the indicated ratios for each triangle: a

tan θ

sin α

iii

cos θ

tan α

sin θ

cos α

iii

cos θ

sin α

B 17

θ°

α°

Find the value of tan θ.

25p 24p

7 Ex 2

Find the side length, rounded to two decimal places: a

b 29° fm

7 cm

27°

f cm

4 cm

12 cm

44°

25° g cm 16.68 cm a cm

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, ﬁnd the value of b. bm

θ° 12 m

If tan (55°) = , ﬁnd the value of d.

A 2313 m 55° d

Calculate the size of angle θ.

Ex 3

For each triangle, ﬁnd the value of: i

θ 6

θ 3

If d is the distance between the base of the wall and the base of the ladder, ﬁnd the value of d, rounded to two decimal places.

1.88 m

38° dm

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A ladder is leaning at an angle of 44° against a 1.36 m high wall. Find the length of the ladder, rounded to two decimal places. 1.36 m

44°

A sand pile has base angles of 40° and is 10.6 m wide. Find the height of the sand pile, h, rounded to one decimal place.

h cm 40° 10.6 m

A string of lights of length 23.9 m is joined to the top of a tree, as shown in the diagram. If the angle that the string of lights makes with the ground is θ °, ﬁnd θ rounded to two decimal places.

23.9 m 4.1 m

θ°

16 Ex 4

There is a 39° angle from the top of the NJ building, which is 300 m high, to the top of the LS building. The diagonal distance between the tops of the two buildings is 200 m. a

Find x, rounded to two decimal places.

Find the height of the LS building, rounded to two decimal places.

200 m 39°

300 m

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17 Ex 5

Find the side lengths, rounded to the nearest whole number: a

54°

16 cm 42°

26 mm

f cm h mm

x cm

32 cm 52°

x cm

58° h cm

11 cm

82 cm

45 cm

A x

Reasoning 18

An isosceles triangle has equal side lengths of 10 cm and a base of 8 cm as shown. Explain how to calculate the size of angle A, rounded to one decimal place.

10 cm

A°

A° 8 cm

If tan θ = 0.4, explain how to ﬁnd the value of b. b cm

θ°

21.32 cm 7.9 cm

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Problem-solving 20−22

20−23

21−24

, ﬁnd the value of cos θ and tan θ, without calculating the size of angle θ.

, ﬁnd the value of cos θ and sin θ, without calculating the size of angle θ.

Consider the octagon:

Use trigonometry to calculate the size of each internal angle rounded to the nearest whole number.

Conﬁrm your answer to part (a) using the formula to ﬁnd an internal angle of a polygon with n sides:

10 cm

12 cm

A sporting ground is in the shape of an irregular hexagon. a

Find the perimeter of the ground.

Find the size of each of the four corner angles of the ground.

35 m 15 m

41 m

15 m 35 m

Calculate the size of angle DEC.

6 x+7 A

6.03

12 − x C 4 B

ngles of elevation and A depression

After this lesson, you will be able to… • define angles of elevation and depression and understand their real-world applications. • model the relative position of two objects using a right-angled triangle. • identify and apply the three distances (horizontal, vertical, and direct) between two objects when given the angle of elevation or depression. • solve problems involving angles of elevation and depression using trigonometric ratios. 266

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Angles of elevation and depression An angle of elevation is the angle from the lower object to the higher one. The angle of elevation from point A to B is the angle between the horizontal line at A and the line connecting the two points.

θ A

An angle of depression is the angle from the higher object to the lower one. B

The angle of depression from point B to A is the angle between the horizontal line at B and the line connecting the two points.

Both angles are measured with respect to the horizontal.

Exploration Rectangle ABCD has a diagonal drawn on it.

1. What is the value of x? What property did you use to ﬁnd this? Is there another way to conﬁrm this?

x 50°

2. What is the angle of elevation from point C? 3. What is the angle of depression from point A?

50°

4. What do you notice about the angle of elevation and depression?

Notice that the angle of elevation between two points will always be equal to the angle of depression between those two points. This is because they are alternate angles on parallel lines (since all horizontal planes are parallel to each other).

θ θ A

When given the angle of elevation or depression between two objects, we will always be able to model their relative position using a right-angled triangle. B

Dir

ect

dis

θ Dir ect

tan

θ Horizontal

Horizontal

dis

tan

6.03 Angles of elevation and depression mathspace.co

Vertical

B Vertical

There are three distances between two objects: horizontal distance, vertical distance, and direct distance. These will represent the adjacent, opposite, and hypotenuse sides respectively. If any two of these are given, then we can ﬁnd the angle of elevation or depression.

267

Example 1 Find the angle of depression from point B to point D. Use x as the angle of depression and round your answer to two decimal places.

9 A

Create a strategy Identify where the angle of depression is and apply the appropriate trigonometric ratio.

Apply the idea The angle of depression is the angle ∠CBD, with adjacent side BC = 9 and hypotenuse BD = 19. We can let x be the unknown angle. Use the cosine ratio

Substitute the values

Apply inverse cosine on both sides

Evaluate using a calculator and round

The angle of depression from point B to point D is 61.73°.

Example 2 A fighter jet, flying at an altitude of 4000 m is approaching a target. At a particular time, the pilot measures the angle of depression to the target to be 13°. After a minute, the pilot measures the angle of depression again and finds it to be 16°.

13° B

16°

4000 m

Airport

a Find the distance AC, to the nearest metre.

Create a strategy Use a trigonometric ratio that relates the 13° and the opposite.

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Apply the idea With respect to the given angle 13°, the opposite side is 4000, and the adjacent side is AC, so we can use the tangent ratio. Use the tangent ratio

Substitute the values

Multiply both sides by AC

Divide both sides by tan 13°

Evaluate using a calculator and round

The distance AC is 17 326 m. b Find the distance BC, to the nearest metre.

Apply the idea With respect to the given angle 16°, the opposite side is 4000, and the adjacent side is BC, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by BC

Divide both sides by tan 16°

Evaluate using a calculator and round

The distance BC is 13 950 m. c Now ﬁnd the distance covered by the jet in that one minute.

Apply the idea The distance covered by the jet in one minute is given by the length of AB. AB = 17 326 − 13 950

Subtract BC from AC

= 3376 m

Simplify the difference

The distance covered by the jet is 3376 m.

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Example 3 A room is rectangular and measures 5 metres in length and 4 metres in width. The angle of elevation from the bottom left corner to the top right corner of the room is 57°. a Draw a diagram of the room.

Create a strategy

Apply the idea

To draw a diagram of the room, we need to keep in mind that the room is rectangular in shape and the given dimensions are in metres. Additionally, the angle of elevation from one corner of the room to the other is given, which means we are considering a 3D model.

h 57°

b Find d, the distance from one corner of the ﬂoor to the opposite corner of the ﬂoor. Leave your answer in surd form.

Create a strategy Apply Pythagoras’ theorem, c2 = a2 + b2.

Apply the idea d2 = 52 + 42 2

d = 41 d=

Substitute the values

Evaluate the right-hand side Take the positive square root of both sides 5

The distance from one corner of the floor to the opposite corner of the floor is m.

c Find h, the height of the room. Give your answer to two decimal places.

Create a strategy Set up an equation involving the height, h, the diagonal on the ﬂoor, , and the angle of elevation, 57°.

h 57° 5

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Apply the idea Use the tangent ratio

Substitute the values

57°

Multiply both sides by Evaluate using a calculator and round The height of the room is 9.86 m.

Idea summary We can solve problems involving angles of elevation and depression using these steps: 1. Draw a diagram if one was not provided. 2. Label all known information on the diagram, including the right angle. 3. Identify which angles or sides need to be found and in what order. 4. Set up the correct trigonometric ratio and solve for the missing side or angle. 5. Give your answer as a sentence.

Angle of elevation

Horizontal

Angle of depression

6.03 Practice questions Understanding 1

1−3

Explorer

1−3

Adventurer

1−3

Trailblazer

Is angle θ an angle of elevation or depression? Explain how you know.

4 2

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A man standing at the top of the tower at point B is looking at the ground at point C. a

Identify the angle of depression in the ﬁgure.

Identify the angle of elevation in the ﬁgure.

What can be said about the size of the two angles?

θ A

Using the given diagram:

Identify the angle of elevation.

How could you draw an angle of depression from A to C?

Which angle in the diagram is the same size as the angle of depression from A to C?

θ σ

Fluency 4

4−7, 9, 10

4−10

5−11

x is the angle of elevation to point A from point C. a

Find x, rounded to two decimal places.

Find y, rounded to two decimal places.

y x C

5 Ex 1

x is the angle of depression from point B to point C. a

Find y, rounded to two decimal places.

Find x, rounded to two decimal places.

25 B

D y

x 5

The person in the picture sights a pigeon above him. Find the angle of elevation, θ, rounded to two decimal places. 30 m

θ 23 m

A helicopter is 344 m away from its landing pad. If the angle of depression to the landing pad is 32°, ﬁnd x, the height of the helicopter above the ground, rounded to the nearest metre.

32° 344 m

xm Landing pad

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The ﬁnal approach of an aeroplane when landing requires the pilot to adjust the angle of depression to 3°, as shown in the diagram. If the plane is 12 m above the runway and has d m until touchdown, ﬁnd d rounded to the nearest metre.

3° 12 m

10 Ex 2

11 Ex 3

A man stands at point A looking at the top of two poles. Pole 1 has a height of 8 m and an angle of elevation of 34° from point A. Pole 2 has a height of 25 m and an angle of elevation of 57° from point A. a

Find the distance from A to B, rounded to two decimal places.

Find the distance from A to C, rounded to two decimal places.

Find BC, the distance between the two poles in metres. Round your answer to one decimal place.

A fighter jet, flying at an altitude of 2000 m is approaching an airport. The pilot measures the angle of depression to the airport to be 12°. One minute later, the pilot measures the angle of depression again and finds it to be 19°. a

Find the distance AC, rounded to the nearest metre.

Find the distance BC, rounded to the nearest metre.

Find the distance covered by the jet in that one minute.

A room measures 6 m in length and 5 m in width. The angle of elevation from the bottom left corner to the top right corner of the room is 60°. Rounding all answers to two decimal places:

Pole 2

25 m

Pole 1 8m A

12° B

19°

C 2000 m

Airport

h 60°

Find the distance from one corner of the ﬂoor to the opposite corner of the ﬂoor.

Find the height of the room.

Find the angle of elevation from the bottom corner of the 6 m long wall to the opposite top corner of the wall.

Find the angle of depression from the top corner of the 5 m long wall to the opposite bottom corner of the wall.

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Reasoning 12

12−13

13−14

An air traffic controller is communicating with a plane, which is flying at an angle of elevation of θ ° from the control tower. The plane is 10 369 m above the ground and is still 23 444 m from the runway. Viggo claims that the angle of depression that the pilot should fly is equal to 90° − θ °. Explain why Viggo is incorrect and correctly calculate the angle of depression that must be flown.

10 369 m

θ° 23 444 m

Fiona says that the angle θ is not an angle of elevation as it is not measured from the ground. Is Fiona correct? Explain your answer.

27 m

18 m

The angle of elevation from the tip of a tree’s shadow to the sun is 43°. The tree’s shadow is 20 m long. a

Early in the morning and late in the afternoon, when the angle of elevation of the sun is low, what happens to the length of the shadow?

What happens to the length of the shadow when the sun is directly overhead?

Find tan 90° on your calculator. How does this relate to your answer to part (b)?

Problem-solving 15, 17−21 15

15, 18−21, 23

43° 20 m

16, 20, 22−23

Fred is on a ship and observes a lighthouse on a cliff in the distance. The base of the cliff is 651 m away from the ship, and the angle of elevation to the top of the lighthouse from Fred is 35°. a

If the top of the lighthouse is x m above sea level, ﬁnd x, rounded to two decimal places.

If the lighthouse is 30 metres tall, how tall is the cliff that the lighthouse stands on?

Round your answer to two decimal places.

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Buzz is standing 250 m from a building and measures the angle of elevation to the top of the building to be 27°. a

If the difference in height between the top of the building and Buzz’s eye is h m, ﬁnd h rounded to two decimal places.

If Buzz’s eye is 172 cm from the ground, ﬁnd the height of the building, rounded to one decimal place.

From the top of a rocky ledge that is 274 m high, the angle of depression to a boat is 15°. If the boat is d, m from the foot of the cliff, ﬁnd d, rounded to two decimal places.

Amelia measures the angle of elevation to the top of a tree from a point 29 m away from the base to be 31°. Find the height of the tree, h, rounded to the nearest metre.

The angle of elevation from an observer to the top of a tree is 18°. If the distance between the tree and the observer is d m and the tree is known to be 3.53 m high, ﬁnd the value of d, rounded to two decimal places.

At a certain time of the day a light post, 6 m tall, has a shadow of 5.8 m. If the angle of elevation of the sun at that time is θ °, ﬁnd θ, rounded to two decimal places.

A ramp of length 311 cm needs to ascend at an angle between 10° and 20° for it to be safe to use.

311 cm

If the height of the ramp is 152 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

If the height of the ramp is 100 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

If the height of the ramp is 25 cm, ﬁnd the angle the ramp makes with the ground, rounded to two decimal places.

At which height(s) is the ramp safe?

height

Two hot air balloons are floating in different locations, and each is tied to the ground directly below. An observer at each location measures the angle of elevation to the opposite balloon. The observers are 1600 m apart as shown in the diagram: a

Calculate the difference in height between the two balloons. Round your answer to the nearest metre.

Find the direct distance between the two balloons. Round your answer to the nearest metre.

3.38°

1.22° 1600 m

Roald is standing at point P and observes two poles, AB and A CD, of different heights. P, B, and D are on horizontal ground:

From P, the angles of elevation to the top of the poles at A and C are 29° and 18° respectively. Roald is 16 m from the base of pole AB. The height of pole CD is 7 m. a

Calculate the distance from Roald to the top of pole CD, B rounded to two decimal places.

Calculate the distance from Roald to the top of pole AB, rounded to two decimal places.

29° 16 m

45°

18°

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6.04

Bearings

After this lesson, you will be able to… • use the starting direction and acute angle to describe the location of a point. • write bearings using mathematical notation. • apply bearings in 3D problems by drawing relevant triangles and using trigonometric ratios. • draw a diagram of a 3D situation, find distances and calculate angles related to bearings.

Compass bearings In surveying and navigation, bearings are used to identify the location of an object. The four main directions of a compass are known as cardinal directions. They are north (N), east (E), south (S), and west (W).

N NW

NNW

NNE

WNW

ENE

WSW

ESE

SSW

SSE

In between each of these cardinal directions are intermediate directions of northeast (NE), southeast (SE), southwest (SW), and northwest (NW) that are 45° from the two cardinal directions closest. There are smaller divisions shown on the compass as well.

S A compass bearing describes the location of a point using: • The starting direction of either north or south; • The acute angle needed to rotate • The direction to rotate, east or west.

Starting direction

Direction of rotation

Angle rotated by

To describe the position of point B from A we would say: “Start at south, then rotate 53° towards east.”

North

We can write this mathematically as: S 53°E West

East 53°

South

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Interactive exploration

Discover this concept in action online

mathspace.co

Example 1

Consider the point A.

35° O

a For the compass bearing of point A from O, which option shows the correct letters that should be written either side of the angle? A

N⬚E

Apply the idea

N⬚W

S⬚E

S⬚W

Since A is closer to north, we start at north and rotate clockwise towards the east direction. So the letters that should be written are in option A, N ⬚ E. b Complete the compass bearing of point A from O. N⬚E

Create a strategy

Apply the idea

Write the angle measure shown in the ﬁgure. The angle given is measured from the north towards the east, therefore we can write in the angle given.

N 35°E

Idea summary Starting direction

Direction of rotation

Angle rotated by

A compass bearing describes the location of a point using: • The starting direction of either north or south • The acute angle needed to rotate • The direction to rotate, east or west.

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Three-figure (true) bearings A three-ﬁgure bearing is: • Measured from north (N) • Measured in a clockwise direction • Written using three ﬁgures T is often, but not always, used to describe a true bearing. If the angle measure is less than 100° it would be written as 040° or 040° T. To use true bearings to describe the location of a plane at point B from the airport at point A: • Place the centre of a compass on the starting point, in this case the airport. • Starting at north, rotate clockwise until we get to the line AB. • Write the angle as the true bearing of point B. The true bearing of B from A is 127° or 127° T.

North

West

127° East B

South

The diagram describes the bearing of P from O. Rotating clockwise from north, we get an angle of 55°.

North

West

55°

P East

Since this measure is less than three digits, we put 0 in front so the true bearing of P is 055°.

South

For the true bearing of O from P, the true bearing would be 180 + 55 = 235°.

North

West O

55°

East 180°

South

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Example 2 Determine the true bearing of point A from point O. a

35° O

Create a strategy

Apply the idea

Measure the bearing clockwise from north, and write it using three digits.

The angle between the north direction and the line OA measured clockwise is 35°. So the true bearing is 035° T.

42° O

Apply the idea The angle between the north direction and the line OA measured clockwise is 360 − 42 = 318°. So the true bearing is 318° T.

Example 3 Consider the following diagram: A

B 42°

52°

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a Find the true bearing of point A from point C.

Create a strategy We can find the angle between AC and north. Then, we can subtract that from 360°.

Apply the idea

52°

To find the angle that AC makes with true north, we can extend the north line from point C to segment AB. This will create a right-angled triangle.

Recall the acute angles in a right-angled triangle sum to 90°, so the angle that AC makes with true north is 90 - 42 = 48°.

B 42° 48°

Starting from point C and rotating clockwise from north gives us an angle of 360 - 48 = 312°. This means the true bearing is 312° T.

b Find the compass bearing of point A from point C.

Apply the idea

Reflect and check

From C to A it is north and west which means it is N 48°W.

Notice going from A to C would be the opposite direction and would be S 48°E.

Example 4 The position of a ship S is given to be 20 kilometres from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates. Find the ship’s coordinates to one decimal place.

N x y

049° 20 km

Create a strategy For the x-coordinate, use the sine ratio given by sin θ = use the cosine ratio given by cos θ =

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, and for the y-coordinate,

Apply the idea In the diagram, we are given θ = 49°, opposite = x, and hypotenuse = 20. Substitute the values Multiply both sides by 20 Evaluate and round In the diagram, we are given θ = 49°, adjacent = y, and hypotenuse = 20. Substitute the values Multiply both sides by 20 Evaluate and round The ship’s coordinates are (15.1, 13.1).

Idea summary True bearings are: • Measured from north (N) • Measured in a clockwise direction

• Written using three ﬁgures • Usually written with a T at the end

Bearings in 3D A problem in three dimensions may use bearings to describe the situation. It is helpful to ﬁrst draw a diagram if one is not given to us. Then, we can solve as usual for problems in 3D, by drawing out triangles in 2D.

Example 5 A miner is mining 500 m below the surface. If you walk 300 m north and then 400 m west from the oﬃce, you will be directly above the miner. a Draw a diagram of the situation in 3D and the related triangles in 2D.

Create a strategy We can start by drawing the surface using the bearings and then placing the miner below the surface. It is helpful to label the vertices.

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Apply the idea The whole scenario can be represented in a 3D diagram like the one shown.

N 400 m W

N 300 m Office

500 m

We can break this into two right-angled triangles which have a common side of WO. W

400 m

300 m 500 m O M

b Find the distance from the oﬃce to the point on the surface directly above the miner.

Apply the idea For triangle ONW, we know two sides and have one missing side, let’s use Pythagoras’ theorem to ﬁnd OW. Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify

The distance from the oﬃce to the point on the surface above the miner is 500 m.

Reflect and check We may have noticed this is the Pythagorean triple, 3, 4, 5 multiplied by 100.

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c Find the direct distance from the miner to the Oﬃce, to the nearest metre.

Apply the idea For triangle OWM, we know two sides and have one missing side, so we can use Pythagoras’ theorem to ﬁnd OM. Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify the surd

Evaluate using a calculator and round

The direct distance from the oﬃce to the miner is 707 m.

Reflect and check If we use this rounded value instead of the exact value of introduced.

, there will be some error

d At what angle of depression from the oﬃce would you dig to reach the miner?

Create a strategy We are now trying to ﬁnd ∠WOM. Since this is a right-angled triangle, we can use trigonometric ratios to ﬁnd this angle.

Apply the idea We know all the side lengths in this triangle, so we can use any pair. However, one of them is rounded, so it is best to use the two exact values.

500 m

θ 500 m

707 m

The sides with exact values are opposite and adjacent to the angle q. Use the tangent ratio

Substitute known values

Use inverse tan ratio

Evaluate using a calculator

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Reflect and check Instead of using trigonometric ratios, we could have recognised that this is an isosceles right-angled triangle, so both acute angles would be 45°.

Idea summary Although bearings refer to directions on the surface, we may use them in 3D problems where a given point is above or below ground. We can draw relevant triangles in 2D to solve one problem at a time.

6.04 Practice questions Understanding 1

1−4

Explorer

Adventurer

1−4

Trailblazer

Write the compass bearing for point P from point O. a

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Write the true bearing of P from point O. a

N 26° W P

N P

N 64° W

P S 26° W

S 64° W

Which of these diagrams correctly displays WNW? A

N NW

WNW

SE S

N NW

WNW E

WNW

What is the equivalent true bearing for these compass bearings? a

Southwest

Northwest

West-Northwest

South-Southeast

Northeast

Southeast

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5, 6, 8, 9, 11−13, 15, 18−20

Fluency Ex 1, 2

5, 6, 8, 9, 11−13, 16, 18−20

Find the true bearing of point A from O.

Find the compass bearing of point A from O.

N A

5, 7, 8, 10, 14, 16, 17, 20

30° O

41° A

A 40° O

O 35° A

Determine the true bearing for: a

Point B from point A.

Point C from point A.

Point C from point B.

B 37°

48°

Determine the true bearing for: a

Point A from point B.

Point A from point C.

Point B from point C.

B 47°

38°

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Ex 3

Find the true bearing of point A from point C.

Find the compass bearing of point A from point C.

B 49°

39°

Find the true bearing of point B from point C.

Find the compass bearing of point B from point C.

B 37°

45°

Find the true bearing of point C from point A.

Find the compass bearing of point C from point A.

B 42°

53°

Find the true bearing of point C from point B.

Find the compass bearing of point C from point B.

B 38°

49°

If a person walks at a compass bearing of SW, and then turns around to go back in the direction they came, what direction are they heading now?

An aeroplane sets off on a bearing of 051° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

A boat sets off on a bearing of 125° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

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If the bearing of the clearing from the town is a°, ﬁnd a rounded to the nearest degree.

N 5.9 km 3.2 km

Clearing

a°

O Town

Luke sailed for 116 km on a bearing of 231°.

Find w, the horizontal distance in kilometres he has sailed from his starting point, rounded to one decimal place. w 231° 116 km

Shortly after take-off, a plane is 42 km south and 57 km west of the airport in Sydney that it left from. a

Find the size of the angle marked b, rounded to one decimal place.

Find the compass bearing of the plane from the airport.

What is the true bearing of the plane from the airport?

O b°

Plane

18 Ex 4

288

42 km

57 km

The position of a ship S is given to be 20 km from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates.

Find the ship’s x-coordinate rounded to one decimal place.

Find the ship’s y-coordinate rounded to one decimal place.

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Sydney

049°

20 km

20 Ex 5

On an orienteering course, Beth runs 1000 metres north from point A to point B, then turns east and runs to point C. If the true bearing of C from A is 062° T and the direct distance Beth must run to get back to point A is d metres, ﬁnd d to the nearest metre.

N B 1000 m

C 062° d m

From a point 15 m due north of a tower, the angle of elevation to the top of the tower is 32°. a

Find the height of the tower h. Round your answer to two decimal places.

Find θ, the angle of elevation to the top of the tower from a point 20 m due east of the tower. Round your answer to the nearest degree.

North

32° hm 15 m

20 m

East

Reasoning 21

21a–c

21b–d, 22

21cd, 22, 23

Given that a, b, c and d are angles between 0° and 90°: a

If a compass bearing is N a°W, write an expression for the true bearing.

If a compass bearing is S b°W, write an expression for the true bearing.

If a compass bearing is S c°E, write an expression for the true bearing.

If a compass bearing is S (90 − a) °E, write an expression for the true bearing.

Explain why the overall horizontal distance from A to C is equal to 12 cos 45° + 16 cos 30°.

Write an expression using sin which will calculate the overall vertical distance from A to C.

Use Pythagoras’ theorem to calculate the direct distance from A to C.

N 45° 12 km

Find the bearing of A from C.

Find the direct distance from A to C.

Find the bearing of A from C.

N 60°

16 km

N N

B 72°

145°

600 m 800 m

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Problem-solving 24, 25, 27, 29, 30

24−27, 30

24, 26−28, 31, 32

During a group hike, a hiker walked 11 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 11 km NW of the base camp. What compass bearing must the hiker have followed during the night?

During a rescue search, a helicopter ﬂew west from point X to point Y, then changed course and ﬂew 10.7 km north to point Z. If point Z is on a bearing of 335° T from point X:

Find the size of ∠Y XZ.

Find b, the distance from point Y to point X, rounded to one decimal place.

Calculate d, the distance the helicopter must ﬂy between point Z and point X, rounded to one decimal place.

N Z

In remote locations, photographers must track their position from their base. One morning a photographer sets out from base, B, to the edge of an ice shelf, S, on a bearing of 055°. She then walked on a bearing of 145° to point P, which is 916 m due east of base.

b c

290

335°

145° S B

∠BAC

N B

∠DCE

Find the distance the boat is from its starting point as a straight line from A to E. Find θ °, the true bearing of the boat from its starting point, to the nearest degree.

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916 m

Find the size of the angles: i

A ﬁshing boat in search of large schools of ﬁsh sails from point A for 26 km on a bearing of 205° to point C. It then sails a further 24 km on a bearing of 226° to point E. Rounding all answers to one decimal place: a

X b km

Rounding all answers to one decimal place: a State the size of ∠BSP. b Find the distance BS. c Find the distance SP. d If the photographer walks directly back to her base from point P, determine the total distance travelled. 27

d km

10.7 km

A drone travels due east for 2.2 km and then travels on a bearing of S 32°E for 5.9 km. a

Given that the compass bearing is a°, write the bearing needed to return to the starting point, in terms Start of a.

Find x, the distance between the end point and the starting point of the drone’s ﬂight. Round your answer to two decimal places.

Find a rounded to the nearest degree.

N 2.2 km

x km

32° 5.9 km

A yacht sailed in a direction so that its ﬁnal position was 248 km west and 225 km south of its starting point. a

Draw a diagram to represent the situation.

Find the true bearing on which the yacht sailed, rounded to one decimal place.

Find the total distance the boat has sailed, rounded to one decimal place.

Three television presenters are practicing their navigation skills before heading off on an expedition to a remote location. Belinda, at point B, is positioned 17.6 m south of Amelia at point A. Carl, at point C, is due east of Belinda and on a bearing of S 38°E from Amelia. a

Draw a diagram to represent the situation.

If Amelia and Carl are d m apart, ﬁnd d to one decimal place.

A yacht travelled a distance of 30 km from port A at a bearing of 110°, then another 40 km to port C at a bearing of 130°. What is the distance between ports A and C?

A and B are two positions on level ground. From an advertising balloon at a vertical height of 740 m, A is observed at a bearing of 043° and B at a bearing of 133°. The angles of depression of A and B, as viewed from the balloon, are 30° and 20° respectively. Rounding all answers to the nearest metre: a

Find the distance from point A to the point directly below the advertising balloon.

Find the distance from point B to the point directly below the advertising balloon.

Find the distance between points A and B.

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6 Chapter review 1

Find the length of the unknown side as an exact value: a 17

A right-angled triangle has a hypotenuse of length 26 and another side of length 24.

A right-angled triangle has a one side of length 6 and another side of length 5.

Could a triangle with side lengths of 6 mm, 8 mm, and 13 mm be a right-angled triangle? Explain.

Rachel hikes south of her starting point for 834 m, then turns and walks 691 m east, before stopping for a lunch break. She then travels south again for 427 m before arriving at her ﬁnal destination.

What is the shortest distance between where Rachel started and where she stopped for lunch? Round your answer to two decimal places.

What is the shortest distance between where Rachel started and where she ﬁnished her journey? Round your answer to two decimal places.

Find the value of f, rounded to two decimal places: a

b 27°

10 cm

f cm

29° 5m

For △DEF, ﬁnd the exact value of : a

sin θ

θ 13 F

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Jason is ﬂying a kite that is attached to the end of a 25.2 m length of string. The angle between the string and the vertical is 40°. If Jason is holding the string 1.2 m above the ground: a

Find the angle of elevation from Jason to the kite.

Find x, rounded to two decimal places.

Find the height h of the kite above the ground, rounded to two decimal places.

25.2 m

40° hm

1.2 m

Find the value of the side length, rounded to the nearest whole number: a

20 cm

58° 23 mm

h mm

67°

x cm

10 cm

The person in the picture sights a pigeon above him. Find the angle of elevation, θ, to two decimal places.

24 m

θ 17 m

An aeroplane is ﬂying at an altitude of 3000 m and approaching an airport. The pilot measures the angle of depression to the airport to be 14°. One minute later, the pilot measures the angle of depression again and ﬁnds it to be 16°. a

Find the distance AC, to the nearest metre.

Find the distance BC, to the nearest metre.

Find the distance covered by the jet in that one minute, to the nearest metre.

14° B

16°

3000 m

Airport

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At a certain time of the day a light post, 5 m tall, has a shadow of 9.4 m. If the angle of elevation of the sun at that time is θ°, ﬁnd θ rounded to two decimal places.

Dave is standing 270 m from a building and measures the angle of elevation of the top of the building to be 19°.

If the difference in height between the top of the building and Dave’s eye is h m, ﬁnd h rounded to two decimal places.

If Dave’s eye is 161 cm from the ground, ﬁnd the height of the building rounded to one decimal place.

Find the true bearing of point A from O.

Find the compass bearing of point A from O.

N A

35° O

O 38° A

Find the true bearing of point A from point C.

Find the compass bearing of point A from point C.

B 38°

49°

If a person walks at a compass bearing of NE, and then turns around to go back in the direction they came, what direction are they heading now?

A boat sets off on a bearing of 121° T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

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Uri sailed for 155 km on a bearing of 215°.

Find w, the distance in kilometres he has sailed west from his starting point, rounded to one decimal place. w 215° 155 km

During a group hike, a hiker walked 12 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 12 km NW of the base camp. What compass bearing must the hiker have followed during the night?

Did you know?

Trigonometry plays a key role in medical imaging, like CT scans and MRIs! These machines use trigonometric calculations to create detailed cross-sectional images of the body. By measuring angles and distances within the scans, trigonometry helps doctors see precise internal views, making it easier to diagnose and treat patients accurately. Without trigonometry, it would be challenging to create such clear, layered images of tissues and organs. Thanks to this powerful tool, medical professionals can detect issues early and provide more targeted treatments.

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Big ideas • Physical objects can be modelled with 3D geometric figures whose properties can be applied to solve real-world problems. • Estimation and accuracy are fundamental concepts for practical decision-making and interpreting measurements across diverse real-world contexts. • Scale factors and ratios provide a framework for comparing and relating measurements, dimensions and volumes across different scales and contexts.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10M01 solve problems involving the surface area and volume of composite objects using appropriate units VC2M10M04 use mathematical modelling to solve practical problems involving direct and inverse proportion and scaling of objects; formulate problems and interpret solutions in terms of the situation, including the impact of measurement errors on the accuracy of results; evaluate and modify models as necessary, and report assumptions, methods and findings VC2M10N01 recognise the effect of using approximations of real numbers in repeated calculations and compare the results when using exact representations VC2M10AM01 solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids

Measurement Chapter outline 7.01 7.02 7.03 7.04

7.05 7.06 7.07

Surface area of prisms and cylinders Volume of prisms and cylinders Surface area and volume of composite solids Investigation: Rainfall collection Estimation and accuracy Investigation: Instruments and methods Investigation: Scale models Optional: Spheres Optional: Cones Optional: Pyramids Chapter 7 review

298 308 318 331 335 344 347

349

If you unrolled a tennis ball, how big do you think its surface area would be? Roughly the same as a playing card, a piece of toast, or this textbook?

7.01 Surface area of prisms and cylinders After this lesson, you will be able to… • find the surface area of right prisms. • find the surface area of cylinders. • solve problems involving the surface area of right prisms and cylinders.

Surface area of right prisms Surface area is the total area of the surface of a solid. To ﬁnd the surface area of a prism, identify the shapes of the different faces, and then add the area of all faces.

Example 1 Consider the rectangular prism with a width, length and height of 5 m, 7 m and 15 m respectively.

5m 7m

Find the surface area. 15 m

Create a strategy We need to add the areas of the 3 pairs of equal rectangular faces: front and back, left and right, top, and bottom.

Apply the idea Front and back = 2 × (15 × 5) 2

= 150 m

Sides = 2 × (15 × 7) 2

= 210 m

Top and bottom = 2 × (7 × 5) 2

= 70 m

Evaluate Multiply the area by 2 Evaluate Multiply the area by 2 Evaluate

Surface area = 150 + 210 + 70 2

= 430 m

298

Multiply the area by 2

Add the areas Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 2 The dimensions of a trapezoidal prism are given in the diagram (not to scale):

10 cm 6 cm

6 cm

5 cm 8 cm

14 cm

Find the surface area of the trapezoidal prism.

Create a strategy A trapezoidal prism consists of two trapezium faces and four rectangular faces.

The total surface area of the prism is the sum of the areas of all the faces.

Apply the idea Surface area = 112 + 2 × 48 + 80 + 2 ×

Area of four rectangles then two trapeziums

= 112 + 96 + 80 + 24 × 5 cm2

Evaluate the brackets and product

= 112 + 96 + 200 cm2

Evaluate the product

= 408 cm

Evaluate

Idea summary To ﬁnd the surface area of a prism, identify the shapes of the different faces, then add the areas of the faces.

Surface area of cylinders A cylinder has a curved face so it’s useful to remember what the net of a cylinder looks like to work out the surface area. A cylinder has two ends in the shape of a circle, and a rectangular piece that creates the curved surface. The total surface area (SA) of a cylinder can be expressed as: SA = 2π r2 + 2π rh

2π r

where r is the radius and h is the height of the cylinder. 7.01 Surface area of prisms and cylinders mathspace.co

299

Example 3 This cylinder has a height of 74 cm and base radius of 21 cm. Find the total surface area in square centimetres. Round your answer to two decimal places. 74 cm

21 cm

Create a strategy The total surface area of a cylinder is given by the formula 2π r2 + 2π rh, where r is the radius and h is the height of the cylinder. We can substitute the given radius and height into the formula to ﬁnd the total surface area.

Apply the idea Surface area = 2π r2 + 2π rh

Surface area of cylinder

= 2π × 21 + 2π × 21 × 74 = 12 534.95 cm

Substitute the values

Evaluate the expression

Idea summary

Surface area of a cylinder = 2π r2 + 2π rh r is the radius h is the height of the cylinder

7.01 Practice questions Understanding 1

Explorer 1−3

Adventurer 1−3

Trailblazer 1−3

A cube has a side length of x cm. a

How many faces does the cube have?

What is the area of one face?

Write an expression for the surface area of the cube.

x cm

300

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

To ﬁnd the surface area of this pentagonal prism, the area of how many faces must be added together?

For each 3D shape, identify the number and type of 2D shapes that make up the net. a

Cube

Cylinder

Square-based pyramid

Hexagonal prism

4, 6acdf, 7a, 8ab, 10, 11ab, 15a, 17−21

Fluency

4, 6bd–f, 7b, 8cd, 10, 11cd, 12a, 13, 15b, 17, 21, 22

Find the surface area of a cube with side length 5 cm.

Find the side length of a cube that has a surface area of 384 cm2.

Find the surface area:

Ex 1

4, 5, 6c–f, 7b, 8cd, 9, 11d, 12b, 13, 14, 15b, 16, 21, 22

b 9.9 m

12 m 8.7 m

4m 7.5 m

5 cm

4 cm

23 cm 6 cm

4 cm

11 cm

9 cm

10 cm 5 cm

3 cm

5 cm

16 cm

4 cm

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301

For each triangular prism: i

Find the value of x.

10 cm

x cm

4 cm

Find the surface area.

16 cm

18 cm 12 cm

3 cm

8 Ex 2

Find the surface area of the prisms. Round answers to two decimal places. a

10 cm 11 cm

5 cm

10 cm

3 cm

5 cm

13 cm

9 cm

4 cm

8 cm

12 m

16 cm

3 cm 17 cm 4m

For this triangular prism:

Find the exact length of the unknown side, k.

Find the exact surface area of the triangular prism.

If the value of k was rounded to the nearest metre before calculating the surface area, by how much would the surface area be overestimated? Round your answer to two decimal places.

10 m

For this rectangular prism: 5

5 5 5 5

5 8

b 302

Determine whether the net matches the rectangular prism.

iii

8 8 8

Find the surface area of the rectangular prism.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7 5 7 5 7 8 7

7 5 7 7

11 Ex 3

Find the surface area of each of these cylinders rounded to two decimal places. a

b 6 cm

93 cm 12 cm

30 cm

9m 8m 10 m

Find the area of the curved rectangular face of these open-ended cylinders, rounded to two decimal places. a

2m 3m

For this cylinder:

Find the circumference of the circular base rounded to four decimal places.

Find the area of the curved face of the cylinder rounded to two decimal places.

4m 2m 4m

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303

Find the height, h, of this closed cylinder if its surface area is 25 547 m2 and the radius of each circular face is 38 m. Round your answer to the nearest whole number.

38 m

For each cylinder: i

Find the value of x, rounded to two decimal places.

Find the total surface area, rounded to two decimal places.

4 cm

xm 23 m

13.6 cm

25 m

The area of the circular face of a cylinder is 8281π m2. The total surface area of the cylinder is 25 662π m2. a

x cm

Find the radius of the cylinder.

Find the height of the cylinder.

A cylinder has a height of 20 cm and a diameter of 12 cm. a

Find the exact surface area of the cylinder.

If the value of π was rounded to 3.1 before calculating the surface area, by how much would the surface area be overestimated? Give your answer rounded to two decimal places.

A birthday gift is placed inside the box shown, which has the shape of a rectangular prism.

12 cm 14 cm

Find the minimum amount of wrapping paper needed to cover this box. 5 cm

The roof of a shed is constructed from two rectangular and two triangular sheets of steel, forming a triangular prism with dimensions as shown in the diagram: Find the surface area of the roof of the shed (do not include the underneath rectangular face of the prism). Give your answer rounded to the nearest square metre.

304

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10 m 6m

Laura is building a storage chest in the shape of a rectangular prism. The chest will be 55 cm long, 41 cm deep, and 39 cm high. Find the surface area of the chest.

Find the surface area of the brickwork for this cylindrical silo. Assume that there is a brick roof but no ﬂoor. Give your answer rounded to two decimal places.

2.24 m

6.68 m

A swimming pool has the shape of a trapezoidal prism 14 m long and 6 m wide. The depth of the water ranges from 1.2 m to 2.5 m, as shown in the diagram. The walls and base of the pool are going to be tiled. Calculate the total area inside the pool that is to be tiled. Give your answer rounded to one decimal place.

Reasoning 23

23−25

14 m

1.2 m

2.5 m

25−27

Here are three formulas: • SA = 2π r2 + 2π rh

• SA = π r2 + 2π rh

• SA = 2π rh

How do you know which formula to use to calculate the surface area of a cylinder? 24

Scott uses the formula SA = 2π r(r + h) to ﬁnd the surface area of a closed cylinder. Lachlan uses the formula SA = 2π r2 + 2π rh. Who is correct?

Quicker Oatmeal Company is choosing between two new packages for their chocolate chip oats. • Option 1 is a canister with a diameter of 6 cm and a height of 10 cm. • Option 2 is a box with dimensions 3 cm × 6.5 cm × 9.5 cm.

Which package should Quicker Oatmeal use in order to minimise the amount of materials used? Explain.

No matter which package Quicker Oats chooses, they will sell it for $4.99 per package. Which package should Quicker Oats use to maximise their proﬁts? Explain.

For the composite shape:

2 cm

List the faces and their dimensions that make up the total surface area.

3 cm

Calculate the total surface area of the composite shape.

3 cm

10 cm

7.01 Surface area of prisms and cylinders mathspace.co

305

Write a simpliﬁed formula to ﬁnd the surface area of a cylinder that has its radius equal to its height.

Problem-solving 28, 30, 33, 35 28

28, 29, 31, 33−35

Find the surface area of the triangular prism, rounded to one decimal place.

29, 31, 32, 34−36

h cm

21 cm 46 cm

16 cm

The Super Cereal Company sells their Frosty O’s cereal in a rectangular box with dimensions 12 cm × 10 cm × 2.5 cm. In order to reduce their environmental impact, Super Cereal wants to reduce the surface area of their boxes while maintaining the proportions of the current packaging. a

List three different scale factors that Super Cereal could use to reduce the total surface area of Frosty O’s boxes by somewhere between 15 cm2 and 25 cm2.

Give the dimensions for a Frosty O’s box that satisﬁes the required reduction in surface area.

A paint roller is cylindrical in shape and has a diameter of 6.8 cm and a width of 31.2 cm.

31.2 cm

Find the area painted by the roller when it makes 300 revolutions. Give your answer in square metres rounded to two decimal places.

306

6.8 cm

This water trough in the shape of a half cylinder is to be painted on the outside with two coats of paint. a

Find the surface area of the outside rounded to two decimal places.

If paint costs $24.50 for a 1 L tin that covers 6.5 m2, ﬁnd the cost to paint the trough.

0.93 m

Both popcorn bags are designed to carry 1680 cm3 of popcorn. Assume there’s minimal wastage of space when the popcorn is packed into each bag. a

Find the height of bag A in centimetres.

Find the height of bag B in centimetres.

Find the surface area of bag A.

Find the surface area of bag B.

Which bag should be used to reduce the amount of paper?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2.88 m

Bag A

7 cm

Bag B

8 cm 10 cm

10 cm

Find the surface area of the trapezoidal prism, rounded to one decimal place.

y cm

3 cm

10 cm

8 cm 13 cm 34

The square prism has a volume of 138.72 cm3 and the side length of the square base is 3.4 cm. Calculate the surface area of the prism.

Jenny wants to make several cans like the one shown. She plans to cut them out of a sheet of material that has an area of 1683 cm2. Determine the number of complete cans she can make.

5 cm

7 cm

If two identical spherical balls with radii of 2.6 cm ﬁt exactly inside a cylinder, ﬁnd the surface area of the closed cylinder. Give your answer rounded to one decimal place.

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7.02 Volume of prisms and cylinders After this lesson, you will be able to… • apply the formula to find the volume of various prisms. • apply the formula to calculate the volume of cylinders. • solve problems involving the volume of prisms and cylinders.

Volume of prisms We can ﬁnd the volume of any prism by multiplying the base area by the height of the prism. The height of the prism is the distance between the two base faces. It’s common to use the word height to represent this dimension as we can always reorient the shape. The formula used to ﬁnd the volume of a prism is given by V = Ah, where A is the base area and h is the height of the prism.

Interactive exploration

Discover this concept in action online

mathspace.co

Example 1 Find the volume of this triangular prism.

5 cm 10 cm 5 cm

Create a strategy

Apply the idea

Use the volume formula V = Ah, where the area of the base is given by

We are given base length, b = 5, height of the triangular base, H = 5; and height of prism, h = 10. Write the formula

308

Substitute the values

Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 2 A prism has a volume of 1080 cm3. If it has a base area of 120 cm2, ﬁnd the height of the prism.

Create a strategy

Apply the idea

Use the volume of a prism formula V = Ah.

We are given V = 1080 and A = 120. Write the formula Substitute the values Divide both sides by 120 Evaluate

Idea summary The volume of a prism is given by:

V = Ah A h

is the base area is the height of the prism

Volume of cylinders We can ﬁnd the volume of a cylinder using the same method, by multiplying the area of the circular base by the height of the cylinder. The formula for the area of a circle is A = π r2, therefore the formula for the volume of a cylinder is V = π r2h, where r is the radius and h is the height of the cylinder.

Interactive exploration

Discover this concept in action online

mathspace.co

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309

Example 3 Calculate the volume of the quarter cylinder. Round your answer to one decimal place.

14 cm

5 cm

Create a strategy Multiply the volume of a cylinder formula by a quarter.

Apply the idea We are given r = 5 and h = 14. Multiply the formula by Substitute the values Evaluate

Idea summary The volume of a cylinder is given by:

V = π r2h r h

310

is the radius of the cylinder is the height of the cylinder

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7.02 Practice questions Understanding

Explorer 1−3

Adventurer

Trailblazer

1−3

Are these statements true or false? a The volume of a prism or cylinder is calculated by multiplying the area of its base by its height. b The volume of a triangular prism can be found by multiplying the base area by one third of the height. c The volume of a cylinder can be determined by multiplying the area of its circular base by its height. d The volume of a rectangular prism can be calculated by multiplying the length, width and height together. e The volume of a cylinder is always greater than the volume of a rectangular prism with the same base area and height.

Explain the similarities and differences in ﬁnding the volume of a rectangular prism and the volume of a cylinder.

Which of these prisms would have a volume that would be found by calculating 4 × 3 × 5? A

B 4 cm 3 cm

5 cm

4 cm

D 5 cm

5 cm 4 cm 3 cm

3 cm

4 cm

7.02 Volume of prisms and cylinders mathspace.co

311

Fluency 4

4, 5ace, 7, 8a–e, 9, 11, 12

What is the shape of the prism’s base?

What is the area of the prism’s base?

iii

What is the volume of the prism?

5ace, 7, 8b–f, 9, 11, 12

5bdf, 6, 7, 8e–h, 10, 11

b 15 cm

10 cm

6 cm 5 cm 18 cm 6 cm

3 cm

5 Ex 1

Find the volume of these prisms: a

A = 38 cm2

A = 52 cm2 10 cm

3 cm

7 cm

11 cm

3 cm

4 cm

3 cm

14 cm 5 cm

10 cm 12 cm

6 cm 312

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

A prism has a volume of 880 cm3. If it has a height of 11 cm, ﬁnd the base area of the prism.

A prism has a volume of 1200 cm3. If it has a base area of 240 cm2, ﬁnd the height of the prism.

Ex 2

8 Ex 3

Find the volume of these solids, rounding your answers to one decimal place. a

5 cm

10 cm

5 cm

17 cm

6 cm 8m 12 cm

3 cm

10 cm 13 cm 3 cm

16 cm 7 cm 19 cm

8 cm

7.02 Volume of prisms and cylinders mathspace.co

313

Find the volume of each of the cylinders. Round your answers to one decimal place. a

A cylinder with a radius of 6 cm and height of 15 cm.

A cylinder with a radius of 7 cm and height of 15 cm.

A cylinder with a diameter of 2 cm and height of 19 cm.

Find the volume of the ﬁgure shown, rounded to two decimal places.

5 cm

15 cm

A cylindrical spa has a radius of 1.8 m and the water is 160 cm deep. Find the volume of water in the spa in litres. Round your answer to two decimal places.

A cylindrical candle has a radius of 46 mm and a height of 0.13 m. Find the volume of the candle in cubic centimetres. Round your answer to two decimal places.

Reasoning 13

13, 14

13−15

Oliver is helping his father build a wooden toy box in the shape of a rectangular prism for his younger brother. The planned dimensions for the box are 1.8 m in length, 1.1 m in width, and 0.9 m in height. They need to calculate the volume of the box to know how much space it will take up. a

Calculate the volume of the box using the exact dimensions.

Oliver’s father prefers to work with rounded dimensions to make the measurements easier. He decides to round each measurement to the nearest metre. Calculate the volume of the box using these rounded dimensions.

Compare the two volumes. Which is greater, and by how much?

A right prism has a rectangular base and three dimensions: length, width and height. How does the volume of a right prism change if we double the length and width while keeping the height the same?

Edrian’s science class is conducting an experiment that involves ﬁlling a cylindrical container with water. The container has a radius of 2.5 m and a height of 1.9 m. The students need to know the exact volume of the container to make sure they have enough water for the experiment.

314

Calculate the volume of the container using the exact value of π and the given dimensions rounded to two decimal places.

One of the students argues that using 3.14 for π will make the calculations easier. Calculate the volume of the container using 3.14 as the approximation for π rounded to two decimal places.

Compare the volumes calculated with the exact value of pi and the approximation. Which is greater, and by how much?

Discuss the potential implications of using rounded or approximated values in these real world situations.

Determine the capacity of the container in millilitres.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Problem-solving 16, 17, 20, 23−26 16

16−18, 21, 22, 24, 26, 27

19, 25−30

A rectangular swimming pool has a length of 27 m, width of 14 m and depth of 3 m. Find the capacity of the swimming pool in litres, if 1 m3 = 1000 litres.

An ice cream store has two different types of cones. One type uses a pointy cone while the other type has a ﬂat base, made up of two cylinders. Both types have a hemispherical scoop on top and are ﬁlled with ice cream. Both cones have the same thickness. Types of ice cream cone Point base cone

Flat base cone 6 cm

3 cm

3 cm 1 cm

8 cm

6 cm

4 cm

Vidar wants to have as much ice cream as possible. Determine which type he should get.

Yolanda wants to have as much cone as possible. Determine which type she should get.

There are two types of cylindrical soup cans available for Ray to purchase at his local store. One has a diameter of 16 cm and a height of 18 cm while the other has a diameter of 18 cm and a height of 16 cm. Which type of can holds more soup?

Sanford Seminole Aquatic Center has a swimming pool that is 25 metres by 55 metres, and 2.5 metres deep. Morcom Aquatic Center has a swimming pool with a volume of 6187.50 cubic metres.

Which pool has the greater volume?

How many times bigger is the volume of the bigger pool than the smaller pool? Round your answer to one decimal place.

Find the exact scale factor between the pool at Morcom and the pool at Sanford Seminole.

Jack’s mother told him to drink 3 large bottles of water each day. She gave him a cylindrical bottle with height 17 cm and radius 5 cm. a

Find the volume of the bottle. Round your answer to two decimal places.

Assuming he drinks 3 full bottles as his mother suggested, calculate the volume of water Jack drinks each day. Round your answer to two decimal places.

If Jack follows this drinking routine for a week, how many litres of water would he drink altogether? Round your answer to the nearest litre.

5 cm

17 cm

7.02 Volume of prisms and cylinders mathspace.co

315

Find the capacity of the ﬁsh tank in litres: 50 cm 30 cm

25 cm

A cylindrical swimming pool has a diameter of 5 m and a depth of 1.8 m. a

How many litres of water can the pool contain? Round your answer to the nearest litre.

Express this amount of water in kilolitres.

Find the area of the trapezium-shaped block of land.

During a heavy storm, 63 mm of rain fell over the block of land. Find the capacity of water that landed on the property in litres.

10 m 17 m

25 m

The roof of a cottage has the shape of a triangular prism. Find the volume of the roof.

10 m

A ﬁsh tank designed to sit in the corner of a room has the shape of a right-angled triangular prism. Determine the capacity of the ﬁsh tank in litres.

50 cm 30 cm 90 cm

316

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Find the volume of the tent in cubic metres:

150 cm 290 cm 200 cm

The roof of a house is flat and rectangular, measuring 16 m by 11 m. a

Find the area of the roof.

Find the volume that could be collected from 20 mm of rain.

How many litres of water is equivalent to the volume collected in part (b)?

Once the water tank is full, a further 5 mm of rain falls on the roof. How many litres of rainfall are not collected as a result of the tank overflowing?

The outline of a rice ﬁeld is shown. a

Find the area of the rice ﬁeld.

Throughout the monsoon season, the rice ﬁeld receives 16 cm of rain. What volume of water, in cubic metres, has fallen on the rice ﬁeld?

35 m 36 m 52 m

45 m 16 m 80 m

The outline of a construction site for the foundations of a large oﬃce building is shown. a

Find the area of the construction site.

Prolonged rainstorms bring 0.21 m of rain to the region and the foundations of the building are ﬂooded. Determine the volume of water of the construction site, in cubic metres.

85 m 85 m

85 m

13 m 84 m

The ﬂoor plan for a house is shown, with measurements given in metres. The annual rainfall over the house is 700 mm. Calculate the potential amount of water that could fall on the house over one year, if 1 m3 holds 1000 litres. Give your answer in kilolitres.

3.00 m

4.00 m 2.60 m

3.20 m

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7.03 Surface area and volume of composite solids After this lesson, you will be able to… • recognise and define composite solids. • calculate the volume of composite solids by adding or subtracting individual volumes. • identify and compute the surface area of composite solids by summing the areas of their faces. • solve problems related to the volume and surface area of composite solids.

Volume of composite solids We can ﬁnd the volume of composite solids by ﬁnding the volume of each of the solids and adding them or subtracting them. To find the volume of this composite solid we can subtract the volume of a simple solid. This composite solid is a rectangular prism with half a cylinder subtracted from it.

If a composite solid is also a prism with a composite shape as its base, we can find its volume the same way we’d find the volume of a prism - by multiplying the area of the base by the height of the prism.

Example 1 This solid consists of a rectangular prism with a smaller rectangular prism cut out of it. Find the volume of the solid. 13 cm 2 cm

6 cm

11 cm

15 cm

Create a strategy Subtract the volume of the smaller rectangular prism from the volume of the larger rectangular prism.

318

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Apply the idea Large prism = 6 × 15 × 13 3

= 1170 cm

Small prism = 11 × 2 × 15 3

= 330 cm

V = 1170 − 330 3

= 840 cm

Multiply the dimensions Evaluate Multiply the dimensions Evaluate Subtract the volume of one prism from the other Evaluate

Example 2 A wedding cake with three tiers rests on a table. The layers have radii of 50 cm, 54 cm and 58 cm as shown in the ﬁgure. If each layer is 21 cm high, calculate the total volume of the cake in cubic 21 cm metres. Round your answer to two decimal places. 21 cm 21 cm

r1 = 50 cm r2 = 54 cm r3 = 58 cm

r2 r3

Create a strategy Add the three volumes and convert the units to cubic metres.

Apply the idea Add the volumes Substitute the radii and heights

Evaluate

To convert to cubic metres we need to divide by 100 . Convert to m3

Evaluate and round

7.03 Surface area and volume of composite solids mathspace.co

319

Example 3 Find the volume of the ﬁgure shown, rounded to two decimal places:

10 m 4m

Create a strategy Add the volume of the square prism to the volume of the half cylinder.

Apply the idea The half cylinder has a diameter of 4 m which means its radius is 2 m. Write the formula Substitute r = 2, h = 10 Evaluate and round Write the formula Substitute s = 4, h = 10 Evaluate Add the volumes Evaluate

Reflect and check Since the composite solid was made by adding simple solids together, its volume is the sum of the volumes of the simple solids.

Idea summary Some composite solids are made by joining simple solids together. We can find the volume of these composite solids by finding the volume of each of the pieces and adding them. In the case where the composite solid is made by subtracting one simple solid from another, we can subtract volumes to find the volume of the composite solid.

320

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Surface area of composite solids The surface area of a composite solid is the sum of the areas of its faces. If the composite solid is a prism, we can find its surface area in the same way that we’d find the surface area of a prism. If the composite solid is not a prism, we can find the surface area by considering the areas of each face of the solid and adding them up.

Example 4 Find the surface area of the house. 3m

6m 9m 5m

Create a strategy Add all the faces together. The diagram shows which faces are equal.

Use Pythagoras’ theorem, c2 = a2 + b2, to ﬁnd the width of the roof panels.

Apply the idea Add the areas and multiply by 2 Substitute the dimensions Evaluate Side walls = 9 × 6 × 2 Multiply the length and width and multiply by 2 = 108 m2 Base = 5 × 9 2

= 45 m

Evaluate Multiply the length and width Evaluate

7.03 Surface area and volume of composite solids mathspace.co

321

Using Pythagoras’ theorem, we can find the width of the roof panels. On the diagram, we can see two right-angled triangles where the shorter sides are a = 3 and

Use Pythagoras’ theorem Evaluate Square root both sides Find the area of the 2 roof sides Evaluate Add all the areas Evaluate

Idea summary The surface area of a composite solid is the sum of the areas of its faces.

Did you know?

Measurement accuracy has transformed sports! In events like swimming and sprinting, times are measured down to the millisecond, making it possible to determine winners by the tiniest margins. Technologies like laser timers and high-speed cameras ensure that every race is as fair and precise as possible. This level of precision also helps athletes improve by tracking their progress in minute detail, pushing the boundaries of human performance. Measurement advancements not only add excitement for fans but also give athletes the data they need to break records and set new standards. 322

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7.03 Practice questions Explorer 1, 2, 3a–e, 4ad

Understanding

Adventurer 1, 2, 3a–e, 4ab

What is a composite solid? Give an example.

Explain what you would do to:

Calculate the surface area of a composite shape.

Calculate the volume of a composite shape.

Trailblazer 1, 2, 3c–f, 4bc

Austin is testing out his new 3D printer by creating some geometric objects. What basic shapes are used to make each of these composite solids? a

5 cm

7 cm

3 cm

8 cm 2 cm

7 cm

8 cm

7 cm

5 cm

4 cm

10 cm

2 cm

4 mm 4 mm

7 cm

3 mm

4 mm

5 cm

7 mm

4 km

5 mm 4 mm 3 mm 5 mm

8 km

7.03 Surface area and volume of composite solids mathspace.co

323

Identify the solids used to make up each of these composite solids. a

b 4 cm 7 cm

15 m 7m

11 cm

8 cm

9 mm 16 mm

11 cm

10 mm

6 cm 4 cm

8 cm

Fluency 5 Ex 1

53 mm

36 mm

5, 6ad, 7, 8a–e, 9a–d

5, 6bd, 7, 8b–f, 8, 9c–f

5bc, 6bc, 8d–h, 9d–h, 10

Find the volume of these composite solids, rounded to two decimal places. a

8 cm

12 cm 3 cm

3 cm

4 cm

10 cm

7 cm 15 cm 3 cm

10 cm

12 cm 20 cm 7 cm 8 cm

5 cm

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16 cm

Find the volume of these composite solids, rounded to two decimal places. a

4 cm 7 cm

b 11 cm

8 cm

8 cm 15 cm

5 cm

9 cm

4 cm 2 cm 9 cm

12 cm 2 cm

7 Ex 2

2 cm

A wedding cake has three tiers. The layers have radii of 51 cm, 55 cm and 59 cm. If each layer is 20 cm high, calculate the total volume of the cake in cubic metres. Round your answer to two decimal places.

51 cm 20 cm

20 cm 55 cm 20 cm

59 cm

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8 Ex 3

Find the volume of these composite solids, rounded to two decimal places. a

b 15 m

10 m 8m

12 m

19.5 m

9m 11 cm

8m 6 cm

17 m

8 cm

4 cm

f 6 cm

3 cm

3 cm 12 cm

16 cm 12 m 6m

h 10 cm 14

14 cm 5 cm 24 20

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10 cm

15 cm

9 Ex 4

Find the surface area of these composite solids, rounded to two decimal places. a

7 cm

12 m

16 cm 6m

1 cm

5 cm

13 cm

7 cm

4 cm 10 cm

9 mm

2 cm

16 mm 3 cm

10 mm

3 cm 53 mm

36 mm

10 cm

h 12 cm

1 cm

7 cm

4 cm

15 cm

5 cm

1 cm

By considering the water bottle as a composite solid, estimate its surface area.

2 cm 1 cm

4 cm

20 cm

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Reasoning 11

11, 12

11, 13

Brenton wants to know the volume of the basketball hoop stand at school, including the base, pole, backboard, and hoop. He’s conﬁdent that he can measure the dimensions of the different parts of the stand, but isn’t sure how to break up the volume into simple solids.

Using the image of the basketball hoop stand, identify the simple solids that make up the stand and describe the dimensions that Brenton would need to measure to ﬁnd their volume. 12

A triangular tunnel is made through a rectangular prism.

3 cm 6 cm 3 cm

8 cm

7 cm

ind the volume of the rectangular prism and subtract the volume of the triangular F prism.

Find the area of the shaded face and use Volume of Prism = A × h.

Explain the process of finding the surface area of the solid formed, including the inside of the tunnel.

The larger prism has had two identical holes carved out of it, each of which is a rectangular prism. All measurements are in metres. a

328

Find the volume of the solid formed using two different methods: i

Explain how to find, and then calculate, the volume of the remaining solid. Round your answer to two decimal places. Explain how to find, and then calculate, the surface area of the remaining solid.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

6.5 5

14 6.5

Problem-solving 14ab, 16, 17 14

14ab, 15−17

14, 15, 18−20

A swimming pool is the shape of a trapezoidal prism.

51 m

Find the volume of the pool in m .

If the pool is three quarters full, what is the volume of the unfilled space of the pool?

If the distance of the water level to the top of the pool is h m when it is is three quarters full, ﬁnd h.

1.4 m

10 m

3.3 m 50.7 m

A hollow pipe needs painting on both inside and outside surfaces. a

Find the external surface area of the curved surface.

Find the total surface area of the two end pieces.

Find the internal surface area.

Calculate the total surface area of the pipe.

24 cm

11 cm 10 cm

The lid of this treasure chest is found to be exactly one half of a cylindrical barrel. Find the surface area of the chest. 30 cm

42 cm

59 cm

The diagram shows the design for a marquee (tent). The material to make the marquee costs $44 per square metre.

Find the area of the front of the marquee.

Find the surface area of one of the side walls (not including the roof).

Find the surface area of the entire roof.

Find the total surface area. Do not include the ﬂoor.

What is the total cost of material to make the marquee?

10 m 8m

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A company manufactures nuts shaped like regular hexagonal prisms, with cylindrical bolt holes cut out of the centre.

0.5 cm

1.0 cm 1.2 cm

The total surface area of a nut before the bolt hole is drilled is 14.7 cm2. Find the surface area of a single nut after the bolt hole is drilled out, including the inside area of the hole.

Each nut that is manufactured requires a zinc coating to prevent corrosion. If 1 kg of zinc is enough to coat a surface area of 1 m2, how many complete nuts can be coated with 1 kg of zinc?

A swimming pool is composed of a trapezoidal prism joined to a half cylinder. a

Find the volume of the pool in cubic metres.

How many complete litres of water can fit in the pool?

If the pool is ﬁlled to a height 10 cm below the top, how many complete litres of water are in the pool?

After construction works at a neighbouring property, a crack opens in the bottom of the pool and water begins to leak from the pool. It is observed that the height of the surface of the water in the pool is decreasing by 7 cm each week. How much water is leaking out each week?

Assuming water continues to leak at this rate, at the end of which week will the pool become empty?

0.8 m

A steel shed is to be constructed with a rectangular cut-out at the front for the entrance. a

Determine the surface area of the shed, assuming the floor is separate.

Construction of the shed requires an additional 0.1 m2 of sheet metal for each 1 m2 of surface area, due to overlaps and wastage. How many complete square metres of sheet metal are required to construct the shed?

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7m 3m

2.5 m 10 m 3m 4m

If the steel sheets cost $18 per square metre, calculate the total cost of the steel required to build the shed.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

INVESTIGATION

Rainfall collection Investigate the rainfall that can be saved from a rooftop.

Objectives • To solve problems involving surface area, volume, and design restrictions • To calculate the amount of rainfall that can be collected based on a roof’s surface area • To identify an appropriate tank size and location for water collection and storage

Rooftop and surface area The task You are a farmer in Hahndorf, South Australia constructing a new building and want an optimal rooftop design for rainwater harvesting. Background information Average annual rainfall in the region: 280 mm Assume that 80% of the rain that falls on the roof can be effectively collected (the rest is lost due to evaporation, overflow, or other factors) The footprint of the building is a rectangle measuring 10 metres in length and 7 metres in width Different building materials, with different costs, may be used for the different designs. You have three roof types to choose from:

Flat roof

Gable roof

Round roof

You must consider: 1. The height of the roof (from the top of the walls to the highest point) should not exceed 5 metres for any design to ensure the house remains aesthetically pleasing and structurally sound. 2. The designs should ensure that there is proper water runoff without pooling. Explore five different roof options, including multiple gable style with different slopes. For each option: • Calculate the surface area. • Estimate the volume of rainwater that can be harvested over one year. • Identify some benefits and drawbacks for that option. Investigation: Rainfall collection mathspace.co

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In the online investigation, there is a link that shows how much goes into determining what size of gutters and downspouts should be used. In particular, it talks about what happens if a roof is more or less steep and how that impacts how much water will overflow if the right size gutters aren’t used.

Investigation: Rainfall collection

Investigate online

mathspace.co

For the gable style roof, alter the pitch to find optimal slope. Determine which slope provides the highest surface area and therefore the maximum rainwater collection potential. Discuss whether there’s a point at which increasing the slope stops being beneficial or becomes impractical.

Investigate Answer these questions after completing the procedure. 1.

Describe a roofing option that is very efficient for harvesting rainwater based on your calculations.

Using the linked blog post, determine what size of gutters and downspouts would be necessary to catch as much of the water as possible.

Why might overflow cause more issues than just not catching the water?

Present your findings in a detailed report, including diagrams of each roof shape, calculations, and which option you would choose.

Rainfall collection The task Having optimised the rooftop design for maximum rainwater collection in your country town, you now need to install a water storage tank. Rainwater harvesting isn’t just about collection; it’s also about storage and efficient usage. Property design Shown are three different layouts for a property. Footprint option 1: A rectangular block with a large garden at the back measuring 20 m × 10 m. A three-metre wide path runs along the right boundary fence for access to the garden. Most of the land near the boundary fences is shaded by neighbouring properties.

30 m

20 m

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Footprint option 2: A square block with a backyard pool measuring 8 m × 5 m. Limited space near the house, but more space near the boundary fences. 25 m

25 m

Footprint option 3: An irregularly shaped block with a detached garage measuring 6 m × 4 m and a small garden at the back.

15 m

28 m

22 m

Work within these constraints: 1. Water is used in many different ways and may need to be treated or come from a different source to be able to drink it, especially in cities. 2. Wet season is from November to April (Assume 80% of the annual rainfall occurs in this period). 3. The maximum height of tank if it is placed near the house must be 2.5 metres. 4. The maximum height of tank if it is placed near a boundary fence is 1.8 metres. 5. The wall height of the home is 2.5 metres. 6. The cost of piping is $10 per metre. Water usage A bath: 150–300 litres Shower: 50–100 litres Flushing a toilet: 6–20 litres Cooking and drinking: 15–30 litres Washing dishes by hand: 5–20 litres Dishwasher: 20–40 litres Washing machine: 40–80 litres

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Analyse property constraints and consider storage needs: • For each property footprint, determine potential locations for the rainwater tank. • Estimate the length and cost of piping needed from the roof to the tank’s location for each property. • Calculate the total volume of rainwater that can be collected during the wet season. • Based on the average daily water usage, estimate the total water consumption during the dry season. • Identify the required storage capacity to ensure sufficient water supply throughout the dry season. Design the optimal tank. Using the constraints of maximum height for tanks near the house and near the boundary fence, determine the optimal dimensions (diameter and height) of a cylindrical tank for each property. Consider introducing variations like underground tanks or slimline/oblong tanks for properties with limited space.

Investigate Compile your findings in a detailed report, including diagrams of each property footprint, potential tank locations, calculations, and your recommendations for each property.

Discussion

334

Water conservation and management have been practised by First Nations Australians for thousands of years. Research the tools and practices they use and how they have changed over the years. Consider how the arrival of non-Indigenous Australians affected access to water for First Nations Australians. In particular, where do water rights currently stand?

Choose one of these aspects of rain collection to research further and then share your finding with your group: gutter and downspout overflow, water damage from poor drainage, drinking from rainwater tanks, tank shape, tank overflow, cost, or ask your teacher if you have another idea you’d like to explore.

If you do not have a water storage tank at your school, is there an optimal place that one may be installed? Or, if you do have a water storage tank at your school, could its size or placement be optimised? How or how not?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7.04 Estimation and accuracy After this lesson, you will be able to… • explore rounding and how it influences the accuracy of calculations, especially in large datasets or multi-step computations. • differentiate between rounding and truncating numbers. • define the term ‘accuracy’ in the context of measurement and discuss factors that can influence it.

Effects of rounding on accuracy Rounding numbers affects the accuracy of any calculation we do with those numbers. The effect of rounding becomes especially important when dealing with large sets of data or calculations involving multiple steps. Each instance of rounding can introduce a “rounding effect” that can cumulatively impact the ﬁnal result’s accuracy. When banks calculate how much interest to pay customers, it is routine to truncate rather than round. For example, they may calculate that you are due to receive $465.239 in interest. They would not round to $465.24 but would truncate to $465.23. Maintaining high precision in numbers can lead to time-consuming calculations. The minimal decrease in accuracy due to rounding is sometimes an acceptable trade-off for the simplicity it provides.

Example 1 The heights of five plants are measured. The measurements are as follows (in centimetres): Plant 1 2 3 4 5

Height 25.467 cm 30.234 cm 28.789 cm 32.512 cm 29.105 cm

a What is the total height of the plants without rounding or truncation?

Create a strategy Add the given heights of the plants.

Apply the idea 25.467 + 30.234 + 28.789 + 32.512 + 29.105 = 146.107 cm

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b What is the total height truncating to 2 decimal places?

Create a strategy

Apply the idea

Truncate each height to 2 decimal places, then add the truncated values.

Truncate each measurement: 25.467 → 25.46 30.234 → 30.23 28.789 → 28.78 32.512 → 32.51 29.105 → 29.10 Add the truncated heights: 25.46 + 30.23 + 28.78 + 32.51 + 29.10 = 146.08 cm

c What is the total height rounding to 2 decimal places?

Create a strategy

Apply the idea

Round each height to 2 decimal places, then add the rounded values.

Round each measurement: 25.467 → 25.47 30.234 → 30.23 28.789 → 28.79 32.512 → 32.51 29.105 → 29.11 Add the rounded heights: 25.47 + 30.23 + 28.79 + 32.51 + 29.11 = 146.11 cm

d How do the truncated and rounded values compare to the actual values?

Create a strategy Compare the values found from previous parts.

Apply the idea Exact total height: 146.107 cm Total height with truncation: 146.08 cm Total height with rounding: 146.11 cm The truncated value is further away from the actual total height than the rounded value is. Therefore, the rounded total height is a more accurate approximation of the exact total height compared to the truncated total height.

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Idea summary Use the most accurate figures you have for calculations and perform any rounding necessary at the end. Rounding before a calculation can make a significant difference to the answers, especially if: • The rounding is all in the same direction; for example all rounded up. • There are several quantities involved in the calculation. The more rounded quantities that are used in calculations, the larger the potential variances that can arise. Truncation makes small differences to individual accounts, but over long periods of time, the amounts can become significant.

Accuracy of measurement devices We use different devices for measurement, including a ruler for length, a weighing scale for weight, a thermometer for temperature and a clock for time. The accuracy of these devices is important in providing reliable results. Accuracy refers to how close a measured value is to the true or actual value. The quality and design of the measurement device can inﬂuence the accuracy but no measuring device is perfectly accurate, there will always be some degree of uncertainty in every measurement. This uncertainty can be due to factors such as the limitations of the device itself, environmental conditions or even human error in reading the measurements.

Example 2 Lila and Alex are working on a geometry project where they measure various geometric quantities using different tools. Their measurements sometimes lead to slight discrepancies. Calculate the percentage error of Lila’s measurements, to two decimal places, if Alex’s measurements are used as the actual values for the given scenarios. a The perimeter of a rectangular garden where Lila measures the sides as 68.2 m and 42.9 m, while Alex measures them as 68.4 m and 43.0 m.

Create a strategy Calculate the perimeter of the rectangular garden based on Lila and Alex’s measurements using the formula P = 2 × (length + width), then calculate the percentage error between their measurements.

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Apply the idea Perimeter according to Lila: PLila = 2 × (68.2 + 42.9) = 222.2 m

Substitute the values Evaluate

Perimeter according to Alex: PAlex = 2 × (68.4 + 43.0) Substitute the values = 222.8 m

Evaluate

The percentage error is given by:

b The area of the circular pool where Lila measures the radius as 25.8 m and uses 3.14 as an estimate of π. Alex measures the radius as 26.0 m and uses the button π on the calculator.

Create a strategy Calculate the area of the circular pool based on Lila and Alex’s measurements using the formula A = π r2 then calculate the percentage error between their measurements.

Apply the idea Area according to Lila: ALila = 3.14 × (25.8)2 2

= 2090.11 m

Substitute the values Evaluate

Area according to Alex: AAlex = π × (26.0)2 2

= 2123.72 m

Substitute the values Evaluate

The percentage error is given by:

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c The surface area of the rectangular box on which Lila measures the dimensions as follows: length 48.6 cm, width 37.5 cm, and height 29.8 cm. Alex’s measurements are slightly different: length 48.8 cm, width 37.6 cm, and height 29.9 cm.

Create a strategy Calculate the surface area of the circular pool based on Lila and Alex’s measurements using the formula SA = 2(wl + hl + hw) then calculate the percentage error between their measurements.

Apply the idea Surface area according to Lila: SALila = 2 × (37.5 × 48.6 + 29.8 × 48.6 + 29.8 × 37.5)

Substitute the values

= 8776.56 cm

Evaluate

Surface area according to Alex: SAAlex = 2 × (37.6 × 48.8 + 29.9 × 48.8 + 29.9 × 37.6)

Substitute the values

= 8836.48 cm

Evaluate

The percentage error is given by:

d The volume of a cubic container where Lila measures the length, width, and height as 71.4 cm each. Alex measures them as 71.6 cm each. Round your answer to two decimal places.

Create a strategy Calculate the volume of the cubic container based on Lila and Alex’s measurements using the formula V = side3, then calculate the percentage error between their measurements.

Apply the idea Volume according to Lila: VLila = (71.4)3

Substitute the values 3

= 363 994.344 cm

Evaluate

Volume according to Alex: VAlex = (71.6)3

Substitute the values 3

= 367 061.696 cm

Evaluate

The percentage error is given by:

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Idea summary Measuring devices can include rulers, weighing scales, thermometers, clocks, etc. These devices are not always accurate due to: • Limitations of the device • Environmental conditions • Human error

7.04 Practice questions Understanding 1

Explorer 1−3

Adventurer 1−4

Trailblazer 1−5

Are these statements true or false? a

Truncation and rounding errors can signiﬁcantly affect the ﬁnal results in a series of calculations.

In financial calculations, rounding or truncating repeatedly over time has no impact.

The greater the number of decimal places in a calculation, the more accurate the result.

Truncation and rounding are the same things when it comes to calculations involving real numbers.

List all numbers with two decimal places that round to 6.4.

What is the largest number with two decimal places that would round to 5.4?

Three two-digit numbers with two decimal places are first rounded to the nearest tenth and then added together. Describe the size of the error that could result.

What is the smallest number with three decimal places that would give an answer of 12.72 when rounded to two decimal places?

Fluency 6

340

6ab, 7, 8

6−8

Tony is calculating the surface area of the solid shown, where the height and diameter of the cylinder are both 5 cm. a

Find the smallest value that each of the measurements could be.

Find the largest value that each of the measurements could be.

Find the smallest possible surface area for the solid, rounded to the nearest whole number.

Find the largest possible surface area for the solid, rounded to the nearest whole number.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

6−8 5 cm

8 cm

6 cm

The weights of five boxes of fruits are measured (in kilograms).

Ex 1

8 Ex 2

Box

Weight

12.783

15.482

11.926

13.643

14.217

What is the total weight without rounding or truncation?

What is the total weight truncating to 2 decimal places?

What is the total weight rounding to 2 decimal places?

How do the truncated and rounded values compare to the actual values?

Jasper has a ruler that is marked in cm. When he measures something with it, he looks carefully to see whether the object is less than halfway between two marks, in which case he rounds down, or more than halfway between two marks, in which case he rounds up. When measuring the rectangle shown, he would record the length as 4 cm. 1

Betty has a ruler that is marked in both cm and mm, so that she can get lengths in cm to one decimal place. Calculate the percentage error, to one decimal place, in Jasper’s measurement compared to Betty’s measurement for: a

The perimeter of a rectangle which Betty measures as being 5.3 cm long and 4.4 cm wide.

The area of a circle where Betty measures the radius as 4.3 cm, given that Betty uses the button π on the calculator and Jasper uses 3.1 as an estimate of π.

The surface area of a rectangular prism which Betty measures as 7.8 cm long, 5.7 cm wide and 4.1 cm high.

The volume of the rectangular prism in part (c).

Reasoning 9

9, 10

9−11

10−12

A company uses a faulty weighing scale that adds 0.1 kg to every measurement. How might this error inﬂuence their cost calculations if they are dealing with large quantities of materials?

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Caleb measures the length of the classroom to be 7 metres. Ida says the same length is 6.8 metres and Lavinia gives her answer as 6.85 metres. a

Explain how all three people could have different answers for the same measurement.

Write down the level of accuracy being used by each person.

Are all their answers correct? Explain your reasoning.

A bank truncates interest rates at the third decimal place for daily calculations on a savings account. Over a year, how could this affect the total interest received by a large number of customers?

Alex and Liam are physics students conducting an experiment to measure the kinetic energy of a ball rolling down a ramp. They know the formula for kinetic energy is 1 mv2, where m is 2 the mass and v is the velocity. Alex has an old scale which only measures mass to the nearest 10 g (0.01 kg). Liam has an accurate scale which gives the mass in kilograms to three decimal places. To measure the velocity, Alex has a simple timer that gives the time it takes for the ball to roll a certain distance to the nearest second, whereas Liam has a radar gun that gives the velocity in m/s rounded to two decimal places. In one trial, where the ball rolls 5 metres, Liam measures the mass of the ball as 0.365 kg and the velocity as 2.24 m/s. a b

Find the measurements that Alex will use in his calculations. Calculate the kinetic energy of the ball for: i

Alex

Liam

Using the answers from (b), calculate the percentage error in Alex’s answer.

The students are then told that they can use the same data to calculate the momentum of the ball using the formula p = mv. Calculate the momentum of the ball for: i

342

Alex

Liam

Using the answers from (d), calculate the percentage error in Alex’s answer.

What do you notice about the answers to parts (c) and (e)? Explain why this is the case.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Problem-solving 13 13

13, 14

13−15

Keepsafe Family Bank pays a small amount of interest on their customers’ accounts when the accounts are in credit (the account holder has money in the bank). The interest is added to the account daily, based on the balance in the account at the end of the day. Before they add the interest to the account, they truncate the amount, removing anything less than one cent, so if they calculated the interest to be $2.5387, the customer would receive $2.53, and the bank keeps $0.0087 or 0.87 cents. If the bank has 5 million customers with accounts in credit and the average amount truncated is $0.005, how much money does the bank keep from these interest payments in a standard 365-day year?

The dating of artefacts by Argon-Argon suggests that those found in volcanic deposits at a site called Tower Hill dated from 36.8 ± 3.8 thousand years ago. a

Find the earliest date (the longest time ago) that the artefacts can be from, measured in thousands of years.

Find the latest date (the shortest time ago) that the artefacts can be from, measured in thousands of years.

Express 3.8 as a percentage of 36.8, rounded to one decimal place.

Prior to 1999, the estimated age of the universe was 13.5 Byr ± 6.5 Byr (Byr means billion years). a

The absolute error compared to 13.5 Byr is 6.5 Byr. Find the percentage error, rounded to two decimal places.

Due to improvements in astronomical data, the current estimate of the age of the universe is 13.5 Byr, with an absolute error of just 200 million years. Find the percentage error in the current estimate, rounded to two decimal places.

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INVESTIGATION

Instruments and methods Investigate the effect of instruments, methods, and truncation on real-world scenarios.

Objectives • To compare measurements using different tools • To understand the impact of rounding and truncation on calculations • To describe settings where measurement errors may impact research results

Scales for a recipe Anzac biscuits We will analyse the instruments used to measure ingredients in a recipe for Anzac biscuits. Originally known as “soldiers’ biscuits”, and made by Australian and New Zealand families and sent to their soldiers abroad during World War I because their ingredients didn’t spoil easily. Today, these oat-based treats commemorate Anzac Day (25 April), honouring the Australian and New Zealand Army Corps (ANZAC) who served in the Gallipoli Campaign. Anzac biscuit ingredients | Serves 30 135 g rolled oats 225 g plain flour 225 g sugar 127.5 g desiccated coconut 60 mL golden syrup 1.5 teaspoons bicarbonate of soda 3 tablespoons boiling water Predict with a classmate how sufficient a bathroom scale or a kitchen scale would be for the measurements of oats, flour, sugar, or coconut. If possible, measure out the ingredients using: A bathroom scale A kitchen scale

Investigate After completing the procedure, answer these questions.

344

Discuss the differences in the ingredients for Anzac biscuits when using a bathroom scale or a kitchen scale.

Do any ingredients in the recipe require less precision? Is there a way that we could adjust the recipe so that a bathroom scale may be used for some of the ingredients? Explain.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Effects of rounding Producing Anzac biscuits Let’s consider a large company that produces Anzac biscuits. Cylindrical mixing tanks may be used in the process for combining ingredients. These tanks alone are just one component of a much larger and more complex industrial baking system. V = π r2h Use the formula for the volume of a cylinder in different ways to calculate the amount of ingredients that will fit in 8 mixing tanks that each stand 1.65 metres and are as wide as they are tall. There are a total of 1000 litres in a cubic metre. First, calculate the volume of the mixing tanks using these criteria: • Truncate the measurements of the tank to one decimal place. • Approximate π as 3.14 in your calculation. • Truncate the size of the tank before converting from cubic metres to litres. Second, calculate the volume of the mixing tanks using these criteria: • Approximate π as 3.14 in your calculation. • Round the dimensions of the mixing tank to two decimal places. • Round the size of the tank to two decimal places before converting from cubic metres to litres. Third, calculate the volume of the mixing tanks using these criteria: • Use π in your calculation. • Use exact values and then rounding after converting from cubic metres to litres.

Investigate Answer these questions after completing the procedure. 1.

Compare the use of truncation to the use of rounding or using exact values in your calculations.

Why do these differences matter in the process of producing biscuits in an industrial complex? Explain.

Other measurement instruments Packaging Anzac biscuits A company must include the cost of their product packaging in their expenses. Companies may use technology, data analytics, and logistics strategies to determine the perfect size and material for their products. Suppose that you run a smaller business that sells Anzac biscuits, and you are determining the right packaging for your product.

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Your business will package your homemade Anzac biscuits in a cylindrical container, as shown: A designer whom you have contracted delivers various-sized cylindrical packages for the biscuits at your business. Consider different ways that you might check the size of these packages: 1. You use a pair of calipers to measure the diameter of the cans. 2. You find the circumference of each can using string 3. You use a ruler to measure the diameter

Investigate Answer these questions after completing the procedure. 1.

How do you know if you are measuring the diameter and not a chord (line connecting two points on the circumference)?

What are the benefits and drawbacks of each of the measurement tools and strategies?

If the cost of the container is determined by the surface area of the material, which calculation from the three in the procedure would you use?

Discussion

346

Give an example of real-world scenario that could be impacted by instruments and methods of measuring. Why is the effect on this scenario important?

With the advancement of technology, we often rely on digital tools for measurements. Discuss the pros and cons of this dependence. What non-digital measurement tools have you seen or used? Are there situations where traditional methods might be more reliable?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

INVESTIGATION

Scale models Investigate scale and ratios in scale drawings.

Objectives • To convert between scale measures and actual measurements • To assess whether scale drawings meet design restrictions • To adjust scale drawings to meet design restrictions

Materials •

Ruler

•

Pen or pencil

•

Scientific calculator

Scale models Design your blueprint You have a small business in a city. You will be designing the blueprint of your building and parking lot while following requirements by the National Construction Code (NCC) and Australia Standard. The blueprint must be drawn to scale and meet the given building code. For this project, consider four design elements: doorways, parking spaces, toilet facilities, and electrical outlets and switches. Doorways Number of exits: There must be at least two doorways for exit. Width: Doorways should have a minimum clear opening width of 850mm. Parking spaces Standard parking bay: About 2.4 m wide and 5.4 m long. Accessible parking bay: At least 3.2 m wide with an adjacent shared area of at least 2.4 m wide and 5.4 m long. Toilet facilities Accessible toilet: Minimum dimensions are approximately 900 mm by 2300 mm. Width: Doorways should have a minimum clear opening width of 850 mm. Electrical outlets and switches Distance from water sources: Outlets in areas like bathrooms should be placed at least 300 mm away from water sources. Distance between outlets: There should be an electrical outlet at most 3 m apart throughout the building.

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Choose a type of business to design the building for. Some ideas are: • Bakery • Bike shop • Chemist • Clothing shop • Milk bar • Restaurant Use the requirements for your building and parking lot to design a blueprint. Any furniture or objects added to the blueprint must also be drawn to scale.

Investigate Answer these questions after completing the procedure. 1.

Have you considered accessibility needs in your design beyond the building code? How can you make your building better for everyone?

Trade blueprints with a classmate and analyse their layout, confirming mathematically that they have met all the building requirements and that their blueprint is drawn to scale.

Make any additions or changes to your blueprint after reviewing with a classmate.

Discussion

348

Why do you think each of the building requirements in the investigation may be mandated by NCC?

Research further building requirements through NCC or Australia Standard. How could you improve your blueprint to further meet other codes?

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7 Chapter review 1

Find the surface area: a

16 m 6 cm

d 13 cm

7 cm

5 cm

12 cm

26 cm 8 cm

10 cm

8 cm

Find the height, h, of a closed cylindrical tower given its surface area is 28 000 m2 and the radius of its circular base is 40 m. Round your answer to the nearest whole number.

Fruit and Nut Co. is deciding between two new packaging designs for their raisins. • Design A is a cylinder with a diameter of 10.2 cm and a height of 11.5 cm. • Design B is a rectangular prism with dimensions 4.4 cm × 11 cm × 12.9 cm. Which design should Fruit and Nut Co. select if they want to use the least amount of material for the packaging? Justify your answer.

For the cylinder: 5 cm

15.7 cm

x cm

Find the value of x, rounded to two decimal places.

Find the total surface area, rounded to two decimal places.

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Find the volume of these prisms: a

b 5 cm

A = 35 cm2 10 cm

14 cm 4 cm

17 cm

4 cm

10 cm 8 cm

15 cm 7 cm

A storage box has a volume of 990 cm3. If its height is 9 cm, determine the base area of the storage box.

Find the volume of these solids, rounding your answers to two decimal places: a

11 cm 20 cm 12 cm

8 cm

A city park has a rectangular playground that measures 19 m by 12 m. a

If the playground is to be covered with a layer of sand for safety reasons, how many cubic metres of sand would be needed for a 15 mm thick layer across the entire playground?

If the workers mistakenly spread an additional 3 mm thick layer of sand across the playground, how many extra cubic metres of sand were used?

Find the volume. Round your answers to two decimal places where necessary. a

9 cm

12 cm

11 cm

9 cm

15 cm 5 cm

7 cm

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5 cm

For each ﬁgure, ﬁnd to two decimal places: i

The volume

6 cm

The surface area

9 cm 6 cm

6 cm

A wedding cake has three tiers. The layers have radii of 15 cm, 20 cm and 25 cm. Each layer is 15 cm high. Calculate the total volume of the cake, rounded to two decimal places.

15 cm

Find the amount of yellow icing needed to cover the top tier of the cake, rounded to two decimal places.

15 cm

Find the amount of white icing needed to cover the middle layer of the cake, rounded to two decimal places.

18 cm

6 cm

15 cm

20 cm 15 cm 25 cm

Find the amount of pink icing needed to cover the bottom layer of the cake, rounded to two decimal places.

The diagram shows the design for a shed. a

Find the total surface area of the shed. Do not include the ﬂoor.

4m 2

If paint costs $24 for a 1 L tin that covers 6.5 m , ﬁnd the cost to paint the shed, rounded to the nearest whole dollar.

8m 11 m 9m

Kareem, Alea, and Trina are calculating the volume of the solid shown. Kareem uses 3.14 for pi in his calculations, Alea uses

for pi in her

calculations, and Trina uses the exact value of pi in her calculations. a

Whose estimation of the volume will be most accurate?

Whose estimation with be least accurate?

Find the difference between Alea’s estimation and Trina’s estimation, rounding each answer to three decimal places.

10 cm 4 cm

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An online streaming service tracks the amount of time a person spends watching a video, rounded down to the nearest minute. Over a month, how might this affect the total viewing time calculations for a large number of users?

A bakery uses a timer that is consistently 30 seconds slow. How might this error impact the quality of the baked items?

Before 2005, scientists believed the oldest fossilized plant was 500 Myr ± 25 Myr (Myr means million years). a

Find the oldest age of the plant fossil.

Find the youngest age of the plant fossil.

Given the absolute error in relation to 500 Myr is 25 Myr, determine the percentage error, rounding to two decimal places.

With the discovery of new fossils, the current estimation for the age of the oldest plant fossil is 500 Myr with an absolute error of just 10 million years. Determine the percentage error of this newer estimation, rounded to two decimal places.

Did you know?

The Burj Khalifa, the tallest building in the world, stands at an astonishing height of 828 metres with 163 floors. If scaled down by a factor of 1:100, it would measure 8.28 metres—still taller than a typical 9-storey building, as each storey is usually about 3 metres high. Architects use scale models like this to test designs for wind resistance, structural stability, and aesthetic proportions before construction begins. During construction, engineers had to account for the “stack effect,” where air pressure differences due to height cause a suction force inside the building. Precise measurements and mathematical modelling helped solve this challenge. 352

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“Not everything that can be counted counts, and not everything that counts can be counted.” –Albert Einstein

Big ideas Probability helps us analyse the chance that an event will occur and provides us with tools to make decisions about the future based on known information.

Curriculum Outcomes VC2M10A05 substitute values into formulas to determine an unknown and rearrange formulas to solve for a particular term VC2M10P01 use the language of ‘if … then …’, ‘given’, ‘of’ and ‘knowing that’ to investigate conditional statements and identify common mistakes in interpreting such language, and describe and interpret situations involving conditional probability; design and conduct simulations using digital tools to model conditional probability and interpret results VC2M10P02 describe the results of two- and three-step chance experiments, both with and without replacements, assign probabilities to outcomes and determine probabilities of events; investigate the concept of independence VC2M10AA02 devise and use algorithms and simulations to solve mathematical problems VC2M10AP01 explore counting principles, and factorial notation as a representation that provides efficient counting in multiplicative contexts, including calculations of probabilities

Probability Chapter outline 8.01 Review Venn diagrams and two-way tables 8.02 Independent and dependent events 8.03 Conditional probability 8.04 Mixed probability concepts Investigation: Simulation 8.05 Optional: Geometric probability Chapter 8 review

356 368 382 394 410 413

Scientists use probability methods to determine the likelihood of space debris hitting land.

8.01 Review Venn diagrams and two-way tables After this lesson, you will be able to… • identify different probability notations and use the probability formula to calculate the likelihood of events. • apply the concepts of intersections and unions in solving probability problems. • interpret and construct Venn diagrams and two-way tables. • convert between two-way tables and Venn diagrams to represent information.

Probability notation Recall that probability is the likelihood of an event (set of outcomes) occurring. The sample space is the set of all the outcomes. To calculate the probability of an event occurring, divide the number of favourable outcomes satisfying the event by the total number of outcomes.

The probabilities of all the possible events in the sample space will add to 1. If we have an event, A, the complement of the event is given by A′. This is the set of outcomes that is not in set A. The probability of the complement is found by: P ( A′) = 1 − P ( A) The intersection of two sets is the outcomes both sets have in common. The probability of both A and B occurring is given by P ( A and B) or P ( A ∩ B). The union of two sets is the combination of outcomes in each set. The probability of either A or B occurring is given by P ( A or B) or P ( A ∪ B).

Example 1 In an experiment, a number is chosen randomly from the list of numbers shown: {2, 3, 5, 6, 7, 10, 12, 14, 15, 16, 19, 20} • Event A = an odd number is chosen • Event B = a multiple of 4 is chosen a Which of the following has the largest probability? A

B′

A∪B

A∩B

Create a strategy Since the sample space is the same for both events, compare the number of outcomes of each option and choose the one with the greatest number of outcomes. 356

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Apply the idea B′ = {2, 3, 5, 6, 7, 10, 14, 15, 19}

Numbers that are not multiples of 4

A ∪ B = {3, 5, 7, 12, 15, 16, 19, 20}

Numbers that are odd or multiples of 4

B = {12, 16, 20}

Numbers that are multiples of 4

A ∩ B = {}

No numbers are odd and a multiple of 4

A = {3, 5, 7, 15, 19}

Odd numbers

B′ has the greatest number of outcomes, so the correct answer is option A.

b Which of the following has a value of 0? A

P( A ∩ B)

P( A′ ∪ B)

P( A ∩ B′)

P( A′ ∩ B′)

Create a strategy Find the probability of each option. Use the formula:

Apply the idea There were 12 numbers in the original list, so this will be the denominator in our probabilities. No numbers satisfy A ∩ B Evaluate 7 numbers are not odd or multiples of 4 5 numbers are odd and not multiples of 4 4 numbers are not odd and not multiples of 4 The correct answer is option A.

Idea summary The notation used for concepts in probability are: • • • • •

A — an event. A′ — the complement of event A. P ( A) — the probability of event A occurring. P ( A ∪ B) — the probability that either A or B occurs. P ( A ∩ B) — the probability that both A and B occurs.

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Two-way tables and Venn diagrams A two-way table and a Venn diagram are both ways to present information from two sets. We can easily convert a two-way table into a Venn diagram, or the other way around, by matching their corresponding elements.

A′

B′

Two-way table of events A and B

Venn diagram of events A and B

Exploration Match each Venn diagram to the two-way table with the same highlighted region. 1. Which two models highlight event A? 2. Which two models highlight event A′? 3. Which two models highlight event A ∩ B? 4. Which two models highlight event A ∪ B? A

A′

B B′

Venn 1 A

Table 1 B

A B B′

Venn 2

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Table 2

A′

Exploration (Continued) A

A′

B B′

Venn 3 A

Table 3 B

A′

B B′

Venn 4

Table 4

For events A and B with sample space S, we can use our notation to show different areas of a Venn diagram and the related two-way table. S

The sample space is the list of possible outcomes that could result from the event or experiment. The notation for sample space as a list is S = {…}.

S A

An event is shown as the entirety of a circle.

S A

The complement of an event is shown as the area outside of the event’s circle. The complement of event A can be written as A′, Ac,

or ~ A.

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S A

The intersection of events is the overlap in the two circles. The notation for the intersection of events A and B is A ∩ B.

S A

The union of events A and B is shown by shading anything contained by the two or more event circles. The notation for the union of events A and B is A ∪ B.

Example 2 Students in Irene’s class were asked if they owned a dog and asked if they owned a snake. The following two-way table shows that information. Owns a snake Doesn’t own a snake

Owns a dog 2 13

Doesn’t own a dog 3 11

Construct a Venn diagram that represents the information provided in the two-way table.

Create a strategy The Venn diagram should have two overlapping circles for each type of pet.

Apply the idea • The 2 that own a dog and a snake will go in the intersection. • The 3 that own a snake but not a dog will go in the snake circle outside the intersection. • The 13 that own a dog but not a snake will go in the dog circle outside the intersection. • The 11 that do not own a snake or a dog will go outside the circles. Dog owners

Snake owners

3 11

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Example 3 Two events are deﬁned as: • Event A: Plays guitar • Event B: Plays piano People attending a concert were surveyed and the results are shown: B B′

A A′ 20 40 90 50

a Find the probability that a randomly selected person does not play guitar.

Create a strategy The probability that a randomly selected person does not play guitar is represented by P ( A′). First, we need to ﬁnd the number of people that do not play guitar. Then, we need to divide it by the total number of people that were surveyed.

Apply the idea

Reﬂect and check

The number of people who do not play guitar is: A different way to ﬁnd the answer would be to ﬁnd the probability that a randomly selected 40 + 50 = 90 person does play guitar and subtract that The number of people surveyed is: number from 1. 20 + 90 + 40 + 50 = 200 The probability that a randomly selected person does not play guitar is:

b Find P ( A ∩ B).

Create a strategy P ( A ∩ B) is the probability that a randomly selected person plays both piano and guitar.

Apply the idea The number of people who play both piano and guitar is 20. The number of people surveyed is 200. The probability that a randomly selected person plays both piano and guitar is:

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Example 4 The outcomes for events X and Y are shown in the Venn diagram. X 12

Y 82

a Find P (Y ).

Create a strategy

Apply the idea

The number of outcomes in Y are the outcomes in the intersection plus the outcomes that are only in Y.

The number of outcomes in Y is:

The total number of outcomes is the sum of all the outcomes in the sample space.

The total number of outcomes is:

82 + 41 = 123

53 + 12 + 82 + 41 = 188 The probability of event Y is:

b Find P (X ∪ Y )′

Create a strategy The number of outcomes in X ∪ Y are the outcomes in both circles. The probability of the complement is P (X ∪ Y )′ = 1 − P (X ∪ Y ).

Apply the idea The number of outcomes in event X or Y is: 12 + 82 + 41 = 135 The probability of the union of the events is

The probability of the complement of the union of the events is: Equation of complement

Substitute

Evaluate

Reﬂect and check Notice that the number of outcomes that are neither in X nor Y is 53. This means

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Idea summary We can easily convert a two-way table into a Venn diagram, or the other way around, by matching their corresponding elements.

8.01 Practice questions Understanding 1

Explorer 1, 3−5

Adventurer

Trailblazer

2−5

Two events are deﬁned as: • Event A: It will rain tomorrow. • Event B: There will be a storm tomorrow. Describe what the notation P (B) represents.

For these events: • Event A: Getting an odd number when a die is rolled. • Event B: Getting an even number when a die is rolled. Write the probability notation to describe the probability of event B in three different ways.

For these events: • Event A: A card randomly selected from a deck is red • Event B: A card randomly selected from a deck is a queen Write the probability notation to describe the probability of a randomly selected card being both red and a queen.

170 tennis players were asked whether they would support equal prize money for the women’s and men’s draw.

Support Males Females

Find the number of favourable outcomes for P (Males ∩ Support). 5

A student creates this diagram of their favourite animals: • Event F: Selecting a four legged animal • Event S: Selecting an animal with stripes a

Do not support

Draw a Venn diagram and shade the region that represents the favourable outcomes for the probability P (F ).

Draw a Venn diagram and shade the region that represents the favourable outcomes for the probability P (S′).

List the outcomes in F ∩ S

35 71

Turtle Hippo

Zebra

Lion

Tiger

Clown Fish

Crow

Kangaroo

Elephant

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Fluency Ex 1

6, 9, 10, 12, 13, 15

6, 7, 9, 10, 12−15

6, 8−11, 14, 16

In an experiment, a number is chosen randomly from the list of numbers shown: 1, 3, 5, 7, 9, 10, 15, 20, 25, 30, 35, 40 • Event A: An odd number is chosen. • Event B: A multiple of 5 is chosen. a

Which of these events has the largest probability? A

B′

A∪B

A∩B

P ( A′ ∩ B′)

Determine whether the probability of these events is zero. i

P ( A ∩ B)

P ( A′ ∪ B)

iii

P (( A ∩ B)′)

In an experiment, a card is chosen randomly from a standard deck of 52 cards. • Event A: A face card (King, Queen, or Jack) is chosen. • Event B: A diamond card is chosen. a

Which of these events has the largest probability? A

B′

A∪B

A∩B

P ( A′ ∩ B′)

Determine whether the probability of these events is zero. i

P ( A ∩ B)

P ( A′ ∪ B)

iii

P (( A ∩ B)′)

If a student from the library is randomly selected, we might select a student studying literature (L) or statistics (S). All of the students are studying one of these subjects, and no one is studying both. The probabilities are: • P (L) = 0.4 • P (S) = 0.6 a

Find the value of P (S ∩ L).

For this Venn diagram:

Find the value of P (S ∪ L). Rugby League Rugby Union

179

185

146

Complete the table of values. Play Rugby League Play Rugby Union Don’t play Rugby Union

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Don’t play Rugby League

Ex 2

100 random people in Australia were surveyed, examining their carbon footprint and the city they lived in. The people were then categorised as living in either an urban or a regional location, and whether each person had a high carbon emission or a low carbon emission. High Emission Low Emission Total

Urban 37 24 61

Regional 13 26 39

Total 50 50 100

Construct a Venn diagram that represents the information provided in the two-way table. 11

A vet has 28 pets visit their practice in a day. The pets are categorised based on whether they have been vaccinated and whether they have been microchipped.

Vaccinated

Construct a two-way table based on the Venn diagram.

Microchipped

Ex 3

A healthy living initiative asked people to describe how often they go to the gym.

Male 33 18

How many people were surveyed?

If one person is chosen at random, what is the probability that they are a male who frequently goes to the gym? Give your answer as a fraction.

If one person is chosen at random, what is the probability that they rarely go to the gym? Give your answer as a fraction.

In a study, some people were asked whether they lie. A partially completed two-way table of the results is shown. Two events are deﬁned as: • Event L: The person lies. • Event C: The person is a child.

Female 34 35

Frequently Rarely

Complete the table.

Find P (C′).

L C C′

L′

15 10

Total 25

Total

A group of students were asked about their siblings. The 2 categories show if they have at least 1 brother (B), and if they have at least 1 sister (S): a

Use probability notation to describe the probability that a selected student has at least 1 sibling.

Calculate the probability that a selected student has at least 1 sibling.

Describe what the notation P (S′) represents.

Have a sister Have a brother

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Ex 4

A group of students were asked why they skipped breakfast. The two reasons given were that they were “not hungry” and they were “too busy”. Two events are defined as:

Not hungry

• Event H: Selecting a student that was not hungry • Event B: Selecting a student that was too busy

Find the value of P (H).

Find the value of P (H ∩ B).

Find the value of P (B′).

Ian makes a Venn diagram about the possible pets he could get from the local shelter. Ian is told that a pet will be selected at random.

Find P (F′).

Find P (D ∩ F′).

Reasoning 17

Dogs

• Event D: The pet selected is a dog. • Event F: The pet selected is ﬂuffy. a

Too busy

Fluffy

9 4

17, 18

17−19

For these two events: • A: Paul wins the badminton tournament. • B: Paul wins the golf tournament. Write the probability notation to describe the probability that Paul wins either the badminton tournament or the golf tournament, but not both. Explain your reasoning.

A group of tourists were asked whether they spoke Mandarin (M ) or Spanish (S). a

How many people speak both languages?

If one person is chosen at random, ﬁnd the probability that they speak neither language.

If one person is chosen at random, ﬁnd the probability that they speak only one of the languages. Explain your reasoning.

M M′

S 58 15

In an experiment, there are only two possible events A and B. If A occurs, B does not occur and vice versa. Are these probabilities true? Use a diagram to justify your answers.

366

S′ 10 17

P ( A ∩ B) = 1

P ( A ∩ B) = 0

P ( A′) = P (B)

P ( A ∪ B) = P ( A)

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P ( A ∪ B) = 1

P ( A) = P (B)

Problem-solving 20, 21 20

20−22

This table describes the departures of trains out of a train station for the months of May and June:

21−23

May June

Departed on time 123 124

How many trains departed during May and June?

Find the probability that a train selected at random departed on time and in May. Give your answer as a percentage rounded to one decimal place.

If a train is randomly selected from the trains that departed in June, find the probability that it departed on time. Give your answer as a decimal rounded to three decimal places.

Find the probability that a train selected at random from the 2 months was delayed. Give your answer as a fraction.

In a survey, 150 individuals were asked whether they enjoyed drinking tea and coffee. A partially completed two-way table of the results is shown. The events are defined as: • Event T: The person likes tea. • Event C: The person likes coffee.

If the probability that a randomly selected individual enjoys drinking both tea and coffee is 22

Delayed 32 47

, what is the number of people in set T ∩ C?

C C′

T′ 29 37

Verniece asked some people in her community whether they were pescatarian or not. 29 responders said they were pescatarian, of which 8 were children. 14 children said they were not pescatarian, and 11 adults said they were not pescatarian. a

Construct a model based on the results of Verniece’s survey.

Find the probability that a randomly selected responder is pescatarian.

Find the probability that a randomly selected adult is pescatarian.

In a factory, 45% of the employees only work on the assembly line, 30% only work in quality control, and 20% work both on the assembly line and in quality control. a

Construct a model to represent this data.

An employee is selected at random from the factory. i

ind the probability that the employee has neither worked on the assembly line nor F in quality control.

f the employees that work on the assembly line, ﬁnd the probability that the O selected worker also works in quality control.

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8.02 Independent and dependent events After this lesson, you will be able to… • differentiate between independent and dependent events. • recognise the role of replacement in probability scenarios. • calculate probabilities of events occurring with and without replacement. • apply mathematical formulas to determine the probability of combined events. • analyse real-world examples to identify whether events are independent or dependent. • use tree diagrams to visualise sequences of events and calculate probabilities.

Independent and dependent events Two events are independent if the outcome of each event does not affect the outcome of the other event. For example, landing on heads and landing on tails when tossing a fair coin are independent events because the first toss has no effect on the second toss. If two events are independent, the probability of both events occurring is: P ( A ∩ B) = P ( A) × P (B) The converse statement is also true. This means that if P ( A ∩ B) = P ( A) × P (B), then A and B are independent. Two events that are not independent are called dependent. For example, receiving a first place prize and receiving a second place prize in a race are dependent events. Once someone has won the first place prize, the probability of winning the second place prize will increase for those who are still in the race.

Example 1 A fair standard six-sided die is rolled 691 times. a If it lands on a six 108 times, what is the probability that the next roll will land on a six?

Create a strategy We can use the formula:

Apply the idea The previous results do not affect the outcome of the next roll. There are 6 numbers on a die and 1 of them is a six.

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b Is the outcome of the next roll independent of or dependent on the outcomes of previous rolls?

Apply the idea The outcome of the next roll is not affected by the outcomes of previous rolls. So the outcome of the next roll is independent.

Example 2 Two events A and B are such that P ( A ∩ B) = 0.3 and P ( A) = 0.5. Calculate P (B) if events A and B are independent.

Create a strategy

Apply the idea

Use the fact that if the events A and B are independent, then P ( A) × P (B) = P ( A ∩ B).

Write the equation Substitute the values Divide both sides by 0.5 Evaluate

Idea summary Two events are independent if the outcome of the each event does not affect the outcome of the other event. Two events that are not independent are called dependent. To check if two events are independent you can use: P ( A ∩ B) = P ( A) × P (B)

Replacement

Probability

A probability tree diagram is useful in tracking two events that happen one after the other. At each point, it branches out to all the possible events that could occur from that point. The probabilities of each event is written on the branches. For example, selecting two cards from a deck of cards, one after the other, are events that occur in sequence. There are two ways this can happen: • Draw a card, then put it back in the deck before selecting another card • Draw a card and keep it, then select another card from the deck

H Branch

0.1 Outcome 0.9 T

The ﬁrst method is described as “with replacement” because the card is placed back into the deck. The second method is called “without replacement” because the card is not placed back into the deck.

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Exploration The jar of marbles shown contains 7 red marbles, 3 blue marbles, and 10 green marbles. 1. A red marble is drawn from the jar, then placed back into the jar. How many marbles are in the jar? 2. A green marble is removed from the jar. How many marbles are left in the jar? 3. Suppose one marble is drawn and placed back into the jar, then a second marble is drawn. Let event A represent drawing a red marble and event B represent drawing a green marble in the second draw. Are these events independent or dependent? Explain. 4. Suppose one marble is drawn, then a second marble is drawn. The ﬁrst marble was not replaced. Are events A and B independent or dependent? Explain.

When selecting objects from a group “with replacement”, selections are independent. Each time you select a card, you have the same probabilities. R R

For example, selecting a red card from a standard deck will be

every time if you replace the card.

B R

B B

When selecting objects from a group “without replacement”, selections are dependent. Each time you select a card, you change the probabilities for the next selection. R R

For example, the probability of drawing a red card from a standard deck is

B R

If we draw a red card and do not replace it, there are only 25 red cards left and only 51 cards left in the deck. The probability of selecting a second red card is

. This can be seen in the top branches of the tree

diagram.

B B

This shows the probability of the second event is dependent on what colour card was drawn ﬁrst.

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Example 3 A number game uses a basket with 10 balls, all labelled with numbers from 1 to 10. One ball is drawn at random, then another ball is drawn at random. a What is the probability that the ball labelled 2 is picked once if the balls are drawn with replacement?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea This could happen in two ways: 1. P (2) × P (2′): The ball labelled 2 is picked ﬁrst and a ball not labelled 2 is picked second. 2. P (2′) × P (2): A ball not labelled 2 is picked ﬁrst and the ball labelled 2 is picked second. Find the probability of picking the ball with 2

Find the probability of not picking the ball with 2 Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

b What is the probability that the ball labelled 2 is picked once if the balls are drawn without replacement?

Create a strategy Find the probability for each way that this could happen, then add them together. For the second event, the total number of balls will need to decrease.

Apply the idea Similarly to part (a) this could happen in two ways. However, for the probability of the second ball, the total number of outcomes will be 10 − 1 = 9 since the ball was not replaced. Add the probabilities of the two ways

Substitute the probabilities

Evaluate and simplify

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Example 4 Three marbles are randomly drawn without replacement from a bag containing 6 red, 6 yellow, 4 white, 6 black and 4 green marbles. Find: a The probability of drawing 3 white marbles.

Create a strategy Remember that the number of white marbles and the total number of marbles in the bag decrease by 1 every time a white marble is drawn.

Apply the idea There are 4 white marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

b The probability of drawing 3 black marbles.

Create a strategy Remember that the number of black marbles and the total number of marbles in the bag decrease by 1 every time a black marble is drawn.

Apply the idea There are 6 black marbles and a total of 26 marbles inside the bag. Multiply the probabilities

Evaluate

c The probability of drawing zero green marbles.

Create a strategy Remember that the number of non-green marbles and the total number of marbles in the bag decrease by 1 every time a non-green marble is drawn.

Apply the idea There are 22 non-green marbles inside the bag. Multiply the probabilities

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Evaluate

d The probability of drawing at least one red marble.

Create a strategy Use the fact that P (at least one red marble) is the complement of P (no red marbles).

Apply the idea There is a total of 20 non-red marbles inside the bag. Multiply the probabilities

Evaluate

We now calculate P (at least one red marble): Subtract P (at least one red marble) from 1

Evaluate

Example 5 Bart is purchasing a plane ticket online. He notices there are only 4 seats remaining, 1 of them is a window seat (W) and the other 3 are aisle seats ( A). Bart purchases the ticket first. His friend, Joh, gets on the computer and purchases a ticket immediately after. The seats are randomly allocated at the time of purchase. a Fill in the probabilities matching the edges of the probability tree for the seat Bart receives and the seat his friend receives:

A W

W d

b A

Create a strategy Use the fact that the ﬁrst event is Bart receiving his seat allocation and the second event is Bart’s friend (Joh) receiving a seat allocation.

Apply the idea

There are 3 aisle seats and 1 window seat. This means that when Bart is assigned a seat, the probability of getting a d A

window seat is , and the probability of getting an aisle seat is .

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A 1

d A

A A

If Bart gets a window seat, there are only aisle seats left when Joh purchases his ticket. So the probability that Joh gets an aisle seat is 1.

If Bart gets an aisle seat, there are 2 aisle seats and 1 window seat left when Joh purchases his ticket. W

The probability that Joh gets a window seat is , and the probability that he gets an aisle seat is .

A A

So, the probabilities matching the edges of the probability tree for the seats are , and

b What is the probability that Bart’s friend, Joh, has an aisle seat?

Create a strategy Find the probability for each way that this could happen, then add them together.

Apply the idea A W

A A

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The second set of branches that finish on a node labelled A correspond to Joh having an aisle seat. That is, WA and AA. The probability of each event occurring is found by multiplying previous probabilities along each branch.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

This means that the probability that Joh has an aisle seat is given by:

Evaluate the multiplication Evaluate the addition

c What is the probability of Joh receiving an aisle seat if Bart has a window seat?

Create a strategy

Apply the idea

Use the probabilities found in part (a).

If Bart receives a window seat, only aisle seats are left. So, the probability of Joh receiving an aisle seat is 1.

Idea summary Events “with replacement” occur when the item drawn is placed back into the group before each selection. Each selection is independent of the others. Events “without replacement” occur when the item remains outside the group after selection. Each selection is dependent of the others. The probabilities of each selection will change depending on previous selections.

8.02 Practice questions Understanding 1

Explorer 1−3

Adventurer 1, 2, 4, 5

Trailblazer 1−5

Identify whether these events are independent or dependent: a

A card is randomly selected from a normal deck of cards, and then returned to the deck. The deck is shuﬄed and another card is selected.

Each student is allowed to randomly pick an item from the teacher’s prize bag.

A coin is tossed, the result noted, and then it is tossed again.

Ursula takes a bus to the station and then immediately gets on a train to work. Is the probability of her missing the train independent or dependent on her missing her bus?

Dean has a box that contains ﬁve blue marbles and four red marbles. He takes a marble from the box, then takes a second marble from the box. Determine whether these events are dependent or independent if: a

He replaces the ﬁrst marble before taking the second.

He does not replace the marble before taking the second. 8.02 Independent and dependent events mathspace.co

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A bag contains 3 pink balls and 2 orange balls. You draw one ball, note its colour, then draw another ball without putting the first ball back. a

Find the missing probability for drawing a second pink ball on the tree diagram.

Which outcome can be found by multiplying

Which outcome(s) has a probability of

Are the events of drawing the balls independent or dependent?

P O

by ?

On the island of Guam, the probability that it rains during rainy season is P (R). This is shown in the tree diagram. a

What do the two sets of branches represent?

Does the weather on the first day affect the chance of rain on the second day?

O O

Which two events have the same probability?

R 0.6

Ex 1

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6, 7, 9, 11−13, 15−17, 19, 22

6, 7, 9−13, 15−17, 19, 22

0.6 0.4

0.6

0.4 S

Fluency

0.4

OO OP

PO PP

S R

6, 8−10, 12, 14−16, 18, 20, 21, 23

On a roulette table, a ball can land on one of 18 red or 18 black numbers. a

If it lands on a red number on the ﬁrst go, ﬁnd the probability that it will land on a red number on the second go.

Are the successive events of twice landing on a red number dependent or independent?

From a standard pack of cards, 1 card is randomly drawn and then put back into the pack. A second card is then drawn. a

Find the probability that neither of the cards are spades.

Find the probability that at least 1 of the cards is a spade.

Are the two draws dependent or independent?

Two standard dice are rolled. One is red and one is white. a

Are the results from the red die and the white die dependent or independent?

Calculate the probability that: i

The same number is rolled.

The sum of the two numbers exceeds 9.

iii

The red die is 4 and the white die is 6.

The red die is even and the white die is odd.

The sum of the two numbers is less than 2.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Ex 2

Two events A and B are such that P ( A ∩ B) = 0.1 and P ( A) = 0.2 Calculate P (B) if events A and B are independent.

Two events A and B are such that P ( A ∩ B) = 0.04 and P ( A) = 0.4 If P (B) = 0.1, state whether the events A and B are independent.

A number game uses a basket with 8 balls, all labelled with numbers from 1 to 8. Three balls are drawn at random, with replacement. Find the probability that the ball labelled 4 is picked:

Ex 3

Exactly once

Exactly twice

Exactly three times

Christa randomly selects two cards from a normal deck of cards. Calculate the probability, in each of the following, in the case that the cards are selected:

With replacement

Without replacement

The ﬁrst card is a Queen of diamonds and the second card is a 10 of spades.

Both cards are 10s.

The ﬁrst card is a Queen and the second card is a King

The ﬁrst card is red and the second card is not a spade.

Mae deals two cards from a normal deck of cards, without replacement. a b

Are the events of drawing Card 1 and drawing Card 2 dependent or independent? Calculate the probability that she deals: i

Two 10s

Two red cards

iii

Two diamonds

A 10 of spades and an Ace of diamonds, in that order

A 10 of clubs and an Ace of spades, in any order

Valentina randomly selects three cards from a normal deck of cards. Calculate the probability if the cards are selected: i

With replacement

The cards are ﬁve of clubs, King of clubs, and Jack of spades, in that order.

The cards are all red.

The ﬁrst card is the 2 of clubs, the second card is a spade and the third card is red.

The cards are all spades.

None of the cards is an 8.

Without replacement

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Ex 4

Three marbles are randomly drawn without replacement from a bag containing 5 red, 4 yellow, 5 white, 4 black and 5 green marbles. Find the probability of drawing: a

Three white marbles

Three black marbles

Zero green marbles

Zero yellow marbles

At least one red marble

In a game of cards, suppose a hand contains a 10, a Jack, a Queen, a King and an Ace. Two cards are drawn from the hand at random, in succession and without replacement. Find the probability that: a

The Ace is drawn.

The King is not drawn.

The Queen is the second card drawn.

A number game uses a basket with 5 balls, all labelled with numbers from 1 to 5. Three balls are drawn at random, without replacement. Find the probability that: a

The ball labelled 4 is picked.

The ball labelled 4 is picked and the ball labelled 1 is also picked.

From a set of 10 cards numbered 1 to 10, two cards are drawn at random, in succession and without replacement. Find the probability that: a

Both numbers are even.

One is even and the other is odd.

The sum of the numbers is 12.

Three fair coins are tossed. Find the probability of obtaining: H H T H H T T H H T T

H T T

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At least 1 head.

TTH in that order.

THH in that order.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

HHH HHT HTH HTT THH THT TTH TTT

A basketball player is practicing his free throws. He makes the shot with a probability of and misses with the probability of . The tree diagram shows all the possible outcomes and probabilities. Find the probability that he will: Hit Hit Miss Hit Hit Miss Miss Hit Hit Miss Miss

Hit Miss Miss

Make all three free throws.

Miss at least one free throw.

Make at least one free throw.

James owns four green jackets and three blue jackets. He selects one of the jackets at random for himself and then another jacket at random for his friend. Find the probability that:

G G

James selects a blue jacket for himself.

Both jackets James selects are green.

B B Ex 5

Han owns four green ties and three blue ties. He selects one of the ties at random for himself and then another tie at random for his friend. a

Write the probabilities for the outcomes on the edges of the probability tree diagram.

Calculate the probability that: i

Han selects a blue tie for himself.

Han selects two green ties.

GG GB

BG BB

G G

B G

B B

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An ice cream shop offers one flavour of ice cream at a discounted price each day. There are 6 flavours for the owner to choose from that could be discounted. 3 of the flavours are sorbet and the other 3 are gelato. The three sorbet flavours are raspberry (R), lemon (L) and chocolate (C). The three gelato flavours are vanilla (V ), mint (M ) and chocolate (C). Each decision has equal probability. a

Write the sample space for which ice cream type and ﬂavour may be chosen.

Find the probability that the lemon sorbet is selected.

Determine the probability that the ﬂavour will be chocolate.

Reasoning 24

24, 25

R S

C V M G C

24, 26, 27

In a lottery there are 37 balls. a

Find the probability of the 22 ball being drawn ﬁrst.

A ball is discarded after it has been drawn. If ball number 22 is drawn on the ﬁrst go, ﬁnd the probability of ball number 15 being drawn next.

Is the probability of each successive ball drawn the same as the probability of the ﬁrst ball drawn?

Are the draws dependent or independent events? Explain your reasoning.

Two cards are to be chosen from a pack of cards numbered 1 to 11. Explain how to ﬁnd the probability of drawing a 1, 2 or 3 ﬁrst, and after replacing the card, drawing a 9, 10 or 11.

Recall that for mutually exclusive events, P ( A ∩ B) = 0. For two events A and B, where P ( A) ≠ 0 and P (B) ≠ 0, explain why the events cannot be both independent and mutually exclusive.

Using the Venn diagram shown, explain why events A and B are independent. A

6 4

Problem-solving 28−30, 33 28

380

29, 31, 33

28, 29, 31, 32, 34

Consider the word WOLLONGONG. If three letters are randomly selected from it without replacement, ﬁnd the probability that: a

The letters are W, O, L, in that order.

The letters are O, N, G, in that order.

All three letters are O.

None of the three letters is an O.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Sarah has a box of Favourites chocolates. In this box there are 30 chocolates, 5 of which are Turkish Delights. Sarah takes and eats a chocolate without looking until she gets a Turkish Delight. a

Find the probability she only eats one chocolate.

Find the probability she eats only two chocolates, rounded to the nearest percent.

Find the probability she eats ﬁve chocolates, rounded to the nearest percent.

As she eats more and more chocolates, state whether the probability of the next chocolate being a Turkish Delight is getting higher or lower.

How many chocolates must Sarah have to eat to be certain that the next chocolate will be a Turkish Delight?

Tom has shuﬄed a standard pack of 52 playing cards. He draws one card from the pack at a time, and then puts it aside. He will stop when he draws the 9 of diamonds. a

Find the probability he has to draw exactly 1 card.

Find the probability he has to draw exactly 2 cards.

Find the probability he has to draw exactly 3 cards.

Find the probability Tom will stop after the 44th card.

As more cards are drawn that aren’t the desired card, does the probability that the next card selected is the 9 of diamonds increase or decrease?

A bag contains 3 green marbles and 4 purple marbles. Two marbles are drawn at random from the bag. a

If the ﬁrst marble is replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree. If the ﬁrst marble is not replaced before the second marble is drawn, complete the table with the probabilities matching the edges of the probability tree.

G A

P G

B P

GG GP

Determine the impact that replacing the marble before drawing a second has on the probability of these events: i

Drawing two green marbles

Drawing two purple marbles

iii

Drawing marbles of different colours

Edge

Probability

Sureya has three races to swim at her school swimming carnival. The probability that she wins a particular race is dependent on whether she won the previous races, as summarised: • The chance she wins the first race is 0.7 • If she wins the first race, the chance of winning the second is 0.8 • If she loses the first race, the chance of winning the second is 0.4 • If she wins the first two, the chance of winning the third race is 0.9 • If she lost the first two, the chance of losing the third race is 0.9 • If she won only one of the first two races, the chance of winning the third is 0.6 a

Construct a probability tree to represent all outcomes in this situation.

Calculate the probability Sureya won all three races, rounded to three decimal places.

Calculate the probability Sureya won the third race, rounded to three decimal places. 8.02 Independent and dependent events mathspace.co

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For breakfast, Moriah has something to eat and drinks a hot drink. She will either eat toast or cereal and then will drink tea or Milo.

• The chance of Moriah making toast is 0.7 • The chance of Moriah drinking Milo is 0.4 a

Construct a tree diagram illustrating all possible combinations of food and drink Moriah can have for breakfast and their associated probabilities.

Find the probability Moriah drinks tea and eats toast.

Find the probability Moriah drinks tea or eats toast.

The proportion of scholarship recipients at a particular university is

. The number of

students at the university is so large that even if a student is removed, we can say that the proportion of scholarship recipients remains the same. If three students are selected at random: a

Construct a probability tree showing all the possible combinations of recipients and non-recipients.

Find the probability that at least one of the students is a scholarship recipient.

Find the probability that at least one of the students is a non-recipient.

Find the probability there is at least one recipient and one non-recipient in the selection.

8.03 Conditional probability After this lesson, you will be able to… • calculate probabilities using the conditional probability formula. • interpret two-way tables, tree diagrams, and Venn diagrams in the context of conditional probability. • solve real-world problems involving conditional probability. • identify the mathematical properties of conditional probability in independent events.

Conditional probability We are often interested in probabilities under certain conditions. For example, if an even number is rolled on a die, what is the probability that it is a six? Conditions will reduce the sample space we are concerned with.

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Exploration A group of people were asked whether they went on holiday last summer. The results are provided in the given table: Teacher Not a teacher Total

Holiday 22 32 54

No holiday 26 20 46

Total 48 52 100

1. When randomly selecting a person from all those surveyed, how many people are we selecting from? 2. When randomly selecting a person from those who are not teachers, how many people are we selecting from? 3. Explain why your answers to the previous questions are different. 4. What is the probability of randomly selecting a person that is not a teacher and went on holiday? 5. Of those who are not teachers, what is the probability of randomly selecting a person that went on holiday? 6. Show that the probability from question 5 is equal to the probability from question 4 divided by the probability of randomly selecting someone who is not a teacher from the people surveyed.

For two events A and B, the probability of event A occurring given that event B has occurred has the notation P ( A∣B), which is read as the “probability of A given B”. This is known as conditional probability and is calculated by:

Note that if we have a Venn diagram or a two-way table, then we may use the number of favourable and total outcomes instead of their probabilities:

If we know the conditional probability, we can calculate the probability of A and B using: P ( A ∩ B) = P ( A∣B) × P (B) or P ( A ∩ B) = P (B∣A) × P ( A) For independent events, the probability of the second event does not depend on the first event, so P ( A∣B) = P ( A).

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Example 1 Find P( A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P (B∣A) = 0.8 and P (A) = 0.5

Create a strategy Use the formula P (B∣A) =

Apply the idea Write the formula

Substitute known values

Multiply both sides by 0.5

Evaluate

Example 2 Use the two-way table to ﬁnd the probabilities. A A′ Total

B′

11 21 32

10 8 18

Total 21 29 50

a Find P ( A∣B).

Create a strategy Use the formula P ( A∣B) =

Apply the idea To use the formula for conditional probability, we need to ﬁnd P ( A ∩ B) and P (B). From the table, we see P ( A ∩ B) =

and P (B) =

. Write the formula

Substitute known values

Evaluate

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Reﬂect and check We can also ﬁnd P ( A∣B) without the formula. Since we are given B as the sample space, the total number of outcomes is 32. Since the total number of favourable outcomes in the sample space is A = 11, then P ( A∣B) =

b Find P ( A∣B′).

Create a strategy The formula for this conditional probability is P ( A∣B′) = P ( A ∩ B′) and P (B′).

, so use the table to ﬁnd

Apply the idea From the table, we see P ( A ∩ B′) =

and P (B′) =

Write the formula

Substitute known values

Evaluate

Simplify

Reﬂect and check Checking our answer, the total number of outcomes is B′ = 18 and there are A = 10 favourable outcomes in B′, so P ( A∣B′) =

c Find P ( A′∣B′).

Create a strategy From the previous part, we know the total number of outcomes is B′ = 18. Next, we use the table to ﬁnd how many outcomes of A′ are in B′.

Apply the idea From the table, we see there are A′ = 8 favourable outcomes. So, Substitute known values

Simplify

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Reﬂect and check Using the formula to check our answer: Write the formula

Substitute P ( A′ ∩ B′) =

Evaluate

Simplify

and P (B′) =

Example 3 A basketball team has a probability of 0.8 of winning its first season and 0.15 of winning its first season and its second season. What is the probability of winning the second season, given they won the first?

Create a strategy Use the formula P (B∣A) =

with:

• Event A: Winning the 1st season • Event B: Winning 2nd season

Apply the idea From the given information we know that: P (Win 1st ∩ Win 2nd) = 0.15 and P (Win 1st) = 0.8. We want to ﬁnd P (Win 2nd∣Win 1st): Start with the conditional probability formula

Substitute the probabilities

Simplify

Example 4 A flight departs from Melbourne to Sydney. The probability that the flight departs on time, given the weather is fine in Melbourne is 0.9, and the probability that the flight departs on time, given the weather is not fine in Melbourne is 0.7. The probability that the weather is fine on any particular day in July is 0.4.

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a By constructing a tree diagram or otherwise, ﬁnd the probability that the ﬂight from Melbourne to Sydney departs on time in a day in July.

Apply the idea We can construct the following tree diagram for this situation:

Fine 0.4

0.9 0.1

0.7

0.6 Not Fine

0.3

On time Late

The ﬂight can be on time in two different ways depending on whether the weather is ﬁne: P (On time) = P (Fine) × P (On time∣Fine) + P (Not fine) × P (On time∣Not fine) P (On time) = 0.4 × 0.9 + 0.6 × 0.7 = 0.78

Substitute the probabilities Evaluate

Reflect and check A two-way table could have been used instead, but the values in the table would need to be calculated first. From the information given, we know: • P (Fine) = 0.4 • P (Not fine) = 0.6 • P (On time∣Fine) = 0.9 • P (On time∣Not fine) = 0.7 The intersection of Fine and On time is: P (Fine ∩ On time) = P (On time∣Fine) × P (Fine) = 0.36 The intersection of Not fine and On time is: P (Not fine ∩ On time) = P (On time∣Not fine) × P (Not fine) = 0.42 Filling in the table and finding the total of the On time column gives us: Fine Not fine Total

On time 0.36 0.42 0.78

Late 0.04 0.18 0.22

Total 0.4 0.6 1

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b Find the probability that the weather is ﬁne in Melbourne given that the ﬂight departs on time on a day in July.

Create a strategy We can use the formula: P ( A∣B) =

Apply the idea Write the formula Substitute the probabilities Evaluate the numerator Multiply by Divide by common factor of 6

Idea summary For two events A and B, the probability of A occurring given that B has occurred is given by:

If we know the conditional probability, we can calculate the probability of A and B using either: P( A ∩ B) = P( A∣B) × P(B) or P( A ∩ B) = P(B∣A) × P( A).

8.03 Practice questions Understanding 1

Adventurer

1−3

Trailblazer 1−4

Are these statements true or false? a

If P ( A) = 0.5, P (B) = 0.3 and P ( A ∩ B) = 0.15, then the events are dependent.

Using the two-way table shown, P (Maths∣Fail) = Pass Fail Total

388

Explorer

Maths 25 15 40

Physics 7 12 19

Total 32 27 59

In a Venn diagram, if events A and B do not overlap, this means P ( A∣B) = 0.

The phrase “The probability of B given A” means that event A has already occurred and we want to know the probability that event B will occur next.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

If one event is “a student drives to school” and a second event is “the weather is rainy”, write the formula for ﬁnding the probability of selecting a student who drives to school given the weather is rainy. Fill in the blanks with “drive” or “rain”:

If P (B∣A) = 0.2, what additional piece of information is needed to ﬁnd P ( A ∩ B)? A

P ( A)

P (B)

P ( A ∪ B)

P ( A∣B)

There are two positions available at a company and the applicants have been shortlisted down to 6: Patricia, Nadia, Amy, Aaron, Lachlan, Jimmy. The two people to ﬁll the position will be picked randomly. • Event A: Patricia is chosen • Event B: Aaron is chosen a

Describe the probability P ( A∣B).

Is P ( A∣B) equal to, less than, or greater than P ( A ∩ B)?

Fluency

5−9, 11−13, 15, 18, 20

5−9, 11−14, 16, 18, 20

6−8, 10, 13, 14, 16, 17, 19−21

Find P (B∣A) given these probabilities for an experiment in which A and B are two possible events: P ( A ∩ B) = 0.48 and P ( A) = 0.6

Ex 1

Find P ( A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P ( A∣B) = 0.6 and P (B) = 0.3

Ex 2

Use the two-way table to ﬁnd the probabilities.

P ( A)

P (B)

P ( A ∩ B)

P ( A ∪ B)

P ( A′ ∩ B′)

P ( A ∩ B′)

P ( A∣B)

P (B∣A)

P ( A′∣B)

P (B∣A′)

P ( A′∣B′)

P (B′∣A′)

Using the Venn diagram, ﬁnd: a

P ( A)

P (B′)

P ( A∣B)

P (B∣A)

P ( A∣B′)

A A′ Total

B′

5 7 12

6 2 8

Total 11 9 20

B 0.2

0.2

0.4 0.2

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A group of people were asked what type of exercise they like to do. The results are shown in the Venn diagram.

walk

A person is chosen from the group at random.

Find the probability that the person walks.

Find the probability that the person runs, knowing that they also walk.

At a university there are 816 students studying ﬁrst year engineering, 497 of whom say football is their favourite sport (set F ). Of the 348 students studying Civil Engineering (set C), 237 of them say football is their favourite sport.

Ex 3

run 10

C w

Find the value of: i

iii

Of the students whose favourite sport is not football, find the probability that a randomly selected student does not study Civil Engineering.

A player is rolling 2 dice and looking at their difference, that is, the largest number minus the smaller number. They draw a table of all the possible dice rolls for 2 dice and what their difference is: a

List the sample space for the difference of two dice.

Find the probability the dice will have a difference of 2.

Find the probability the dice will have a difference of 2 given that one of the dice rolled is a 3.

1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 5 4 3 2 1 0

A card is randomly drawn from a standard 52 pack of cards. Find the probability that it is a Jack, Queen, King or Ace if: a

No additional information is known.

We know it’s a 10, Jack or Queen.

We know it’s a 9 or a Queen.

We know it’s not 2, 3, 4 or 5.

A software development team has a probability of 0.8 of successfully completing their first project and 0.55 of successfully completing both their first and second projects. Find the probability of successfully completing the second project, given they completed the first one successfully.

In a population, 35% of people have brown eyes, 45% of people have blonde hair, and 10% of people have both brown eyes and blonde hair. A person is chosen randomly from the population. Find the probability that they have:

390

Blonde hair, knowing that they have brown eyes

Brown eyes, knowing that they have blonde hair

Two dice are rolled, one after another. Find the probability, in simplest form, of rolling: a

A pair of fours, given the ﬁrst die is a four.

A pair with a sum of 7 or more, given the ﬁrst die is a two.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Ex 4

Four identical balls labelled 1, 2, 3 and 4 are in a bag. Two balls are randomly drawn from the bag in succession and without replacement. Find the probability that: a

The ﬁrst ball is labelled 4 and the second ball is labelled 2.

The sum of the numbers on the two balls is 5.

The second ball drawn is 1 given that the sum of the numbers on the two balls is 5.

Two cards are randomly drawn in succession and without replacement from a deck of cards numbered from 1 to 20. Find the probability that the second card is: a

An even number given that the ﬁrst card is a 10.

Less than 5 given that the ﬁrst one is a 14.

A number divisible by 5 given that the ﬁrst card is a 15.

A bus service operates from Town A to Town B. The probabilities shown were found with regard to departure times and traffic in June: • The probability that the bus departs on time, given the traffic is light in Town A is 0.8 • The probability that the bus departs on time, given the traffic is heavy in Town A is 0.7 • The probability that the traffic is light on any particular day in June is 0.4 Find the probability that:

The bus service from Town A to Town B departs on time on a day in June.

The traﬃc is light in Town A given that the bus departs on time on a day in June. Give the answer as a fraction in simplest form.

A museum has a new software system that determines whether a painting is fake or authentic. The system has an accuracy of 75%. Art experts have previously determined that 5% of paintings that the museum examines are fakes. Find the probability that:

The software correctly identiﬁes a painting as fake.

The software incorrectly identiﬁes an authentic painting as fake.

The software will identify a painting as fake.

A painting that is identiﬁed as fake is actually fake, rounded to two decimal places.

In a survey, 531 people are asked whether they watch My Kitchen Rules (MKR) or Masterchef (MC). Of the total respondents, 177 people watch both and 65 watch neither. The number who watch MKR is twice the number who watch both. a

Find the number of people who only watch MKR.

Of the people who watch MC, ﬁnd the probability that a randomly selected person also watches MKR.

Of those who don’t watch MC, ﬁnd the probability that a randomly selected person watches neither.

The probability of two independent events, A and B are P ( A) = 0.8 and P (B) = 0.5 Determine the probability of: a

Both A and B occurring

Neither A nor B occurring

A or B or both occurring

B occurring but not A

A occurring, given B occurs

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Reasoning 22

22, 23

23, 24

Given the conditional probability formula,

show that, P ( A∣B) × P (B) = P (B∣A) × P ( A) 23

Use the Venn diagram shown, where B is a subset of A, to ﬁnd: a

P ( A∣B)

P (B∣A)

P ( A′∣B′)

P (B′∣A′)

Explain why the probability of A occurring given B has already occurred is always certain.

A B 5

A Venn diagram is shown in which the size of the regions are proportional to the number of elements in that region.

A B

Which conditional probability is greater, P ( A∣B) or P (B∣A)? Justify your answer.

Problem-solving 26−28 25

25, 26, 28

25, 26, 28, 29

Edwin watches three episodes of TV each night. He begins with News or Current Affairs, then the next two shows are either Comedy, Horror or Animation. He never watches more than one Horror show, and if he watched the News, he will follow this immediately with a Comedy. a

Construct a tree diagram of all possible options.

If, at every stage, the possible outcomes of each choice are equally likely, ﬁnd the probability Edwin watches two Comedies, given that he watched the News.

When Fred gets ready for work in summer, he ﬁrst decides whether it will be a hot day (H) or not (N ) and then decides whether to wear a tie (T ) or to just dress casually (C). The chance of Fred deciding it will be a hot day is 0.7. If he decides it will be a hot day, there is a 0.2 chance he will wear a tie. If he decides it will not be a hot day, there is a 0.85 chance he will wear a tie. a Construct a probability tree for this situation. b Calculate the probability Fred decided it was a hot day given that he wears a tie. Round your answer to two decimal places.

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For breakfast each morning, Marge eats porridge, toast or cereal. With that, she will either drink orange juice, tea or hot chocolate. Find the probability that Marge will: a Eat toast and drink tea b Eat cereal or drink orange juice c Eat porridge given that she drinks hot chocolate

Two hundred people were questioned about whether they voted for Labor, Liberal or Greens last election and who they’ll vote for this election.

Labor (last election) Liberal (last election) Greens (last election) Total a b c d

Labor (next election) 26

Liberal (next election) 40

1 5 10

Total 70 90 200

Complete the table. Find the probability that a randomly selected person will vote Labor next election given that they voted Liberal last election. Find the probability that a randomly selected person did not vote for the Greens last election given that they will vote Green next election. If a person voted Labor or Liberal last election, ﬁnd the probability they’ll vote Liberal next election.

A class of 13 students had their height, shoe size and 400 m race time collected. The results are shown in the table: a Let T be the event that a randomly chosen student from the class has a race time greater than 90. Find the value of P (T ). b Let H be the event that a randomly chosen student from the class has a height greater than 130. Find the value of P (H). c Find the value of P (T∣H). d Let S be the event that a randomly chosen student from the class has a shoe size greater than or equal to a size 10. Find the value of P (H∣S). e

140

Greens (next election)

Which of the events appear to be dependent, H and S or H and T?

Athlete Aaron Tobias Lisa Georgia Ben Patricia Vanessa Sandy Adam Amelia Rosey Neil Roald

400 m Height time (sec) (cm) 94 137 62 165 89 100 64 155 98 100 127 102 84 147 98 144 54 154 84 147 81 126 104 134 67 127

Shoe size 8 10 11 7 10 11 7 8 7 8 12 9 8

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8.04 Mixed probability concepts After this lesson, you will be able to… • use Venn diagrams, two-way tables, and tree diagrams for organising data. • perform probability calculations using appropriate formulas. • interpret results in the context of the problem.

Mixed probability concepts Many situations in probability can be modelled by a Venn diagram, an array or two-way table, and/or a tree diagram to organise the information, determine the size of different groups, and do calculations. Probability notation can also help us to organise the information given in a question, and these formulas can help us calculate different probabilities. If two events are independent: P ( A ∩ B) = P ( A) × P (B) When two events are dependent, the probability of each event will change, depending on the first event. P ( A ∩ B) = P ( A∣B) × P (B) For conditional probability:

For mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) For non-mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

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Example 1 A bag contains the letters A, B, and C. A letter is randomly drawn out of the bag without replacement and placed on the table. This is done three times.

ABC ACB BAC BCA CAB CBA

a Find the probability that A is drawn before B is drawn.

Create a strategy Count the number of branches where A appeared before B.

Apply the idea

Reﬂect and check

The sequences where A is drawn before B are ABC, ACB, and CAB. There are 3 desired outcomes and 6 total outcomes possible.

Notice each sequence has a probability of

Therefore, the probability is .

Since there are 3 possible sequences, the probability of drawing A before B is:

b Find the probability that two A’s are drawn.

Create a strategy

Apply the idea

Note that the letters are not replaced after being As there is only one A inside the bag, it is drawn. impossible to randomly draw two A ’s without replacement. So the probability is 0.

c Find the probability the word CAB is spelled out in that order.

Create a strategy Consider the number of outcomes that spell out CAB.

Apply the idea Only one of the six outcomes spells CAB. The probability is .

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Reflect and check We can check the answer by thinking about the probability of drawing C first, then A, then B. The probability of drawing C first is . For the second draw, there are 2 letters left. The probability of drawing A next is . There is only one letter left for the third draw, so the probability of drawing B third is 1. So the probability of drawing the letters CAB in that order is:

d Find the probability that B is drawn last, given that A is drawn ﬁrst.

Create a strategy The probability of selecting B last given A was drawn ﬁrst can be calculated using the formula P (B last∣A first) =

Apply the idea There is only one way to select B last and A ﬁrst, so P (B last ∩ A first) = . For the ﬁrst letter, A is one of the three letters, so P ( A first) = . Start with the conditional probability formula

Substitute known values

Evaluate the division

Simplify

Reﬂect and check With conditional probability, the given condition determines the total number of outcomes. If we know A was drawn ﬁrst, there are only 2 possible outcomes: drawing B last or drawing C last. Since drawing B last is one of those outcomes, the probability is .

Example 2 At a restaurant, 50 people are asked about whether they would like to add rice (R) or broccoli (B) to their meal. Some of the results are shown in the Venn diagram.

14 15

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a Find the number of people in R ∩ B′.

Create a strategy This is the blank section of the Venn diagram. We can subtract the amounts in the diagram from the total number of people surveyed.

Apply the idea

Reﬂect and check

The number of people in R ∩ B′ is

We can now complete the Venn diagram.

50 − 9 − 14 − 15 = 12

n (R ∩ B′) = 12 12

14 15

b Find P (B∣R).

Create a strategy Use the formula P (B∣R) =

Apply the idea Using the Venn diagram from part (a), we know the number of people in B ∩ R is 9 and the number of people in R is 9 + 12 = 21. Write the formula Substitute known values

Reﬂect and check We can check our answer using the conditional probability formula and

, so Start with the conditional probability formula

Substitute known values

Evaluate the division

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c Find P (R∣B′).

Create a strategy Use the formula

Apply the idea Using the Venn diagram from part (a), we know the number of people in R ∩ B′ is 12 and the number of people in B′ is 12 + 15 = 27. Start with the conditional probability formula

Substitute known values

Evaluate the division

Simplify

Reﬂect and check Checking our answer, we can look at the Venn diagram to conﬁrm that of those who did not choose broccoli (27 people), 12 of them chose rice.

Example 3 600 people were questioned as to whether they read novels in paperback form or on an e-reader. • 120 said they do both. • 180 said they read paperbacks. • The probability that a person who didn’t read paperbacks also didn’t read on an e-reader was 30%. a Complete the table for these results: Paperback

Not paperback

E-reader Not e-reader

Create a strategy Add a total row and column to the table to make it easier to ﬁnd missing values. Paperback

Not paperback

E-reader Not e-reader Total

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Total

Apply the idea The first piece of information goes in the first cell of the table. The second piece of information helps us fill in the first column and last row.

E-reader Not e-reader Total

Paperback 120 60 180

Not paperback

Total

420

600

The third piece of information can help us ﬁll in the rest of the table. Of the people who didn’t read paperback books, 30% did not read on an e-reader. 30% × 420 = 126

E-reader Not e-reader Total

Paperback 120 60 180

Not paperback 294 126 420

Total 414 186 600

b Calculate the probability that a randomly selected person reads novels on an e-reader but not in paperback.

Create a strategy Use the table from part (a) to ﬁnd the number in the row for E-reader and the column for Not paperback, E ∩ P′.

Apply the idea In the table from part (a), we see 294 people read novels on an e-reader but not in paperback. Divide desirable outcomes by total outcomes

Simplify

c Of those that read on e-readers, what percentage read in both formats? Round your answer to the nearest percent.

Create a strategy Divide the number of people who read on both formats by the number of people who read on an e-reader.

Apply the idea The number of people who read on both formats is 120. The number of people who read on an e-reader is 414. Use a calculator to estimate the probability as a percentage, and round to the nearest whole percentage:

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Reﬂect and check We can check this answer by using the formula for conditional probability: P (P∣E) = Start with the conditional probability formula

Substitute known values

Evaluate the division

Convert to a percentage

Evaluate using a calculator and round

Example 4 Vanessa has 12 songs in a playlist. Four of the songs are her favourite. She selects shuffle and the songs start playing in random order. Shuffle ensures that each song is played once only until all songs in the playlist have been played. Find the probability that: a The first song is one of her favourites.

Create a strategy The probability can be calculated using

Apply the idea Vanessa has 4 favourite songs and there are 12 songs in the playlist. The probability that the ﬁrst song is one of her favourites is

b Two of her favourite songs are the ﬁrst to be played.

Create a strategy The probability for the second song will be dependent on the ﬁrst song that is played.

Apply the idea From part (a), we know the probability that the ﬁrst song is one of Vanessa’s favourites is . This leaves 3 favourite songs out of 11 songs that have yet to play. So, the probability that the second song is also her favourite is

Multiplying these probabilities together gives us:

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c At least 1 of her favourite songs is played in the ﬁrst three.

Create a strategy The probability that at least 1 of the songs is her favourite is the complement of the probability that no songs are her favourite.

Apply the idea The probability that the ﬁrst song is not a favourite is

Since there are 7 non-favourite songs left and 11 songs left in the playlist, the probability that the second song is not a favourite is

For the third song, there are 6 non-favourites left and 10 songs left in the playlist. The probability the third song is not a favourite is

The probability that no songs are Vanessa’s favourite is:

The probability that at least one of the songs is Vanessa’s favourite is:

Idea summary Venn diagrams, arrays, and tree diagrams can be used to organise information and help solve problems in probability. We can also use the formulas for complementary events, independent events and conditional probability to solve problems. For independent events: P ( A ∩ B) = P ( A) × P (B) For dependent events: P ( A ∩ B) = P ( A∣B) × P (B) For conditional probability: P ( A∣B) = For mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) For non-mutually exclusive events: P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

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8.04 Practice questions Understanding 1

Explorer 1−3

Adventurer

Trailblazer

1−3

Match each of the Venn diagrams to the correct statement. i

B A

ii A B

0.32

0.08

0.12 0.48

iii

A B

iv A

0.32

0.08

0.48 0.12

Events A and B are mutually exclusive.

Events A and B are independent.

P ( A∣B) = 1

P ( A ∩ B) ≠ P ( A) × P (B)

Match each formula to the probability described. i

P ( A ∪ B) = P ( A) + P (B) − P ( A ∩ B)

P ( A ∩ B) = P ( A) × P (B)

iii

P ( A) + P (B) = 1

P ( A ∩ B) = P ( A∣B) × P (B)

Conditional probability

Complementary events

Non-mutually exclusive events

Independent events

Which statement is equivalent to this statement? Find the probability that a randomly selected person is female, given they are a team leader.

402

Of the females, ﬁnd the probability the person is a team leader.

If a person is a team leader, then ﬁnd the probability that they are female.

Find the probability that a randomly selected person is male knowing that they are a team leader.

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6−8, 10, 11, 13, 15, 17, 18

Fluency 4

4−6, 8, 10−13, 15−18

4−6, 9, 11, 12, 14, 16, 19, 20

Find the values of w, x, y, and z, given that P ( A ∩ B) = 0.3, P (B) = 0.6, and that the events A and B are independent.

Use the Venn diagram to ﬁnd P ( A∣B). Give the answer as a fraction in simplest form.

A A′

B w x

B′ y z

0.3

0.15

0.25 0.3

Ex 1

Three cards labelled 2, 3 and 4 are placed face down on a table. Two of the cards are selected randomly to form a two-digit number. The outcomes are displayed in the tree diagram. a b

3 2 4

List the sample space of two digit numbers produced by this process.

2 3

Find the probability that:

2 appears as a digit in the number.

The sum of the two selected cards is even.

iii

The number formed is greater than 40.

2 is the second digit, given 3 is the ﬁrst digit.

2 4 3

Students were asked what they are allergic to. The table shows the results: Allergic to dairy Not allergic to dairy

Allergic to nuts 14 25

Not allergic to nuts 12 25

If a student is chosen at random, ﬁnd the probability that:

Ex 2

The student is allergic to dairy.

The student has an allergy.

The student does not have an allergy.

The student is also allergic to dairy if he or she is allergic to nuts.

In a survey, 100 oﬃce workers were asked whether they buy a coffee (set C) or lunch (set L) during the work day. Some of the results are shown in the Venn diagram. Find: a

The number of workers in (L ∩ C′)

P ( C∣L)

P ( L∣C′)

L 43

47 7

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Of the 100 students starting kindergarten surveyed, 80% believe in Santa, 41% believe in the Tooth Fairy and 31% believe in both. a

Complete the table: Believe in Santa

Do not believe in Santa

Total

Believe in Tooth Fairy Do not believe in Tooth Fairy Total

Ex 3

100

Of those who believe in Santa, ﬁnd the probability that they also believe in the Tooth Fairy. Write your answer as a percentage rounded to two decimal places.

Of those who believe in either Santa or the Tooth Fairy, what is the probability they believe in both?

Of those who believe in the Tooth Fairy, what is the probability they do not believe in Santa?

Two sets of numbers, A and B, are such that set A contains the even numbers from 1 to 20, inclusive, and set B contains the factors of 48 from 1 to 20, inclusive. a

List the numbers in set A.

List the numbers in set B.

How many numbers are unique (only appear in one set)?

Find the probability that a randomly selected number is in set B, given that it is in set A.

In a survey, 166 children were asked whether they had read or watched at least one of the stories in a popular children’s series: • 62 hadn’t watched any of the movies. • 39 had read and watched at least one of the stories. • 97 had read the book. • 4 had never read nor watched any of these. a

Complete the table for these results: Read the book Didn’t read the book Total

404

Watched the movie 39

Didn’t watch the movie 4 62

Find the probability that a randomly selected student had: i

Read the book but hadn’t watched the movie.

Read the book, given that they had watched the movie.

iii

Watched the movie, given that they hadn’t read the book.

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Total 97 166

In a survey, 1000 people were asked about whether they bought (set B) or grew (set G) their own herbs. The results are listed: • 4% said they do neither. • 40% said they do both. • Of those who bought their herbs, the probability that they also grew herbs was . In this Venn diagram, where x, y and z represent numbers of people: a

Find the value of: i

iii

Find the number of people who only grew their own herbs.

Of those who grew herbs, what is the probability that they also bought herbs?

G z

y x

In a group of 69 people, 45 said they had been to Europe (E), 26 said they had been to Asia ( A) and 8 people said they had been to neither Europe nor Asia. Find: a

P ( A ∩ E′)

The number of people in (E ∩ A)

P ( E∣A)

P ( A∣E′)

At an Italian restaurant, Alessandra orders an entree, main and dessert. She has 2 entrees, 3 mains and 2 desserts to choose from:

• Entrees: calamari, soup • Mains: spaghetti, lasagna, risotto • Dessert: gelato, tiramisu

The possible combination of meals that Alessandra could order is shown in the tree diagram. Spaghetti

Calamari

Lasagna

Risotto

Gelato Tiramisu Gelato Tiramisu Gelato Tiramisu

Spaghetti

Lasagna Soup Risotto

Gelato Tiramisu Gelato Tiramisu Gelato Tiramisu

If each combination of dishes is equally likely, ﬁnd the probability that Alessandra orders lasagna given that she ordered calamari.

Find the probability that Alessandra ordered spaghetti and tiramisu, given that she ordered soup.

Find the probability that Alessandra ordered risotto or spaghetti, given that she ordered soup.

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Two marbles are randomly drawn without replacement from a bag containing 1 blue, 2 red and 3 yellow marbles. a b

Construct a tree diagram to show the sample space. Find the probability of drawing these: i

A blue marble and a yellow marble, in that order

A red marble and a blue marble, in that order

iii

2 red marbles

No yellow marbles

2 blue marbles

A yellow marble and a red marble, in that order

vii A yellow and a red marble, in any order 16

In tennis, if the ﬁrst serve is a fault (out or in the net), the player takes a second serve. A player serves with these probabilities: • First serve in: 0.55 • Second serve in: 0.81

Construct a probability tree showing the probability of the ﬁrst two serves either being in or a fault.

Find the probability that the ﬁrst serve is a fault.

Find the probability that the player makes a double fault (both serves are a fault).

Nine pilots from Star Jet and seven pilots from Air Tiger offer to take part in a rescue operation. If 2 pilots are selected at random: a

Construct a probability tree showing all possible combinations of airlines from which the pilots are selected.

Find the probability that the two pilots selected are from: i

Ex 4

Different airlines

A carton contains a dozen eggs, of which 4 contain a double yolk. If 3 eggs are chosen at random for a cake, find the probability that: a

The same airline

They all have a double yolk.

At least 1 has a double yolk.

From a standard pack of cards, 1 card is randomly drawn by Andrea and then put back into the pack. A second card is then drawn by Bridie. • Event A: Andrea selects a heart. • Event B: Bridie selects a heart. a

Calculate the probability that neither of their cards are hearts.

Calculate the probability that at least 1 of the cards is a heart.

The probability that Joanna, Nadya and Adam get permission to go on their school trip are 0.8, 0.1 and 0.2, respectively. Find the probability that at least 1 of them gets permission.

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Reasoning 21

21−23

23, 24

For P ( A ∩ B): a

Write the formula of independent events.

Write the formula of dependent events.

Which formula, the one from part (a) or the one from part (b), can be used for both independent and dependent events? Explain.

A sample of 20 students, A-T, from the same school were asked whether they had a haircut in the last month (H) and whether they had been sick in the last month (S). The results are shown in the given tables:

Student Hair Sick

A ✓ −

B ✓ ✓

C − −

D ✓ ✓

E − −

F ✓ −

G ✓ ✓

H −

Student Hair Sick

K −

L −

M −

✓

O − −

P ✓ −

Q −

✓

N ✓ ✓

✓

I − −

J ✓ −

R ✓ ✓

S − −

T −

Find the probability that a student: i Had a haircut in the last month ii Had been sick in the last month iii Was sick and had a haircut in the last month

Find the value of P (H) × P (S).

Does the data suggest that getting haircuts and being sick in the last month are independent or dependent? Explain your reasoning.

✓

In a memory game, 16 pairs of identical cards are randomly placed face down. When someone has a go, they turn one card over and then turn a second card over to try to ﬁnd an identical pair. If an identical pair is found, the two identical cards are removed from the pack. a

At the beginning of the game, Kathleen turns a card over. Find the probability of her selecting a card that makes an identical pair.

The ﬁrst two cards Kathleen picks are identical. Is the probability of picking any more pairs of identical cards independent or dependent on the number of pairs picked before? Explain your reasoning.

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Candidates for a professional examination are allowed a maximum of two attempts at the exam. Over a period of time, it has been shown that 60% of the candidates pass on the ﬁrst attempt and, if a second attempt is required, 70% of the candidates pass. Assume that these probabilities apply to a randomly chosen candidate. a

Why does the second set of branches only have one fork?

Given that a candidate passes, calculate the probability that they passed on the ﬁrst attempt. Give the answer as a fraction in simplest form.

0.6

25, 27−29

0.7

0.4

Explain why it is inappropriate to assume that the probabilities apply to a randomly chosen candidate.

Problem-solving 25, 27, 28 25

0.3

26, 28−30

A school consisting of 410 primary students and 410 secondary students offers everyone an optional end of year project to improve their grade. There are various types of projects they can do. A summary of the number of student submissions are shown in the given table: Primary Secondary a b c d

Essay 18 76

Creative story 64 68

Art work 46 73

Other Project 24 45

Did not submit 258 148

Find the probability of selecting student that submitted a project. Find the probability of selecting student that submitted a project given that they are a secondary student. Find the probability that a submitted project is an essay. Find the probability that a submitted project is an essay given that the project was submitted by a primary student.

Find: a

P ( A ∩ B ∩ C′)

P (( A ∪ B) ∩ C′)

P ( B∣C)

P ( A′∣C)

P (( A ∪ B)∣C′)

B 0.17 0.18 0.1 0.07 0.11 0.12 0.2 C

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0.05

Yvonne is struggling to decide what to watch online. She decides to pick one movie at random from the streaming website. A Venn diagram of her options sorts movies into three categories based on their genre: Comedy, Action and Horror.

Comedy 10

Action 4

How many of the movies are horror ﬁlms?

How many of the movies ﬁt into only one genre?

Find the probability of selecting a film that fits into only one genre given that it fits into the horror genre.

6 13 Horror

In a particular city, 55% of the population are adults. 8% of the children and 2% of the adults are left-handed. a

What percentage of the city’s population is left-handed?

If a person is chosen at random and is left-handed, what is the probability that the person is an adult? Give the answer rounded to two decimal places.

The probability of spotting a rare bird species in a national park on any given day is 0.3. If one is spotted one day, the probability of spotting another the next day increases to 0.6, due to ﬂock movement. However, if none are spotted, the probability of spotting one the next day decreases to 0.15. What is the probability of spotting exactly one rare bird over two consecutive days?

In a speciﬁc deck of 10 cards numbered from 1 to 10, the probabilities of drawing each number are in the ratio 1: 2: 3: 4: 5: 6: 7: 8: 9: 10. If two cards are drawn with replacement, what is the probability that they add up to 15?

Did you know?

Probability can explain traffic patterns! The chance of a traffic jam depends on factors like weather, and even the time of day. Probability models incorporate weather data to predict traffic flow. For example, the likelihood of delays is higher during heavy rain, snow, or fog due to reduced visibility and slower driving speeds. Data collected from past traffic incidents during similar weather conditions help refine these predictions. 8.04 Mixed probability concepts mathspace.co

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INVESTIGATION

Simulation Investigate probability situations using simulation.

Objectives To use probability simulations to model problems which may be counterintuitive including the three-door and birthday problems

Materials •

Device with internet access

Simulations Experimental probability approaches theoretical probability as the number of trials increases to infinity. Rather than physically conducting an experiment 10 000 times to show this, we can use tools to simulate an experiment instead.

Running a simulation We can run simulations manually, but often this is not much faster than physically doing the experiments. Using computer programs we can use random numbers and coding to simulate realworld scenarios. There will often be assumptions that need to be coded in.

Birthdays! What do you think is the probability that two people in your class share the same birthday? Why do you believe that?

Interactive exploration

Discover this concept in action online

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Investigate After looking at the simulations: 1.

If we have only two people in a room, what's the probability that they will share a birthday?

As we add more people to the room, do you think the chances of two people sharing a birthday will increase or decrease? Why?

Using the applet, what is the experimental probability of having two or more people with the same birthday in a class of 30?

What group size would give a 50% chance of someone sharing a birthday? Does this surprise you?

What group size would nearly guarantee that there is a shared birthday? Does this seem intuitive?

Write pseudo-code that might be behind the applet or explain how you think it works.

The three door problem The game In a game show, you’re presented with three doors. Behind one is a car, and the other two hide goats. You choose a door. The host, knowing what’s behind each, reveals a goat behind one of the other doors. They then offer you a choice: switch to the other unopened door or stick with your initial pick. Before running the simulation, what do you intuitively expect the outcome to be if you always switch doors versus if you always stay? Why?

Interactive exploration

Discover this concept in action online

mathspace.co

Investigate Consider the questions after completing the simulation. 1.

According to the theoretical probability, is the car more likely to be behind the door you picked or behind one of the doors you did not pick?

How crucial is the assumption that the host always opens a door with a goat behind it? What would happen to the probabilities if the host randomly opened any door without this knowledge?

Using probability trees or other methods, can you show why the results are what they are? Why does one strategy seem to have an advantage over the other?

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Discussion 1.

How can knowing coding make doing simulations more efficient?

To create this simulation, what additional knowledge would have been important for the creator of it to know? https://www.biointeractive.org/classroom-resources/ modeling-disease-spread

Were the results of the simulations always what you expected? Can you think of any counter-intuitive probability problems?

Did you know?

Probability plays a huge role in baseball strategy! Coaches use it to decide when to take risks or play it safe during critical moments. For example, probability can help determine whether a player should attempt to steal a base or stay put, based on factors like the pitcher’s throw speed, the catcher’s reaction time, and the runner’s success rate in similar situations. By calculating these probabilities, teams make smarter, data-driven choices that can turn the game in their favour. It’s not just about instinct; probability helps teams predict which moves are most likely to score runs, making every decision a bit more strategic. In a close game, those small, calculated choices can make all the difference between a win and a loss!

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8 Chapter review 1

Construct a two-way table based on the information in the Venn diagram.

A spinner is divided into 6 equal sections: 3 sections are red and 3 sections are yellow. a

If the spinner lands on yellow on the ﬁrst spin, ﬁnd the probability that the second spin will be yellow.

Are the successive events of spinning yellow twice dependent or independent?

A photo album contains 5 photos, all numbered from 1 to 5. Four photos are drawn at random, with replacement. Find the probability that the photo with the number 2 is drawn: a

At least once

If three letters are randomly selected from the word CIRCUMFERENCE without replacement, ﬁnd the probability that: a

Exactly once

The letters are R, C, U in that order.

All three letters are E.

Find P (A ∩ B) given these probabilities for an experiment in which A and B are two possible events: P ( A∣B) = 0.4 and P (B) = 0.5

Use the two-way table to ﬁnd the probabilities. a

P (A)

P (A ∩ B)

P (B∣A′)

P (A′∣B′)

A A′ Total

Use the Venn diagram to ﬁnd: a

P (B′)

P (A ∪ B)

P ( A∣B)

B′

14 6 20

8 12 20

0.35

Total 22 18 40

0.1

0.2

0.35

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Amy owns ﬁve necklaces and four earrings. She selects one of the items at random for herself and then another item at random for her sister: a

Find the probability that Amy selects a necklace for herself.

Find the probability that both items Amy selects are earrings.

N N

Find the values of j, k, l, and m, given that P (A ∩ B) = 0.4, P (B) = 0.5, and that the events A and B are independent.

B j l

A A′

EN EE

B′ k m

In a survey, 640 participants were asked whether they enjoy watching basketball or volleyball. Of the total respondents, 120 enjoy watching both sports and 70 enjoy neither. The total number who enjoy watching basketball is three times the number who enjoy watching both sports. a

Complete the Venn diagram.

Of those who enjoy watching volleyball, ﬁnd the probability that a randomly chosen person also enjoys watching basketball.

If a randomly selected person does not enjoy watching volleyball, then ﬁnd the probability that they dislike watching basketball too.

120

In a group of 100 people, 60 said they went swimming, 40 said they went hiking, and 10 people said they neither swam nor hiked. a

Complete the table: Went hiking

Did not hike

Went swimming Did not swim Total b

414

Find the probability that a randomly selected person: i

Went swimming and hiking

Went swimming given they went hiking

iii

Went hiking knowing they went swimming

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Total

100

For breakfast, Elijah has something to eat and something to drink. He will either eat an egg sandwich or a muﬃn and will drink orange juice or iced coffee. • The chance of Elijah eating an egg sandwich is 0.6 • The chance of Elijah drinking iced coffee is 0.5

Construct a tree diagram showing all possible combinations of food and drink Elijah can have for breakfast and their associated probabilities.

Find the probability Elijah drinks orange juice and eats an egg sandwich.

Find the probability Elijah drinks orange juice or eats an egg sandwich.

Find the probability Elijah drinks iced coffee given he ate a muﬃn.

In a city, 40% of people prefer tea over coffee, 30% of people enjoy reading newspapers, and 15% of people both prefer tea and reading newspapers. A resident is chosen randomly. Find the probability that they: a

Enjoy reading newspapers given that they prefer tea

Prefer tea, knowing that they enjoy reading newspapers

A box contains 20 candies, of which 6 are sour. If Jenna randomly chooses 3 candies to eat, ﬁnd the probability that: a

All are sour.

At least 1 is sour.

A small magazine asked people from different states to send in a vote on whether they supported Daylight Saving Time. The diagram shows the proportion of people that voted YES or NO in NSW, ACT and QLD: NSW YES NO

ACT YES NO

QLD YES NO

NSW

ACT

QLD

YES

NSW YES

YES

ACT

NSW YES

QLD

NSW

QLD NO

Find the probability that a randomly chosen person voted “YES”.

Of those that are not from NSW, ﬁnd the probability that a randomly chosen person voted “NO”.

Considering just the voters from NSW and ACT, ﬁnd the probability that a randomly chosen person was from NSW and voted “YES”.

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Big ideas • Different representations of data highlight different characteristics of the data. • Comparing and interpreting data sets involves analysing statistical measures and visual representations to gain meaningful insights, make informed conclusions and understand the variation and relationships within the data.

Curriculum Outcomes VC2M10ST01 compare data distributions for continuous numerical variables using quartiles and interquartile range and appropriate data displays including boxplots, histograms and dot plots; discuss the shapes of these distributions in terms of centre, spread, shape and outliers in the context of the data VC2M10ST04 analyse claims, inferences and conclusions of statistical reports in the media and other places, by linking claims to displays, statistics and representative data, including ethical considerations and identification of potential sources of bias VC2M10AST01

calculate and interpret the mean and standard deviation of data and use these to compare data sets; investigate the effect of individual data values, including outliers, on the standard deviation VC2M10AST02

identify measures of spread, and understand their interpretation and usefulness with respect to different data distributions

Univariate statistics Chapter outline 9.01 Representations of data 9.02 Quartiles 9.03 Boxplots 9.04 Cumulative frequency 9.05 Compare data sets 9.06 Samples and populations 9.07 Optional: Standard deviation 9.08 Optional: Comparisons using standard deviation Chapter 9 review

418 436 449 464 476 491

501

In medicine, scientists use standard deviation to assist in understanding drug effectiveness and side effects!

9.01 Representations of data After this lesson, you will be able to… • understand the purpose and usage of different types of data displays. • identify the best representation for a given data set. • create and interpret frequency tables, histograms, bar charts, column graphs, dot plots and stem-and-leaf plots. • apply knowledge of data displays to solve problems related to a data set.

Frequency tables Before displaying data with a graph, we ﬁrst need to organise it into meaningful groups. Frequency tables are helpful to organise data sets. Colour of cars in the school carpark Colour Frequency white 14 red 2 blue 3 black 1 yellow 1

Ungrouped frequency tables are the best choice to organise: • Categorical data Example: colour of cars • Discrete numerical data when there is a small number of possible values Example: number of siblings

Heights of year 9 students Height (cm) Frequency 145 ≤ Height < 150 3 150 ≤ Height < 155 10 155 ≤ Height < 160 8 160 ≤ Height < 165 13 165 ≤ Height < 170 1

Grouped frequency tables are best for: • Continuous numerical data Example: heights • Discrete numerical data when the data can take a large number of possible values Example: attendees at football match

For grouped frequency tables, all intervals must be of equal size, and there should not be gaps between intervals. We can see class intervals in a few different formats: • 1 –5, this class includes 1 and 5. Used for discrete data only. • 1 – < 5, this class includes 1, but not 5. Used for continuous data. • 1 ≤ x < 5, this class includes 1, but not 5. Used for continuous data. This is the most common representation. Assume this to be the case unless it is specifically indicated otherwise. • 1 < x ≤ 5, this class does not include 1, but does include 5. Used for continuous data. To perform calculations with the data set, we usually find the class centre. The class centre is the middle score of each class interval. For the interval 1–5, the class centre would be

Selecting the interval width is important. If the intervals are too narrow or too wide, the shape of the data distribution will not be apparent. As a guide, 6 to 12 intervals will typically be most useful for moderately sized data sets. 418

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Example 1 What would be the most appropriate way of representing data from: a A survey conducted of 1000 people, asking them how many languages they speak? A

Leaving the data ungrouped and constructing a frequency table

Grouping the responses and constructing a frequency table

Create a strategy Consider whether the data will have a wide range of responses.

Apply the idea

Reflect and check

Since we would expect that the responses for An example of a frequency table representing the number of languages people speak to be responses to this survey is shown: fairly limited, e.g., 1 to 6 languages, we can leave Languages spoken Frequency the data ungrouped and construct a frequency 1 462 table. 2 353 The correct answer is option A. 3 98 4 44 5 36 6 7 b A survey conducted of 1000 people, asking them how many different countries they know the names of? A

Grouping the responses and constructing a frequency table

Leaving the data ungrouped and constructing a frequency table

Create a strategy Consider whether the data will have a wide range of responses.

Apply the idea

Reflect and check

The number of countries a person can name An example of a grouped frequency table would be extremely varied. They might know representing responses to this survey is shown: just a few or be able to name over 190 countries. Countries Frequency This means the most appropriate way to 1 − 30 514 represent this data is to group the responses 31 − 60 387 and construct a frequency table. The correct 61 − 90 86 answer is option A. 91 − 120 6 121 − 150 3 151 − 180 2 181 − 210 2 Notice that inequality notation was not used since the data is discrete.

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Idea summary Frequency tables are the best choice to organise categorical data and discrete numerical data when there is a small number of possible values. Grouped frequency tables are best for continuous numerical data and discrete numerical data when the data can take on a large number of possible values.

Histograms, bar charts and column graphs These data displays show the frequency of data values as the length of horizontal bars or vertical columns.

Interactive exploration Discover this concept in action online

mathspace.co

Histograms are similar to column graphs, with vertical columns used to display numerical data, but are used in different contexts.

Data type Gaps

Number of students

Grouping Original data set

Bar chart (column graph) Categorical or discrete with small number of possibilities Gaps between bars to show separate categories Ungrouped data Original data preserved

No gaps between bars to show continuity Grouped data Original data grouped and not preserved

Each student in a class was surveyed and asked about the colour of their eyes.

12 10 8 6 4 2

The data is categorical and the results are displayed in a bar chart as shown.

Green Blue Hazel Brown Eye colour

420

Histogram Continuous or discrete with large number of possibilities

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Number of students

Each student in a class was surveyed and asked their heights.

12 10

The data is numerical and continuous, so the results are displayed in a histogram. The data is continuous data because it can take any value within a range. The height of each column represents the frequency of each data item within each interval.

8 6 4 2 135 140 145 150 155 160 165 Height (cm)

Unless noted otherwise, the left boundary is included in the class interval, while the right boundary is not. In this histogram, 135 is included in the first class interval, but 140 is in the second class interval.

A histogram may have a logarithmic scale when data is unevenly spread over a very wide range of values. We take the logarithm of each value and then group the data using the logarithmic values. The populations of the largest 100 cities in Australia is a great case for a logarithmic scale histogram. With a traditional scale, we can tell there are only a few large cities, but cannot easily read off how many fall in each interval.

100

Frequency

80 60 40 20 0

0 0 0 0 0 00 0 00 0 00 0 00 00 0 0 0 0 0 0 0 10 20 30 40 50 0

Population

When the data is converted to a logarithmic scale, we can see more detail and read off the number of cities with more than one million people more easily.

Frequency

40 30 20 10 0

4 5 6 7 Logarithmic population

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Example 2 In product testing, the time when faults are detected in producing a certain piece of machinery is recorded over the course of a day. The frequency table shows the results. Working hours 0≤t<4 4≤t<8 8 ≤ t < 12 12 ≤ t < 16

Number of faults 10 14 20 16

a Construct an appropriate data display.

Create a strategy The data is grouped continuous numerical data, so a histogram is most appropriate. Use the class intervals to draw the horizontal axis and the maximum frequency to scale the vertical axis, then draw the bars on the histogram.

Apply the idea We can draw a histogram using the class interval boundaries as the labels on the horizontal axis and the frequencies on the vertical axis. Number of faults

Faulty machinery 20 18 16 14 12 10 8 6 4 2 0

b How many faults were recorded throughout the day?

Create a strategy Add the frequency columns to find the total number of faults recorded.

Apply the idea Each value in the frequency column represents a fault that was recorded. Number of faults = 10 + 14 + 20 + 16 = 60

Add the frequencies Evaluate

There were 60 total faults recorded throughout the day.

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4 8 12 Working hours

c How many faults were recorded before 8 working hours had occurred?

Create a strategy

Apply the idea

Check the frequencies where the working hours If we look at the ﬁrst and second rows of the are less than 8. table, we see that there were 10 + 14, which equals 24 faults, so 24 faults happened before the 8 working hours ended.

d How many faults were recorded in the last 4 hours of the working day?

Create a strategy

Apply the idea

Look at the frequency for the last interval, which corresponds to the last 4 hours of the working day.

The last row of the table shows that 16 faults were recorded in the last 4 hours of the working day.

Example 3 The histogram shows the population of 30 different Australian cities plotted on a logarithmic scale.

Population of Australian cities 16 14 Frequency

12 10 8 6 4 2 0

5 6 Log Population

a One city has a population of 282 673 people. What is the log10 of 282 673 rounded to two decimal places?

Create a strategy Enter log10 282 673 into your calculator.

Apply the idea log10 282 673 = 5.451 284 328 08 ≈ 5.45

Evaluate using the calculator Round the answer

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b What population does the number 4 represent on the log population scale?

Create a strategy Since the values along the x-axis are logarithmic results and population is the input, we want to ﬁnd the value for x where log10 x = 4.

Apply the idea log10 (104) = 4 4

10 = 10 000

Definition of logarithms Evaluate

The number represents a population of 10 000 on the log population scale.

c How many cities have a population of at least 1 000 000?

Create a strategy

Apply the idea

Identify where a population of 1 000 000 is represented on the log population scale, then count the relevant frequencies.

We have: log10 1 000 000 = 6 This means that any city in the 6 ≤ c < 7 class interval (denoting c as the log population of the cities) will have at least a population of 1 000 000. Therefore there are 5 cities with a population of at least 1 000 000.

d How many cities have a population of between 10 000 and 1 000 000?

Create a strategy

Apply the idea

Identify where a population of 10 000 and 1 000 000 is represented on the log population scale, then count the relevant frequencies.

We have: log10 10 000 = 4

log10 1 000 000 = 6

This means that any city in the 4 ≤ c < 5 and 5 ≤ c < 6 class intervals will have a population between 10 000 and 1 000 000. In total, there are 11 + 14 = 25 cities with populations within this range.

Idea summary Column graphs are used for categorical or numerical discrete data. They can have vertical or horizontal columns where the heights of the columns represent the frequencies. Histograms are used for numerical continuous data. They appear similar to column graphs but do not have spaces between the columns.

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Dot plots Dot plots are a graphical way of displaying the distribution of numerical or categorical data on a simple scale with dots representing the frequency of data values. They are best used for small to medium size sets of data and good for visually highlighting how the data is spread and whether there are any gaps in the data or outliers.

Days per week spent exercising

In a dot plot, each individual value is represented by a single dot, displayed above a horizontal line. The graph appears similar to a pictograph or column graph with the number of dots representing the total count. • To correctly display the distribution of the data, the dots must be evenly spaced in columns above the line. • The scale or categories on the horizontal line should be evenly spaced. • A dot plot does not have a vertical axis. • The dot plot should be appropriately labelled.

Example 4 Sally is a casual nurse. She used a dot plot to keep track of the number of shifts she did each week for a number of weeks.

Shifts per Week

a Over how many weeks did Sally record her shifts?

Create a strategy Each dot represents 1 week, so count all the dots in the graph.

Apply the idea Number of weeks = 5 + 3 + 1 + 8 + 2 + 5

Add the number of dots in each column

= 24 Evaluate

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b For how many weeks did she work 3 shifts?

Create a strategy

Apply the idea

Count the number of dots above 3 shifts on the dot plot.

The dot plot shows 5 dots at 3 shifts, indicating there were 5 weeks where she worked 3 shifts.

c How many weeks did she work less than 7 shifts?

Create a strategy

Apply the idea

Count the number of dots with less than 7 shifts on the dot plot.

Count all the dots at 3, 4, 5 and 6 shifts. By adding up the dots (5 + 3 + 1 + 8 = 17), we ﬁnd that she worked for a total of 17 weeks with less than 7 shifts.

d When Sally works at least 6 shifts a week, she buys a weekly train ticket. What proportion of the time did she buy a weekly train ticket?

Create a strategy Divide the number of weeks where she worked at least 6 shifts by the total number of weeks.

Apply the idea We count all the dots with a score of 6 or more. Examining the dot plot, we sum the dots, 8 + 2 + 5 = 15, indicating she worked for 15 weeks with at least 6 shifts. The total number of weeks, found in part (a), is 24. The proportion of time Sally bought a weekly train ticket is the time.

, which can be simplified as

Idea summary In a dot plot, each individual value is represented by a single dot, displayed above a horizontal line. The number of dots represents the total number of data values. We can read off the frequencies by counting the dots. They are best used for small to medium size sets of data and are good for visually highlighting how the data is spread and whether there are any gaps or outliers. 0

Days per week spent exercising

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Stem-and-leaf plots A stem-and-leaf plot, or stem plot, is used for organising and displaying numerical data. It is appropriate for small to moderately sized data sets. The graph is similar to a column graph on its side. An advantage of a stem-and-leaf plot over a column graph is the individual scores are retained and further calculations can be made accurately. In a stem-and-leaf plot, the right-most digit in each data value is split from the other digits, to become the ‘leaf’. The remaining digits become the ‘stem’. Stem

Leaf

0 3 6

5 5 6

0 3 6 8

The data values 10, 13, 16, 21, 26, 27, 28, 35, 35, 36, 41, 41, 45, 46, 49, 50, 53, 56, 58 are displayed in the stem-and-leaf plot shown.

6 7 8 1

5 6 9

Key: 2∣1 = 21 • The stems are arranged in ascending order, to form a column, with the lowest value at the top. • The leaf values are arranged in ascending order from the stem out, in rows, next to their corresponding stem. • A single vertical line separates the stem-and-leaf values. • There are no commas or other symbols between the leaves, only a space between them. • To correctly display the distribution of the data, the leaves must line up in imaginary columns, with each data value directly below the one above. • A stem-and-leaf plot includes a key that describes the way in which the stem and the leaf combine to form the data value.

Example 5 The stem-and-leaf plot shows the age of people entering through the gates of a concert in the first 5 seconds.

Stem

Leaf

0 2 3 4 5 7 7 8

2 2 4 7

2 6 8 9 9

4 5

9 Key: 1∣2 = 12

a How many people passed through the gates in the ﬁrst 5 seconds?

Create a strategy Each leaf represents the age of a person, so count the number of leaves.

Apply the idea To ﬁnd the number of people surveyed, we can count the number of leaves in the stem-and-leaf plot, since each data entry will have one leaf.

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There are 6 people with an age between 10 and 19, 8 people in their 20s, 4 people in their 30s, and 1 person in their 50s, so we can add all these numbers to ﬁnd the total number of people. Total number of people = 6 + 8 + 4 + 1 = 19

b What was the age of the oldest person?

Create a strategy

Apply the idea

Find the largest number recorded.

The largest number will be in the largest stem which is 5, and have the largest leaf, which is 9. This stem and leaf make the number 59. The oldest person is 59 years old.

c What percentage of the concert-goers were under 24 years old? Round your answer to one decimal place.

Create a strategy

Apply the idea

Divide the number of people whose ages are less than 24 by the total number of people.

There are 9 people who are less than 24 years old, and the total number of people is 19.

Then, convert to a percentage by multiplying by 100%.

This gives us a fraction of

. Now, we need to

convert the fraction to a percentage. × 100% = 47.4% 47.4% of the concert-goers were under 24 years old.

Idea summary Stem

Leaf

0 3 6

5 5 6

0 3 6 8

6 7 8 1

5 6 9

A stem-and-leaf plot, or stem plot, is used for organising and displaying numerical data. It is appropriate for small to moderately sized data sets. An advantage of a stem-and-leaf plot is that individual scores can be seen. The right-most digit in each data value is split from the other digits, to become the ‘leaf’. The remaining digits become the ‘stem’.

Key: 2∣1 = 21

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Choosing an appropriate data display There are many considerations to be made when choosing an appropriate data display for a data set which has been collected. Histograms, dot plots, and stem-and-leaf plots have their own strengths and drawbacks.

Example 6 Determine the best type of data display(s) for each set of data. a The number of goals scored in each of the weekly school futsal games for the year.

Create a strategy First, decide the size and possible range of values of the data set.

Apply the idea

Reflect and check

The number of games in a season would be around 40 which is not too large. The number of goals per game would not vary greatly, likely between 0 and 5.

It could be difficult to find appropriate bin widths for a histogram without hiding the shape of the data.

A dot plot or bar chart would be appropriate.

b The height of all 196 Olympic gymnasts in the most recent summer games

Apply the idea

Reflect and check

Since there is a large number of data points and the data is continuous, a histogram would be the best graph to summarize the data. Since there might be significant height differences between male and female gymnasts, a histogram will show any gaps in the data.

Technology can ease some of the inefficiencies of graphing data. With technology, we can create and compare different displays to determine which visual best reflects the data.

Idea summary Strengths Histogram Bar chart Dot plot Stem-andleaf plot

Drawbacks

Easily display large or spread out values of a data set; shape, centre and measures of spread are clearly visible. Easily display large data sets; shape clearly visible. Useful for individual data values. Shape is clearly visible. Useful for individual data values. Shape is clearly visible.

Cannot see individual data values Only used for categorical or discrete data Not suitable for large quantities of data Not suitable for large quantities of data

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9.01 Practice questions Understanding 1

Explorer

Adventurer

1−5

Trailblazer

2−6

Lucy counted the number of carrots she harvested from her garden over a few months and recorded the data in a table.

Month 1 2 3 4 5 6 7 8

In which months did Lucy harvest the same number of carrots?

Carrots 55 58 56 72 65 61 72 70

Mr. Hopkins recorded the hair colours of the students in his class in a column graph. Students’ hair colours 10

Number of students

9 8 7 6 5 4 3 2 1 0

Bonde

Brown

Black

Red

Colour

430

Which hair colour is most common?

Which hair colour is least common?

How many more students have red hair than blonde hair?

For the data provided in the table. a

Find the range.

Find the mode.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Score 68 69 70 71 72 73

Frequency 16 41 30 31 49 29

The stem-and-leaf plot shows the ages of 20 employees in a company. a

Find the number of employees in their 30s.

What is the age of the oldest employee?

What is the age of the youngest employee?

Leaf 2

0 1

0 2 4 8 8

3 4 8

1 1

2 8 8 9 2 5

Key: 1∣2 = 12 years old 5

Find the class centre for the class interval 25–45.

How do we determine the frequency of each score or group of scores using: a

Fluency 7

Ex 2

A histogram

7, 9−11, 13, 16

A stem-and-leaf plot

7−12, 16

7−12, 14−16

For the surveys, should the data be grouped or ungrouped when constructing a frequency table? a

A survey of 2000 people, asking them how many pets they own

A survey of 2000 people, asking them how many different animals they can list

A frequency table shows the resting heart rate (in beats per minute) of some people taking part in a study.

Heart rate 45–54 55–64 65–74 75–84

Complete the table.

How many people took part in the study?

How many people had a resting heart rate between 55 and 74?

How many people had a resting heart rate below 65?

Class centre

Frequency 15 25 26 30

The ages of people (in years) when they bought their ﬁrst car from a particular neighbourhood are displayed on the histogram: Age Frequency 16 ≤ Age < 18 18 ≤ Age < 20 20 ≤ Age < 22 22 ≤ Age < 24 24 ≤ Age < 26 26 ≤ Age < 28

Age of first car purchase 20 16 Frequency

Ex 1

A dot plot

12 8 4 0

18 20 22 24 26 28 Age

Complete the frequency table based on the histogram.

How many people were surveyed?

How many people bought their ﬁrst car before they were 22 years old?

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Ex 5

A group of recent college graduates were asked how old they were when they worked their first paid job. Their responses are shown in the dot plot. a

How many people were surveyed?

How many people worked their first paid job at 15 years old?

How many people were younger than 16 when they worked their first paid job?

If a person is at least 16 years old, they can drive themselves to work. What proportion of people surveyed could have driven themselves to their first paid job?

How many students took the test?

What was the highest score on the test?

What test score occurred the most frequently?

Age of first paid job

The scores for a recent spelling test are shown in a stem-and-leaf plot. The scores are out of 100.

12 13 14 15 16 17 18 19 20 Age

Spelling test results 6

2 3 3 6 6 7

0 1

2 3 4 5 6 9 9 9

4 5 7 9

How many marks separate the highest score and the most frequent score?

Minutes taken 1 2 3

How many people took part in the survey?

What proportion of people surveyed took 1 minute to make a decision?

Miss Yen recorded her students’ favourite pets in the column graph. Students’ favourite pets 15

Hamster

Dog

Cat

Fish

Pets

432

Which type of pet is most popular?

Which type of pet is least popular?

How many students picked a ﬁsh as their favourite pet?

How many students were surveyed in total?

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4 5 8

Key: 1∣2 = 12

In a survey, some people were asked approximately how many minutes they take to decide between brands of a particular product. The results are are shown in a frequency table.

Number of students

Ex 4

Bird

Frequency 13 17 12

An amusement park recorded the number of people that visited during a week of the school holidays and displayed the data in a column graph. Visitors to Adventure Land 290

Number of Visitors

270 250 230 210 190 170 150 130 Mon

Tues

Wed

Thurs

Fri

Sat

Sun

Day

Ex 3

Ex 6

On what day did Adventure Land have the most visitors?

How many more visitors were there on Thursday than Sunday?

On what day did Adventure Land have 220 visitors?

The histogram shows the income (in dollars) of 24 employees plotted on a logarithmic scale. a

A employee has an income of $46 974. What is the log10 of 46 974 rounded to two signiﬁcant ﬁgures?

What income does the number 5 represent on the log income scale?

How many people have an income of at least $10 000?

What percentage of people have an income less than $100 000?

How many people have an income of at least $10 000 but less than $10 000 000?

Frequency

12 11 10 9 8 7 6 5 4 3 2 1 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 log income

Determine whether a histogram, dot plot, or stem-and-leaf plot would be most appropriate for these data sets. Explain your answer. a

Heights of students in a cohort size of 300

Marks of students in a class size of 20

The results from a survey of 15 people asking how many pets they have

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Reasoning 17

17, 18

17−19

The frequency table shows the data distribution for the length of leaves collected from a species of tree in the botanical gardens.

Leaf length (mm) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 84

How many leaves less than 60 mm were collected?

What was the most common interval of leaf length collected?

Are leaves more likely to be greater than 60 mm in length or less than 60 mm in length? Explain your reasoning.

For three column graphs:

Tom

John Name

Kate

50 45 40 35 30 25 20 15 10 5 0

Tom

Is the tallest person the oldest? Explain your reasoning.

Is the heaviest person the oldest? Explain your reasoning.

John Name

Kate

Height 140

Is the heaviest person the tallest? Explain your reasoning.

Is the youngest person the lightest? Explain your reasoning.

Is it possible for a 10-year old to be lighter than an 8-year old based on the data displayed? Explain your reasoning.

Centimetres

120

100 80 60 40 20 0

Tom

John Name

Kate

The frequency table shows the average time spent travelling to work for 52 people. a

434

10 9 8 7 6 5 4 3 2 1 0

Mass

Kilograms

Years

Age

Frequency 5 11 19 49 43

Construct a histogram to display the data shown in the frequency table.

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Commute time (minutes) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100 Total

Frequency 15 17 10 6 4 52

Determine whether each of the statements about the data is accurate and explain. i

he data suggests that living close to work is not a priority for people. Roughly equal T proportions of people live less than 40 minutes away as those that live more than 40 minutes away.

he data suggests that people prefer a shorter commute to work. The majority live T within 40 minutes travel, and in general the longer the commute the less people there are in that category.

iii

he data shows that everyone lives within an hour travel from their work, with the T most amount of people living between 20 and 40 minutes away.

he data shows that most people travel to work by car or by walking, since most T travel times are fairly short, and only a few people travel by bus or train.

Problem-solving 20

The histogram shows the length of time, in hours, that Daniel parked his car in a car park each day over a certain number of days. a

Estimate the total number of hours that Daniel’s car was parked in a car park.

Daniel received 152 hours of free parking through his job, and he is charged $4.30 per hour for any additional time. How much was he charged for parking?

20−22

The histogram shows the weights (in kilograms) of 26 zoo animals plotted on a logarithmic scale. a

A monkey has a weight of 45.8 kg. What is the log10 of 45.8 rounded to two signiﬁcant ﬁgures?

What weight (in kg) does the number −2 represent on the log weights scale?

How many animals have a weight of at least 1000 kg?

What percentage of animals have a weight less than 0.1 kg? Round your answer to two signiﬁcant ﬁgures.

How many animals have a weight of at least 0.1 kg but less than 100 kg?

Frequency

20, 22

Number of days

12 11 10 9 8 7 6 5 4 3 2 1 0

0 1 2 3 4 5 6 7 Length of parking time (in hours)

−2 −1 0 1 2 3 log weights

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The table lists the oil production in barrels per day for 10 countries. Construct a frequency histogram of oil production using a log scale. Country

Oil production (barrels per day)

Saudi Arabia

10 625 000

Norway

1 763 000

Ecuador

555 000

Brunei

113 000

France

61 000

Albania

21 000

Finland

13 700

Portugal

7100

Japan

4000

Costa Rica

300

log10 oil production

9.02 Quartiles After this lesson, you will be able to… • identify the lower quartile, median and upper quartile in a data set. • calculate the interquartile range in a data set. • apply the concept of quartiles and IQR to solve problems.

Quartiles The range is a measure of spread based on the minimum and maximum in a data set, but does not tell us about the spread of data within these two values. Range = Maximum − Minimum Maximum : The highest value in the data set Minimum : The lowest value in the data set The median is a measure of centre, and tells us where the middle of the data set is. For a data set with n items, Median is the

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value

To get a better picture of the internal spread in a data set, we can ﬁnd the quartiles of a data set, which can be used for a measure of spread called the interquartile range (IQR). First quartile

Second quartile Q1

2, 3, Minimum

5+5 2 Q1 = 5 Q1 =

Third quartile Q2 10, 11, Median

13,

Fourth quartile

16,

Q3 18,

22, 39, Maximum

16 + 18 2 Q3 = 17 Q3 =

Quartiles are scores at particular locations in the data set-similar to the median, but instead of dividing a data set into halves, they divide a data set into quarters. Minimum Lower quartile, Q1 Median, Q2 Upper quartile, Q3 Maximum

Lowest value At most 25% of the data is below this value 50% of the data lies on either side of this value At most 25% of the data is above this value Highest value

Let’s look at how we would divide up some data sets into quarters now. Make sure the data set is ordered before ﬁnding the quartiles or the median.

Median

First locate the median, between the 4th and 5th scores:

Now there are four scores in each half of the data set, so split each of the four scores in half to ﬁnd the quartiles. Q1

Median

We can see the ﬁrst quartile, Q1 is between the 2nd and 3rd scores. Similarly, the third quartile, Q3 is between the 6th and 7th scores.

For 9 scores, the 5th term is the median. There are four terms on either side of the median. So Q1 is still between the 2nd and 3rd scores and Q3 is between the 6th and 7th scores.

Median

For 10 scores, the median is between the 5th and 6th scores. This time, however, there are 5 scores on either side of the median. So Q1 is the 3rd term and Q3 is the 8th term.

Median

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Each quartile represents 25% of the data set. For a data set with n items, Q1 is the

value

Median is the Q3 is the

value value

Example 1 Here are Ray’s scores from his last 13 rounds of golf played: 66, 66, 68, 68, 70, 78, 80, 84, 106, 116, 126, 130, 132 a What is his median?

Create a strategy Use the formula:

th where n is the total number of scores, to ﬁnd the position of the median.

Apply the idea There are n = 13 scores in the list. Position of median = = 7th Median = 80

Substitute n = 13 Evaluate Choose the 7th score

b What is the lower quartile?

Create a strategy Find the median of the lower half of the scores excluding the median.

Apply the idea The lower half of the scores are: 66, 66, 68, 68, 70, 78. Lower quartile = = 68

Average the middle scores Evaluate

c What is the upper quartile?

Create a strategy Find the median of the upper half of the scores excluding the median.

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Apply the idea The upper half of the scores are: 84, 106, 116, 126, 130, 132. Upper quartile =

Average the middle scores

= 121

Evaluate

Idea summary • Q1 is the ﬁrst quartile (sometimes called the lower quartile). It is the middle score in the bottom half of the data set. • The median, sometimes called Q2, is the second quartile. • Q3 is the third quartile (sometimes called the upper quartile). It is the middle score in the top half of the data set.

Interquartile range The interquartile range (IQR) is the difference between the third quartile and the ﬁrst quartile. 50% of scores lie within the IQR. Since it focuses on the middle 50% of the data set, the interquartile range often gives a better indication of the internal spread than the range does, and it is usually not affected by outliers.

IQR = Q3 − Q1 Q1

Lower quartile

Upper quartile

Interactive exploration Discover this concept in action online

mathspace.co

Example 2 Use this frequency table to: Score 5 14 18 24 32 38 50

Frequency 1 1 3 2 1 5 5

a Find the number of scores.

Create a strategy Add all the frequencies.

Apply the idea Number of scores = 1 + 1 + 3 + 2 + 1 + 5 + 5

Find the sum

= 18

Evaluate the addition

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b Find the median.

Create a strategy Use the formula: median.

th where n is the total number of scores, to ﬁnd the position of the

Apply the idea There are 18 scores in the list. Position of median = = 9.5

Substitute n = 18 Evaluate

Therefore the median is between the 9th and 10th score. Using the frequency table we can see that the 9th score is 38 and the 10th score is 38.

c Find the lower quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the lower half of the scores.

There are 9 scores in the lower half excluding the median. The lower quartile will be the 5th score from the frequency table. Lower quartile = 18

d Find the upper quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the upper half of the scores.

There are 9 scores in the upper half excluding the median. The upper quartile will be the 5th in the upper half, which is the 9 + 5 = 14th score overall. Upper quartile = 50

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e Find the interquartile range.

Create a strategy Use the interquartile range formula: IQR = Q3 − Q1

Apply the idea IQR = 50 − 18

Substitute the quartiles

= 32

Evaluate

Example 3 Answer the questions using the dot plot shown:

a Find the total number of scores.

Create a strategy

Apply the idea Number of scores = 15

Count the number of dots in the dot plot.

Count the dots

b Find the median.

Create a strategy Use the formula:

th where n is the total number of scores, to ﬁnd the position of the median.

Apply the idea Position of median = = 8th

Substitute n = 15 Evaluate the division

The 8th score in the dot plot is on 15. Median = 15

c Find the lower quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the lower half of the scores.

There are 7 scores in the ﬁrst half excluding the median. The lower quartile will be the 4th score in the dot plot. Lower quartile = 9

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d Find the upper quartile of the set of scores.

Create a strategy

Apply the idea

Find the median of the upper half of the scores.

There are 7 scores in the second half excluding the median. The upper quartile will be the 8 + 4 = 12th score in the dot plot. Upper quartile = 16

e Find the interquartile range.

Create a strategy

Apply the idea IQR = 16 − 9

Use the interquartile range formula: IQR = Q3 − Q1

Substitute the quartiles

Evaluate

Idea summary To calculate the interquartile range:

IQR = Q3 − Q1 IQR Q1 Q3

is the interquartile range is the ﬁrst quartile is the third quartile

9.02 Practice questions Explorer

Understanding 1

1−4

Adventurer

Trailblazer

1−4

Find the median for each data set of scores: a

2, 10, 28, 35, 50

8, 9, 11, 12, 13, 15

41, 42, 43, 44, 49, 50, 51, 53, 58

3, 4, 6, 7, 7, 8, 12, 14

For each data set, the median is shown. Find the upper and lower quartiles. a

0, 1, 4, 6 8, 9, 11 Median

22, 24, 25, 29, 32, 34 Median: 27

5, 11, 19, 20, 22 24, 26, 27, 27

14, 15, 16, 16, 17, 18, 19, 20, 24, 25, 30, 36

Median Median: 18.5

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What is the interquartile range for this data set? Q1 = 4

Q3 = 14.5

1, 2, 3, 5, 5, 7, 11, 14, 15, 16 Minimum Median = 6

Match the term with the deﬁnition. i

Highest value

50% of the data lies on either side of this value

iii

At most 25% of the data is above this value

The lowest value

At most 25% of the data is below this value

Minimum

Lower quartile, Q1

Median

Upper quartile, Q3

Maximum

Fluency Ex 1

Maximum

5, 6, 7a–d, 8, 9, 11−14, 17

5, 7c–f, 8−11, 13, 15−17

5, 7e–h, 8, 10, 11, 15−18

The data set shows Rachelle’s scores from her last 13 rounds of golf played: 56, 58, 60, 62, 64, 72, 74, 78, 98, 108, 120, 124, 128

Find her median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

The data set shows Luke’s scores from his last 17 exams: 42, 46, 48, 51, 52, 54, 56, 68, 72, 76, 78, 82, 85, 86, 88, 92, 96

Find his median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

For each set of scores: i

Sort the scores in ascending order.

Find the number of scores.

iii

Find the median.

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

40, 39, 15, 17, 10, 6, 24

−4, −6, −1, 7, 9, 7, 9

8, 20, 19, 4, 15, 14, 10

42, 28, 22, 40, 20, 54, 32, 43

84, 85, 79, 71, 69, 88, 82, 78

102, 115, 110, 113, 100

228, 205, 198, 202, 207, 197

19.5, 29.6, 39.1, 22.4, 15.8, 46.2, 35.4

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8 Ex 2

The table shows the luggage weight, in kilograms, of 30 passengers: a

Find the mean check in weight, rounded to two decimal places.

In kilograms, ﬁnd the:

c 9

Ex 3

Median

Lower quartile

iii

Upper quartile

Frequency 4 5 5 3 4 6 3

Matches 45 46 47 48 49 50 51 52

Frequency 1 2 5 3 7 25 6 2

Phone calls 0 1 2 3 4 5 6 7

Frequency 8 5 10 6 6 7 3 2

In which quartile does the mean lie?

A sample of matchboxes were selected for quality control and the number of matches in each box recorded in a given frequency table. a

How many matchboxes were sampled?

Find the range of the data.

Find the interquartile range.

A group of students were asked how many phone calls they had made the previous day. The information was collected in a given frequency table. a

How many students were surveyed?

Find the range of the data.

Find the interquartile range.

For this dot plot:

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Weight 16 17 18 19 20 21 22

9 10 11 12 13 14 15 16 17 18

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Find the range.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The dot plot shows the ages of customers in a mobile phone store in one day:

Age of customers

Find the lower quartile.

Find the interquartile range.

Find the upper quartile.

The stem-and-leaf plot shows the number of hours students spent studying during an entire semester: a

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Hours of study 6

2 7

0 1

2 2 4 7 9 2 5 7

Key: 5|2 = 52 For this bar chart: a

Organise the data into a frequency table.

Find the median score.

Find the lower quartile score.

Find the upper quartile score.

Find the interquartile range.

5 4 Frequency

3 2 1 0

The column graph shows the number of pets that each student in a class owns. a

Find the lower quartile.

Find the upper quartile of the set of scores.

Find the interquartile range.

Frequency

10 9 8 7 6 5 4 3 2 1 0

8 7 Score

2 3 4 5 6 Pets

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Dylan records the number of ﬁsh he catches for each ﬁshing trip over a period of time:

Frequency

20 18 16 14 12 10 8 6 4 2 0

Number of fish caught

How many ﬁshing trips did he go on?

Find the range in the number of ﬁsh caught.

Find the interquartile range.

This list shows the number of points scored by a basketball team in each game of their previous season: 59, 67, 73, 82, 91, 58, 79, 88, 69, 84, 55, 80, 98, 64, 82

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Find the median value.

Find the lower quartile score.

Find the upper quartile score.

What percentage of games did the team score more than in the upper quartile?

What percentage of games did the team score less than in the lower quartile?

The stem-and-leaf plot shows the batting scores of two cricket teams, the Emerald Eagles and the Sapphire Sharks. Cricket team batting scores a Find the median score for the Emerald Emerald Eagles Sapphire Sharks Eagles. b

Find the median score for the Sapphire Sharks.

Find the range for the Emerald Eagles.

Find the range for the Sapphire Sharks.

Find the interquartile range for the Emerald Eagles.

Key: 6|1|2 = 16 and 12

Find the interquartile range for the Sapphire Sharks.

What are the differences between the distribution of scores for the Emerald Eagles and the Sapphire Sharks?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Reasoning

20−22

20−23

The bar graph shows the marks (out of 10) that students received on a spelling test: a

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

Explain the effect scores less than 7 have on the summary statistics.

28 24 Frequency

19, 22

20 16 12 8 4 0

7 8 Mark

These data sets each contain 10 scores:

Set 1: 10, 12, 14, 16, 18, 20, 22, 24, 26, 28

Set 2: 10, 12, 14, 16, 18, 40, 42, 44, 46, 48

Set 3: 10, 42, 44, 46, 48, 50, 52, 54, 56, 58

Which data set has the largest interquartile range and what does this tell you about the spread of data in this data set?

Identify if each set is symmetrical, skewed-left or skewed-right and justify your response.

There is a test to measure the Emotional Quotient (EQ) of an individual. Here are the EQ results for two groups of 21 people listed in ascending order: Group 1: 90, 90, 91, 92, 93, 94, 95, 95, 95, 97, 99, 100, 108, 114, 116, 116, 117, 118, 118, 122, 129 Group 2: 85, 85, 85, 86, 86, 88, 90, 95, 97, 98, 100, 100, 100, 100, 102, 105, 105, 105, 110, 110, 115

Calculate the range, interquartile range, median, and mean for both data sets.

Compare the mean of both data sets. What does this tell you about the EQ scores of the groups?

Compare the spread of both data sets. What does this tell you about the variability in EQ scores within the groups?

The dot plot shows the number of pets students have, for 25 students.

Number of pets

Find the lower quartile.

Find the percentage of students who have more pets than the upper quartile. Explain why it is not exactly 25%.

Find the upper quartile.

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Is the percentage of of data values between the upper and lower quartiles always exactly 50%? Explain why or why not, including an example to justify your conclusion.

Problem-solving 24, 25a–f 24

24, 25

24−26

In a competition, a contestant must complete 12 challenges earning as many points as possible. Her scores for the ﬁrst 11 challenges are: 38, 43, 45, 66, 67, 82, 92, 102, 105, 108, 119 Determine her score in the 12th round if the lower quartile of all of her 12 scores is 55.

For a week, the airline Fly Air decided to keep track of how many minutes behind schedule each ﬂight departed. 100 results are shown in the dot plot. a

Find the median. What is the meaning of the median in this scenario?

Find the lower quartile.

Find the upper quartile.

Find the interquartile range.

If a ﬂight is delayed for 10 minutes or more, the airline incurs a fee. Based on the dot plot, how many ﬂights did the airline incurred a fee for?

What percentage of ﬂights did the airline incur a fee for?

A rival airline, Fly High, had a median delay time during the same week of 45 minutes.

5 10 15 20 25 30 35 40 45 50 55 60 minutes

How many delayed ﬂights did Fly Air have that were longer than 45 minutes? h

What percentage of Fly Air’s ﬂights had delay times that were longer than Fly High’s median delay time?

To gain a place in the main race of a car rally, teams must compete in a qualifying round. The median time in the qualifying round determines the cut off time to make it through to the main race. Here are some results from the qualifying round: • 75% of teams finished in 159 minutes or less. • 25% of teams finished in 132 minutes or less. • 25% of teams finished between with a time between 132 and 142 minutes. a

Find the cut-off time required in the qualifying round to make it through to the main race.

Find the interquartile range in the qualifying round.

In the qualifying round, the ground was wet, while in the main race, the ground was dry. To make the times more comparable, the ﬁnishing time of each team from the qualifying round is reduced by 5 minutes. Find the new median time from the qualifying round.

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9.03 Boxplots After this lesson, you will be able to… • identify the five-number summary from a boxplot. • construct a boxplot using given data. • interpret outliers and their impact on data representation. • describe the shape of data distribution through a boxplot.

Boxplots Boxplots (also known as box-and-whisker plots) are a way of showing the five-number summary for a data set. The five-number summary is comprised of the following five statistics: Minimum Lower quartile Median Upper quartile Maximum

17 52 69 87 100

Minimum value

Median

Lower Quartile

Maximum value Upper Quartile

100

As you can see, the boxplot consists of a number line, a rectangle, and two horizontal lines (the whiskers). The box represents the middle 50% of the scores and its size tells us the interquartile range. 25% of the scores lie in each quartile. So for this example, 25% of the scores would lie between 17 and 52, 52 and 69, 69 and 87, and 87 and 100.

Minimum and outlier

Upper quartile Q3

Median Q2

Lower quartile Q1

Maximum ∼25%

∼25%

Interquartile range

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449

Outliers are data points which are very far away from the rest of the data set. On a boxplot, the outliers are shown as points or crosses. When there are outliers, the end of the whiskers may not be the minimum or maximum. Boxplot with outliers

Similar to other data displays, we can describe the shape by saying it is symmetrical, skewed right or left, or uniform. Symmetrical

A symmetrical data set is distributed around the centre with a similar frequency on the left and right. 0

10 15 20 25 30 35 Left skew

A left-skewed data set is where the majority of the data points have higher values, with some data points at lower values. It is sometimes called a negative skew. 0

10 15 20 25 30 35 Right skew

A right-skewed data set is where the majority of the data points have lower values, with some data points at higher values. It is sometimes called a positive skew. 0

10 15 20 25 30 35 Uniform

A uniform data set is evenly distributed across all values.

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10 15 20 25 30 35

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Keep in mind: • when describing skewed distributions, it’s better to use the median and interquartile range because they are less impacted by outliers. • when describing symmetrical or uniform distributions, it’s better to use mean and range because they take the values of all data points into account.

Interactive exploration Discover this concept in action online

mathspace.co

Example 1 Create a boxplot to represent the data in the table. Minimum Lower quartile Median Upper quartile Maximum

10 20 40 55 75

Create a strategy

Apply the idea

Construct a boxplot using the data from the table.

0 10 20 30 40 50 60 70 80 90 100

Example 2 For the boxplot shown, fill in the table with the corresponding values: Minimum Maximum Outliers Range IQR

0 4 8 12 16 20 24 28 32 36 40 44

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Create a strategy Use the illustration shown to determine the values for the table. Minimum and outlier

Upper quartile Q3

Median Q2

Lower quartile Q1

Maximum ∼25%

∼25%

Interquartile range

The range is the difference between the minimum and maximum. For the outliers, determine the points beyond the whiskers. For the IQR, measure the distance from the lower to the upper quartile, or the length of the box.

Apply the idea Here’s the complete table:

Minimum Maximum Outliers Range IQR

1 43 1, 41, 43 42 7

Example 3 The stem-and-leaf plot displays the scores of students in a class on an exam.

Leaf 6

Key: 6|1 = 61

a Construct the five-number summary.

Create a strategy

Apply the idea

• To find the minimum value, locate the smallest score in the lowest stem. • To find the maximum value, locate the highest score in the highest stem. • Calculate the median by finding the middle score if the number of scores is odd, or by averaging the two middle scores if even. • Determine the lower quartile by identifying the middle score of the values below the median. • Identify the upper quartile as the middle score of the values above the median.

Here’s the five-number summary:

452

Minimum Lower quartile Median Upper quartile Maximum

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

61 69 73 80 85

b Construct a boxplot for the data.

Create a strategy

Apply the idea

Use the five-number summary in part (a) to construct the boxplot.

Exam scores

c Describe the shape of the boxplot.

Apply the idea The boxplot shows a slight right skew.

Example 4 The boxplot shows the mathematics test results for Mr. Smith’s and Ms. Johnson’s classes. Mr. Smith’s class

Ms. Johnson's class

100

a Calculate the range of Mr. Smith’s class.

Create a strategy

Apply the idea

Use the whiskers of the boxplot to calculate the range.

95 − 70 = 25 Mr. Smith’s class has range of 25.

b Calculate the interquartile range of Ms. Johnson’s class.

Create a strategy

Apply the idea

Use the edge of the boxes to calculate the interquartile range.

85 − 70 = 15 Ms. Johnson’s class has an interquartile range of 15.

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c How much higher was the median of Mr. Smith’s class than Ms. Johnson’s class?

Create a strategy

Apply the idea

Use the line inside the box to determine the median of both classes

85 − 80 = 5 Mr. Smith’s class had a median of 5 marks higher.

d Which class performed better overall? Explain your answer.

Create a strategy

Apply the idea

Use the measures of spread and central tendency to compare the class marks.

Mr. Smith’s class had a higher maximum, median and minimum score than Ms. Johnson’s. Mr. Smith’s class also had a smaller range, indicating more consistent results. Thus, Mr. Smith’s class performed better overall.

Idea summary Boxplots are a graphical representation showing the ﬁve-number summary for a data set.

Minimum value

Median

Lower Quartile

Maximum value Upper Quartile

100

Outliers can be shown as dots or crosses and are far away from the rest of the data set. We can describe the shape based on the distribution of the data set. • • • •

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Symmetrical boxplots are symmetrical about the median. Left-skewed boxplots have the majority of data points with higher values. Right-skewed boxplots have the majority of the data points with lower values. Uniform boxplots have all quartiles the same width.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

9.03 Practice questions Understanding 1

1−4

Adventurer

Trailblazer

1−4

Fill in the missing numbers for each boxplot. a

Explorer

Minimum 20 Lower quartile 28 Median

Upper quartile

Maximum

Minimum 16 Lower quartile 33 Median

Upper quartile

Maximum

Minimum 26 Lower quartile 33 Median

Upper quartile

Maximum

For each five-number summary, find the: i

Range

Interquartile range

Minimum 26 Lower quartile 33 Median 41 Upper quartile 44 Maximum 45

Minimum Lower quartile Median Upper quartile Maximum

16 27 35 42 50

Is each statement true or false for the data set shown in the boxplot? a

The maximum is 16

The minimum is 8

The median is 11

There is an error as the median is not exactly in the middle of Q1 and Q3

The interquartile range is 3

The value 1 is an outlier

2 4 6 8 10 12 14 16 18 20

9.03 Boxplots mathspace.co

455

For the boxplot shown: a

What percentage of scores lie between each of these values? i

Q1 and Q3

Minimum and Q1

iii

Median and Maximum

Q1 and Maximum

Minimum and Q3

In which quartile(s) is the data the least spread out?

In which quartile(s) is the data the most spread out? 5, 6ab, 7, 8, 9abgi, 11a, 12

Fluency 5

5, 6bc, 7, 8, 9cdhi, 10, 11b, 13, 14

Minimum

Maximum

Range

Median

Interquartile range

6cd, 7, 8, 9efhj, 10−14

8 10 12 14 16 18 20

Construct a five-number summary for each boxplot: a

Scores

8 10 12 14 16 18 20

Scores

456

For the boxplot shown, ﬁnd the:

8 10 12 14 16 18 20

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

8 10 12 14 16 18 20

Scores

38 40 42 44 46 48 50 52 54 56 58 60

Ex 1

Construct a boxplot to represent the information in the tables: a

Ex 2

Minimum 5 Lower quartile 25 Median 40 Upper quartile 55 Maximum 65

Minimum Lower quartile Median Upper quartile Maximum

30 31 47 53 58

For the boxplot shown, ﬁnd: a

Minimum

Maximum

Outliers

Range

Interquartile range 0 4 8 12 16 20 24 28 32 36 40 44

Ex 3

For each of the data sets: i

Construct the ﬁve-number summary.

Draw a boxplot.

iii

Describe the shape of the boxplot.

14, 15, 20, 21, 23, 24, 27

21, 25, 29, 31, 32, 34, 39, 40

40, 12, 8, 20, 12, 28, 48, 36, 60

28, 4, 40, 16, 20, 48, 20, 44, 56

42, 43, 48, 54, 59, 51, 53, 50, 57, 45

60, 66, 67, 68, 69, 72, 73, 76, 77, 78, 86

Leaf

2 3 5 8 9

0 4 4 5 7 7

0 2 3 4 6 7

3 3 5 9

2 4 5

20 21 22 23 24 25 26 27 28 29 30 31

A cyclist measured her heart rate immediately after ﬁnishing each event in which she competed. The results are recorded in the boxplot. a

Find and interpret the range.

Find and interpret the interquartile range.

2 3 5 5 Key: 1|2 = 12

Key: 1|2 = 12 i

Leaf

10 11 12 13 14 15 16 17 18 19 20 21

Heart Rate

155 160 165 170 175 180 185 190 195 200

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Use the summary statistics to determine: i

The shape of each data set

The data set with the higher measure of centre

iii

The data set with the higher measure of spread

a Data set A Data set B

Mean 10 15

Median 12 15

IQR 2 5

Data set A Data set B

Minimum 1 10

Q1 4 18

Median 5 20

Range 5 10 Q3 10 21

Maximum 13 22

IQR 6 3

A mathematics test is given to two classes. The marks out of 20 received by students in each class are represented in the boxplots. a

Range 12 12

Class 10P

Complete the table: Class 10P

Class 10Q

Median Lower quartile Upper quartile Range Interquartile range Outliers b

Which class tended to score better marks?

Would the range or interquartile range be better for describing the spread of Class 10Q?

0 2 4 6 8 10 12 14 16 18 20 22

Class 10Q

0 2 4 6 8 10 12 14 16 18 20 22

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The boxplot shows the age at which a group of people obtained their driving licences: a

What is the oldest age?

What is the youngest age?

What percentage of people were aged from 18 to 22?

The middle 50% of responders were within how many years of one another?

In which quartile are the ages least spread out?

The bottom 50% of responders were within how many years of one another?

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Age

Ex 4

The boxplots represent the daily sales made by Carl and Angelina over the course of one month. Angelina's sales

Carl's sales

Calculate the range for Angelina’s sales.

Calculate the range for Carl’s sales.

By how much did Carl’s median sales exceed Angelina’s?

By comparing the middle 50% of sales for each person, identify which salesperson had the more consistent number of sales.

Which salesperson had a more successful sales month? Explain your answer.

Reasoning 15

15, 17−19

15−17, 19, 20

16, 19−22

Ten participants had their pulse measured before and after exercise with results shown in different boxplots. Compare the pulse rates before and after exercise for these participants. Pulse before exercise

Pulse after exercise

45 50 55 60 65 70 75 80 85 90

80 85 90 95 100 105 110 115 120 125

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How does a boxplot help us to understand the centre and spread within a data set?

The test scores of 11 students in Science and Music are listed: • Science: 52, 59, 87, 66, 74, 76, 73, 63, 91, 86, 82 • Music: 87, 63, 60, 43, 74, 82, 66, 61, 88, 60, 84

Construct a boxplot for the Science scores.

Construct a boxplot for the Music scores.

In which subject did students perform better overall? Explain your answer with reference to measures of central tendency and spread.

These boxplots shows the number of points scored by two basketball teams in each of their matches. Gamma Geckos

30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70

Delta Dragons

30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70

Compare the median score of the Gamma Geckos and the Delta Dragons.

Compare the range of scores for the Gamma Geckos and the Delta Dragons.

Compare the interquartile range for the Gamma Geckos and the Delta Dragons.

If the two teams were to play against each other in the next game, determine which team you expect to win. Justify your answer.

The boxplots summarise results from a medical study. The treatment group received an experimental drug to relieve cold symptoms, and the control group received a placebo. The boxplots show the number of days each group continued to report symptoms:

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Control group

Treatment group

8 10 12 14 16 18 20

Describe the shape of the data from the control group.

Describe the shape of the data from the treatment group.

Does the drug have a positive effect on patient recovery? Explain your reasoning.

A boxplot was created for a data set. The value 32 was a typo and should have been 23. a

Can we draw a new boxplot with this information? Explain.

How does this change the range?

How does this change the interquartile range? 10

Create a context where a data set might meet the given conditions. Justify your answer. a

A range between 1 and 4.

A small interquartile range and a large range.

The marks in an end-of-year exam for a class of students are shown: 52, 56, 59, 64, 66, 77, 78, 80, 80, 80, 81, 84, 86, 90, 95, 96 a

Construct a boxplot for the data.

Calculate the interquartile range.

What percentage of marks lie in the range 85 to 96?

Which values do the lowest 75% of scores lie between?

If the score 91 was added to the data set, what percentage of the marks are below Q3? Why isn’t this 75%?

Problem-solving 23, 26bd 23

23−25, 26a–c

Construct a boxplot for the data set: • Range = 14 • Lowest score = 2 • IQR = 8

23, 25, 26a–c

• 25% of scores are below 4 • 50% of scores are above 8

For each of the conditions, use ﬁve of the digits from 1 to 9 with no repetition to complete a ﬁve-number summary: min: ⬚, lower quartile: ⬚, median: ⬚, upper quartile: ⬚, max: ⬚

The smallest possible interquartile range with the largest possible range.

The range is three times the length of the interquartile range.

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Match the histograms on the left to the corresponding boxplots on the right: Histogram A

Box plot 1

0 1

9 10

1 2 3 4 5 6 7 8 9 10

Histogram B

Box plot 2

0 1

2 3 4 5 6 7 8 9 10

Histogram C

Box plot 3

1 2 3 4 5 6 7 8 9 10

Histogram D

0 1

1 2 3 4 5 6 7 8 9 10

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Box plot 4

9 10

Create a boxplot for the data shown in each graph. Monthly salaries of employees 30

Number of employees

20 15 10 5 20 25 30 35 40 45 50 Salaries (in thousands)

Debbie’s blog visitors

Number of days

Scores of students

10 9 8 7 6 5 4 3 2 1 0

100

Number of pages read

10 0

50 100 150 200 Number of visitors

Did you know?

Univariate statistics are vital in education for assessing student performance! Schools often use the mean and standard deviation of test scores to summarise overall achievement and identify areas for improvement. For instance, by analysing the range of scores in a maths exam, teachers can pinpoint whether the class grasped key concepts or if certain topics need to be revisited. 9.03 Boxplots mathspace.co

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9.04 Cumulative frequency After this lesson, you will be able to… • calculate cumulative frequencies from a frequency table. • construct a cumulative frequency table from a given frequency distribution table. • interpret and solve problems involving cumulative frequencies. • construct and interpret cumulative frequency polygons. • apply knowledge of cumulative frequencies and their graphical representations to solve problems.

Cumulative frequency We have seen that a frequency table allows us to group data into intervals and tells us how many data values are in that interval. Cumulative frequency is a ‘running total’ of the frequencies. To calculate it, we add an additional column to the frequency distribution table: Class interval 50 ≤ t < 55 55 ≤ t < 60 60 ≤ t < 65 65 ≤ t < 70 70 ≤ t < 75 75 ≤ t < 80 80 ≤ t < 85 Total

Frequency 5 10 25 26 40 49 28 183

Cumulative frequency 5 5 + 10 = 15 15 + 25 = 40 40 + 26 = 66 66 + 40 = 106 106 + 49 = 155 155 + 28 = 183

• The first value in the cumulative frequency column will always be the same as the first value in the frequency column. • To get the second cumulative frequency value we add the second frequency to the first cumulative frequency value, 5 + 10 = 15. • The second cumulative frequency value tells us there are 15 values in the interval 50 ≤ p < 60. • The third cumulative frequency value tells us there are 40 values in the interval 50 ≤ p < 65, and so on. • The final cumulative frequency value is always equal to the sum of the frequencies. In this case, there are 183 values in the entire data set, represented by 50 ≤ p < 85.

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Example 1 A principal wants to investigate the performance of students at his school in Performing Arts. To do this, he has the marks of each student studying Performing Arts collected into groups and put into a frequency table. Each group of marks is assigned a grade. The frequency table for this is shown. a Complete the cumulative frequency column. Grade E D C B A

Score (x) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100

Class centres

Frequency ( f ) 7 14 32 97 62

Cumulative frequency (cf )

Create a strategy The cumulative frequency of the ﬁrst row is the frequency of that row. For the following rows, add the frequency of each row to the cumulative frequency of the previous row.

Apply the idea The ﬁrst value in the cumulative frequency column is the same as the ﬁrst value in the frequency column, so we have 7. For the second cumulative frequency value, we have 7 + 14 = 21. Repeating the process to get the next cumulative frequency value, we have the completed table: Grade E D C B A

Score (x) 0 ≤ x < 20 20 ≤ x < 40 40 ≤ x < 60 60 ≤ x < 80 80 ≤ x < 100

Frequency ( f ) 7 14 32 97 62

Cumulative frequency (cf ) 7 21 53 150 212

b How many marks were recorded?

Create a strategy

Apply the idea

We can ﬁnd the total frequency in the last entry of the cumulative frequency column.

This value is the same as adding all the entries in the frequency column. Total frequency = 212

c Complete the sentence. ‘Approximately three quarters of the scores recorded are greater than ⬚’.

Create a strategy

This can be reworded to one quarter of the scores are below this value: ⬚’. 9.04 Cumulative frequency mathspace.co

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Apply the idea There are 212 scores, so one quarter of that is

= 53.

The cumulative frequency reaches 53 in the interval 40 ≤ x < 60, so 60 is the corresponding upper boundary score. ‘Approximately three quarters of the scores recorded are greater than 60’.

Example 2 For this table: Score (x) 10 11 12 13 14

Cumulative frequency (cf ) 7 15 18 20 26

a How many scores were there in total?

Create a strategy

Apply the idea

We can ﬁnd the total number of scores in the last entry of the cumulative frequency column.

This value is the same as adding all the entries in the frequency column. Number of scores = 26

b How many scores of 14 were there?

Create a strategy Find the difference between the cumulative frequency at 14 and the cumulative frequency at 13.

Apply the idea There were 26 scores after the 14s were counted and 20 before. Number of 14s = 26 − 20 =6

Subtract the values Evaluate

c How many scores were less than 13?

Create a strategy

Apply the idea

Find the corresponding cumulative frequency of The corresponding cumulative frequency of the the score 12. score 12 is 18, so there were 18 scores which are less than 13.

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Idea summary Cumulative frequency is a ‘running total’ of the frequencies. To calculate it, we add an additional column to the frequency distribution table.

Cumulative frequency graphs

Cumulative frequency

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

A cumulative frequency histogram will always increase in size from left to right. The frequency represented by any particular column will be equal to the difference in height between that column and the one before it.

Cumulative frequency

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

A cumulative frequency polygon, also known as an ogive, is a line representing the cumulative frequency and it joins the top right-hand corner of each column. Here is a frequency histogram and polygon displayed together.

Cumulative frequency distribution Global life expectancy, 2016 200 180 160 140 120 100 80 60 40 20 0 50 55 60 65 70 75 80 85 Life expectancy (years)

Here is a cumulative frequency polygon displayed on its own.

Cumulative frequency

We have already seen how frequency tables can be used to create a histogram. The cumulative frequencies can also be plotted to create another type of chart, called a cumulative frequency graph. This graph is made up of a cumulative frequency histogram and polygon.

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We can use a cumulative frequency polygon to find values such as the median and quartiles which are at 25%, 50% and 75% of the cumulative frequencies.

Example 3

Cumulative frequency

For this cumulative frequency histogram:

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

40 45 Class

a How many scores are there in total?

Create a strategy For a cumulative frequency histogram the total will be the cumulative frequency for the last column.

Apply the idea The 50 – < 55 column has the cumulative frequency of 14. So there 14 scores in total.

Reﬂect and check Notice this is different from a normal histogram where we would add up the frequencies of all of the columns. Since these columns represent cumulative frequency, this is already done.

b How many scores between 50 – < 55 occured?

Create a strategy Find the difference between the cumulative frequency at 50 – < 55 and at 45 – < 50.

Apply the idea There were 14 scores after the 50 – < 55s were counted and 12 before. Number of 50 – < 55s = 14 − 12 =2

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Subtract the values Evaluate

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

c Construct a cumulative frequency polygon for the data.

Create a strategy The cumulative frequency polygon connects the upper right-hand corner of each column.

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Reﬂect and check On its own, the ogive would appear as shown.

Cumulative frequency

Apply the idea

40 45 Class

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

40 45 Class

Idea summary A cumulative frequency polygon, also known as an ogive, is a line graph connecting the upper right-hand corner of each column of a cumulative frequency histogram.

9.04 Practice questions Understanding 1

Explorer 1−3

Adventurer 2−4

Trailblazer 2−4

Complete the frequency table based on the data set: 2.1, 5, 4, 5, 5.15, 5, 3.8, 4, 5.32, 5.5, 1.25, 3, 3.75, 3, 5, 2.82, 4.6, 1, 6, 5, 6.56, 3, 1, 1.6, 6 Score 1≤x<2 2≤x<3 3≤x<4 4≤x<5 5≤x<6 6≤x<7

Frequency

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Complete the cumulative frequency column. Score (x) 1≤x<5 5≤x<9 9 ≤ x < 13 13 ≤ x < 16 17 ≤ x < 20

470

Frequency ( f ) 8 5 8

Cumulative frequency (cf ) 8 21 26

Is each statement true or false? a

The height of the last column tells us the total.

The cumulative frequency must always add up to 100.

To find the cumulative frequency we add the frequency to the previous cumulative frequency.

To find the frequency we subtract the previous cumulative frequency from the cumulative frequency.

A cumulative frequency polygon may have segments that are decreasing.

Fluency Ex 1

Cumulative frequency (cf )

Complete the cumulative frequency table: Score (x) 12 ≤ x < 13 13 ≤ x < 14 14 ≤ x < 15 15 ≤ x < 16

Frequency ( f ) 4 5 9 5 4

5−10, 13

For the frequency table: a

Complete the cumulative frequency column.

Calculate the total frequency.

Approximately half of the scores recorded are greater than what score?

For the given frequency table: a

Complete the cumulative frequency column.

Calculate the total frequency.

Approximately one third of the scores recorded are greater than what score?

5−10, 12, 13

6−14

Score (x)

Frequency ( f )

1–5 6–10 11–15 16–20 21–25 26–30

15 26 18 14 7 2

Score (x)

Frequency ( f )

20–24 25–29 30–34 35–39 40–44 45–49 50–54

7 18 25 12 8 4 1

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Cumulative frequency (cf )

Ex 3

For this table: a

Find the total number of scores recorded.

Find the number of times a score of 13 occurred.

Find the number of times a score less than 15 occurred.

Score (x) 11 12 13 14 15

Cumulative frequency (cf ) 9 13 19 27 30

For this cumulative frequency histogram: Score (x)

Frequency ( f )

20 ≤ x < 25 25 ≤ x < 30 30 ≤ x < 35 35 ≤ x < 40 40 ≤ x < 45 45 ≤ x < 50 50 ≤ x < 55 55 ≤ x < 60 60 ≤ x < 65 65 ≤ x < 70 70 ≤ x < 75

3 3 3 1 2 4 13 13 12 4 2

Cumulative frequency (cf ) 3 6 9 10 12 16 29 42 54 58 60

Class marks 70 60 50 40 30 20 10 0

20 25 30 35 40 45 50 55 60 65 70 75 Marks

Find the total number of scores recorded.

Find the number of times a score between 25 and 30 occurred.

Find the number of times a score between 50 and 55 occurred.

Find the percentage of scores that were less than 35.

For this cumulative frequency histogram: a

How many scores are there in total?

How many scores between 40– < 50 occurred?

Construct a cumulative frequency polygon for the data.

Cumulative frequency

Ex 2

100 90 80 70 60 50 40 30 20 10 0

20 30 40 50 60 70 80 Class

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The frequency table shows the time (t), in minutes, that it takes for people in a certain town to walk to the nearest park. Construct a cumulative frequency polygon for the data. Time (sec) 0 ≤ h < 10 10 ≤ h < 20 20 ≤ h < 30 30 ≤ h < 40 40 ≤ h < 50

Frequency 18 20 17 16 14

The graph shows the cumulative frequency polygon of the mass (in grams) of fish caught in a fishing competition: Construct a frequency histogram to depict the distribution of the mass of the fish.

170 Cumulative frequency 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10

Mass (g)

472

Construct a frequency table for the data represented in the cumulative frequency histogram.

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Cumulative frequency

50 100 150 200 250 300 350 400

20 18 16 14 12 10 8 6 4 2 0

80 85 90 95 100 Score

A pair of dice are rolled 50 times and the numbers appearing on the uppermost face are added to give a score. The results are recorded in the given table. a

What is the lowest possible score when a pair of dice are rolled?

What is the highest possible score when a pair of dice are rolled?

Complete the table by ﬁnding the cumulative frequency values.

Find the number of times a score of 8 occurred.

Find the number of times a score more than 9 occurred.

Find the number of times a score of at most 6 occurred.

Score (x)

Frequency ( f )

2 3 4 5 6 7 8 9 10 11 12

1 2 5 5 5 9 7 5 8 1 2

Cumulative frequency (cf )

The number of sightings of the Northern Lights were recorded across various Canadian locations over a period of 1 month. The list represents the number of sightings at each location: 11, 10, 10, 9, 7, 8, 8, 12, 12, 12, 12, 12, 12, 9, 9, 12, 9, 9, 8, 8 a

Complete the table.

In how many locations were there at least 8 sightings?

In how many locations were there less than 11 sightings?

Number of sightings 7 8 9 10 11 12

Number of locations ( f )

Cumulative frequency (cf )

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Reasoning 15

15, 16

16, 17

A jump rope competition has routines which are judged on a scale of 0 to 10, and scores are rounded to one decimal place. The average of the scores from the middle three judges makes the score for the routine. 50

Cumulative frequency

45 40 35 30 25 20 15 10 5 0

Score

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Explain how the frequency of a certain score can be determined from a cumulative frequency histogram.

What is the most frequent range of scores?

Families were asked how many times they contracted the ﬂu during winter. The information has been partially ﬁlled out in the table. Explain your reasoning for each part of the problem. a

Complete the frequency and cumulative frequency values in the given table.

Find the total number of families that responded.

Find the median number of times a family contracted the ﬂu.

Find the mean number of times a family contracted the ﬂu. Round your answer to two decimal places.

Construct a cumulative frequency histogram for the data. Explain the signiﬁcance of the histogram.

Find the number of families who contracted the ﬂu more than twice.

Score (x) 0 1 2 3 4

Frequency ( f ) 0

Cumulative frequency (cf ) 2

3 2

To find each key feature, would you rather use a histogram or cumulative frequency histogram? Explain your reasoning. a

Total

Frequency for one interval or value

Shape of distribution

Number of data points below a particular value

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Problem-solving 18

18−20

A 1500 m swimmer records her time (min:sec) over several training sessions. Her times are recorded in the histogram. Construct a cumulative frequency table for the data using the given intervals.

Find the total number of training sessions she completed.

Find the number of times she recorded a swim time faster than 16 : 40.

Find the percentage of swims that were less than 16 : 30.

10 9 8 7 6 5 4 3 2 1 0

16 :0 0 16 :10 16 :2 0 16 :3 0 16 :4 0 16 :5 0 17 :0 0

Cumulative frequency

18, 19

Time Recorded

The heights (cm) of 22 students in a class are listed: 164, 167, 158, 159, 150, 166, 150, 146, 149, 161, 164, 163, 152, 161, 157, 157, 153, 157, 156, 165, 162, 161

Construct a cumulative frequency histogram for the data. Use the discrete intervals of 146–150, 151–155, etc.

How many students are taller than 155 cm?

How many students are at most 155 cm tall?

How many students are taller than 150 cm but shorter than 156 cm?

Calculate the centre of the class 151–155.

For each graph, use the cumulative frequency curve to estimate: i

The median score

iii

The upper quartile

100 90 80 70 60 50 40 30 20 10

Score 1500 1900 2300 2700

The lower quartile

14 13 12 11 10 9 8 7 6 5 4 3 2 1

Score 550 600 650 700 750

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9.05 Compare data sets After this lesson, you will be able to… • interpret the five-number summary in a boxplot. • compare the range and interquartile range between two boxplots. • draw conclusions from real-world examples involving parallel boxplots. • compare data sets effectively. • understand the relationship between different graphical representations of the same data set.

Parallel boxplots We have seen that the range and interquartile range can be used to measure the spread of data and both can be seen on a boxplot. Parallel boxplots are used to compare two sets of data visually. When comparing boxplots, we compare the five-number summary and also the range and interquartile range. The two boxplots are drawn parallel to each other. It is important to clearly label each boxplot. Below are parallel boxplots comparing the time it took two different groups of people to complete an online task. 5

Under 30s

Over 30s

14 16 18 Seconds

20 22 24 26 28 30

The boxplots must be drawn on the same scale to compare them.

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Example 1 The two boxplots show the data collected by the manufacturers on the lifespan of light bulbs, measured in thousands of hours. Manufacturer A

Manufacturer B

3 4 5 Thousands of hours

a Complete this table using the two boxplots. Write each answer in terms of hours, remembering to multiply the values on the data display by 1000. Manufacturer A

Manufacturer B

Median Lower quartile Upper quartile Range Interquartile range

Create a strategy • To ﬁnd the lower quartile, median, and upper quartile, ﬁnd the corresponding values of the vertical lines of the box, respectively. • For the range, use the formula: Range = Highest score − Lowest score • For the interquartile range, use the formula: IQR = Q3 − Q1

Apply the idea Since the scale is in thousands of hours, we can multiply the numbers on the scale by a thousand to ﬁnd the number of hours. Median Lower quartile Upper quartile Range Interquartile range

Manufacturer A 4 × 1000 = 4000 2.5 × 1000 = 2500 4.5 × 1000 = 4500 (5 − 1) × 1000 = 4000 (4.5 − 2.5) × 1000 = 2000

Manufacturer B 5 × 1000 = 5000 3.5 × 1000 = 3500 6 × 1000 = 6000 (8 − 1.5) × 1000 = 6500 (6 − 3.5) × 1000 = 2500

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b Which manufacturer produces light bulbs with the best lifespan?

Create a strategy Choose the manufacturer with the greater median.

Apply the idea The data set for Manufacturer B has a median of 5000 hours, while the median of the data set for Manufacturer A is 4000 hours. So Manufacturer B produces light bulb with the best lifespan.

Reflect and check In fact, the best lightbulb produced by Manufacturer A has a lifespan of 5000 hours, which is the same as the median of Manufacturer B. This means that about half of the lightbulbs produced by Manufacturer B have a greater lifespan than all of the lightbulbs produced by Manufacturer A.

Example 2 This boxplots show the number of goals scored by two football players in each season. Sophie

Holly

15 Goals

a Who scored the most goals in a season?

Apply the idea By looking at the endpoints of the right whiskers, Sophie scored 18 goals, while Holly scored 19 goals. So Holly score the most goals in a season.

b How many more goals did Holly score in her best season compared to Sophie in her best season?

Apply the idea Number of goals = 19 − 18 =1

Subtract 18 from 19 Evaluate the subtraction

Holly scored 1 more goal in her best season.

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c What is the difference between the median number of goals scored in a season by each player?

Apply the idea Sophie’s median is 11 and Holly’s median is 10. Difference = 11 − 10 =1

Subtract 10 from 11 Evaluate the subtraction

d What is the difference between the interquartile range for both players?

Apply the idea Use the formula: IQR = Q3 − Q1 Sophie’s IQR = 14 − 7 =7 Holly’s IQR = 15 − 6 =9 Difference = 9 − 7 =2

Substitute the quartiles Evaluate Substitute the quartiles Evaluate Subtract the IQRs Evaluate

Idea summary When comparing boxplots, we compare the ﬁve-number summary values of each data set (the quartiles and minimum and maximum values) and the range and interquartile range.

Different graphs of the same data We can identify the centre and spread using key features. Measures of centre summarise the data set with a single value. We often use this to generalise which group performed better. • Mean • Median • Mode Measures of spread summarise the spread of a data set. We often use this to generalise which group was more consistent. The lower the spread, the more consistent the data. • Range • Interquartile range Different displays help us to identify different key features.

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From frequency displays like histograms, dot plots, and stem-and-leaf plots we can identify: • Shape • Frequencies, grouped or individual • Mode • Outliers From boxplots, we can identify: • Shape • Range and interquartile range • Median • Outliers We should expect then that the shape of data would be the same whether it is represented in a polygon, boxplot or histogram.

Frequency

All representations of symmetric data: 10 9 8 7 6 5 4 3 2 1 0

Histogram of approximately symmetric data

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of symmetric data

Frequency

All representations of positively skewed data: 10 9 8 7 6 5 4 3 2 1 0

Histogram of positively skewed data

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of positively skewed data

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All representations of negatively skewed data: Histogram of negatively skewed data

Frequency

10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

Boxplot of negatively skewed data

Looking at the diagrams above, can you see the similarities in the representations? We can see the skewed tails, where the bulk of the data sits, and general shape. These are some of the features you can use to match histograms and boxplots. We can also look at the data range.

Interactive exploration Discover this concept in action online

mathspace.co

Example 3 Match the boxplot shown to the correct histogram.

9 10

1 2 3 4 5 6 7 8 9 10

Create a strategy

Apply the idea

Look for characteristics of skewed and symmetrical distributions of data.

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Idea summary We should expect that the shape of data would be the same whether it is represented in a boxplot or histogram. Remember that the shape of data can be symmetrical, left skewed or right skewed.

9.05 Practice questions Understanding 1

Adventurer 1−3

Trailblazer 1−3

The histogram and the boxplot display the same data. 10 9 8 7 6 5 4 3 2 1 0

482

Explorer 1, 2

0 10 20 30 40 50 60 70 80 90 100 10 20 30 40 50 60 70 80 90 100

Which data display shows the mode of the data set?

Which data display shows the median of the data set?

Which data display shows the range of the data set?

Which data display shows the number of values between 10 and 20?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Each pair displays the same data set. What is the skew for each pair of histograms and boxplots? a

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 100

10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90

1 2 3 4 5 6 7 8 9 10

Which key features can be read off each display? a

Histogram

Boxplot

Dot plot

Cumulative frequency histogram and polygon

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Fluency Ex 1

4, 5, 8, 10

4−8

5−9

A class took an English test and a Mathematics test. Both tests had a maximum possible mark of 20. The parallel boxplots show the results of the tests:

English

Mathematics

Complete the table using the two boxplots. English

Mathematics

Median Lower quartile Upper quartile Range Interquartile range b Ex 2

In which test did the class tend to score better marks?

The parallel boxplots show the prices, in dollars, of the items on the menu of an upmarket restaurant and the menu of a fast food restaurant. Upmarket Restaurant

Fast Food

Price

484

Which restaurant has the higher median price for the items they sell?

What is the difference between the median prices of the items sold by each restaurant?

Which restaurant has a greater price range for the items on their menu?

What is the price difference between the most expensive items sold by each restaurant?

What amount of the cheapest item at the fast food restaurant could be bought for the same price as the most expensive item at the upmarket restaurant?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Two weightlifters both record their number of repetitions with a 70 kg bar over 30 days. The results are displayed in the boxplots:

Which weightlifter has the more consistent results?

Which statistics determine which is more consistent?

Which statistic is the same for each weightlifter?

Which weightlifter can do the most repetitions of 70 kg?

Repetitions

The residents of two blocks of townhouses were asked the number of pets they own. The frequency of various responses are presented in these dot plots. a

Is the pet ownership a little lower or higher in Block A than Block B?

In Block A, how many pets do most households have?

In Block B, how many pets do most households have?

What is the shape of the data for Block A?

Find the range of the number of pets in Block A.

Which block has more variability in the the number of pets?

Do either sets of scores have an outlier?

Block A

Block B

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Ex 3

Match the histograms to the corresponding boxplots. Histogram A Boxplot 1 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Boxplot 2 Histogram B 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Histogram C Boxplot 3 10 9 8 7 6 5 4 3 2 1 0

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Histogram D Boxplot 4 10 9 8 7 6 5 4 3 2 1 0

486

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 90 100

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Match each cumulative frequency graph to its corresponding boxplot: Cumulative Frequency Graph A

Cumulative Frequency Graph B Cumulative Frequency

Cumulative Frequency 80

20 10

10 Mass (g)

Mass (g)

5 10 15 20 25 30 35 40 45 50 55 60

Cumulative Frequency Graph C

Cumulative Frequency Graph D

Cumulative Frequency

10 Mass (g)

5 10 15 20 25 30 35 40 45 50 55 60

Mass (g) 5 10 15 20 25 30 35 40 45 50 55 60

boxplot 1

boxplot 2

5 10 15 20 25 30 35 40 45 50

boxplot 4

boxplot 3

5 10 15 20 25 30 35 40 45 50

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The parallel boxplots show the distances, in centimetres, jumped by two high jumpers.

Matt

Paul

140

150

160

170

Who had a higher median jump?

Who made the highest jump?

Who made the lowest jump?

180

Distance (cm)

Reasoning

11−14

12−14

Two History classes, each with 20 students, were given a 10 question True/False test. The results for each class are shown. 10 9 8 7 6 5 4 3 2 1 0

Class 1 Number of Students

Number of Students

11, 12

5 6 7 Questions Correct

Class 2

10 9 8 7 6 5 4 3 2 1 0

8 9 10 Questions Correct

Calculate the mean for Class 1 rounded to one decimal place.

Calculate the mean for Class 2 rounded to one decimal place.

Find the range for Class 1.

Find the range for Class 2.

Do you think Class 1 studied for their test? Justify your reasoning.

Do you think Class 2 studied for their test? Justify your reasoning.

Besides studying, what is another factor that may have had an effect on the test results for each class?

Two bookstores recorded the selling price of all their books. The results are presented in the parallel boxplots. Bookstore A prices ($)

Bookstore B prices ($) x 50 488

100

110

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

120

130

140

150

Which bookstore had the more consistent prices? Explain your answer.

Comparing the most expensive books in each store, how much more expensive is the one in store B?

Is each statement true or false? 5% of the books in Bookstore B are at least as expensive than the most expensive 2 book in Bookstore A.

5% of the books in Bookstore B are cheaper than the cheapest book in 2 Bookstore A.

Explain why each pair of histograms and boxplots do not match: a

Salaries earned by employees at a software company are given in a histogram.

Frequency

10 9 8 7 6 5 4 3 2 1 0

65 000

75 000

85 000 Salaries ($)

95 000

105 000

Using the centres, construct a boxplot of the data.

Find an estimate for the interquartile range.

Using the boxplot, approximately what percentage of salaries lie in the range $90 000 to $100 000?

Using the boxplot, what can be said about the highest 25% of salaries? Would this conclusion be different from using the histogram? Explain your answer.

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Problem-solving 15 15

15, 16

Cooper and Marion are racing go-karts. The times (in seconds) for the 12 laps of their qualifying race are shown: • Cooper: 58.9, 46.5, 52.6, 66.6, 58.4, 53.1, 45.0, 52.1, 52.4, 52.7, 44.8, 51.7 • Marion: 47.8, 54.6, 68.5, 68.0, 62.8, 57.2, 54.8, 63.4, 58.1, 64.3, 66.2, 47.1 a

Construct the ﬁve-number summary for each set.

Are there any outliers?

Create a parallel boxplot of the two sets of times with the outlier displayed separately.

Which racer will be in pole position for the ﬁnal race, if it is given to the racer with the fastest qualifying lap time?

Does spinning out on a lap, causing a high outlier, impact the selection for pole position? Explain your answer.

Two friends compete in hammer throw competitions and train together over a season. They compete in 15 competitions and their ﬁnal throw for each competition is shown: • Tim: 29.8, 37.4, 33.9, 38.8, 34.3, 36.5, 34.5, 30.0, 35.2, 38.4, 33.0, 33.2, 39.6, 35.0, 36.9 • Odi: 32.2, 35.4, 34.8, 33.0, 38.4, 26.0, 40.0, 37.2, 39.5, 42.4, 38.6, 42.3, 38.4, 42.8, 37.2 a

Complete the table of statistics, rounded to one decimal place: Minimum Q1 Median Q3 Maximum Mean Range Interquartile range

490

Tim 29.8 33.2 35.0 37.4 39.6

Odi

37.2

Which competitor throws more consistently? Explain your answer.

Identify any outliers.

Who is the better hammer thrower? Explain your answer.

When considering Odi’s average throw is it reasonable to remove the outlier before calculating the mean? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

9.06 Samples and populations After this lesson, you will be able to… • distinguish between a population and a sample in statistics. • identify the key factors that make a sample representative of a population. • understand the differences between a parameter and a statistic. • recognise the importance of sample size in data collection. • distinguish between primary and secondary data, and assess the reliability of secondary data sources. • apply the concepts of samples and populations in practical scenarios.

Samples and populations In statistics, a population refers to every member within any particular group of interest. A survey conducted on every member of a population is called a census. Collecting data from every member of a population is the most accurate way of gathering information, but it is not always possible and can be very expensive. Collecting data from a sample, can be quicker and less expensive. A parameter is a number that summarises data from a population. • Population size: N • Population mean: m A statistic is a number that summarises data from a sample. • Sample size: n • Sample mean: A sample is representative if the characteristics of the sample are close to the population. Key factors that affect if a sample is representative include: • Demographic factors: If the population is diverse in terms of age, gender, ethnicity, or other factors, the sample should reflect this diversity. • Characteristics relevant to the study: If you’re studying a particular trait or behavior, individuals in the sample should exhibit a range of these characteristics. • Random selection: This minimises bias and increases the likelihood of the sample being representative of the population from which it was drawn. If a sample is unbiased and represents the population well, we can use sample survey data to make conclusions about the population. The size of the sample is important. If the sample is too large, it may be too expensive or timeconsuming to collect the data. If it is too small, the sample may not be representative of the population.

Interactive exploration Discover this concept in action online

mathspace.co

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Primary data is data that has been collected firsthand. If you are the researcher, primary data is data that you gather yourself, directly from observation, surveys, experiments, or interviews. Secondary data is data that has already been collected and analysed by someone else. This could be in the form of books, articles, reports, or statistics that you find online or in the library. When using secondary data: • Make sure the population from which the data was collected matches your target population. • Check whether a census or sample was used. If a sample was used, make sure it is representative of the population. • Analyse whether there is any source of bias.

Example 1 A sample of 25 people is drawn from a population. In this sample, the youngest is 18 years old, and the oldest is 64. Which two of these might be populations that this sample was drawn from? A Residents of a retirement home

Employees at a bank

C Students from a primary school

People stuck in traﬃc due to road works

Create a strategy Remember that a sample is a smaller group than the entire population but every member of the sample must be a member of the overall population.

Apply the idea

Reﬂect and check

Which of these populations would be likely to include people of this age group?

To check our answer we can conﬁrm that the other options are incorrect. Option A, residents of the retirement home must be of at least a retirement age (usually 65). While for Option C, the students attending a primary school will only range from ages 4 to 12

The answers are options B and D.

Example 2 A researcher captures 400 fish in a lake, tags them, then releases them. The next day, he captures 1200 fish, of which 100 have tags attached to them. Estimate the size of the fish population.

Create a strategy The proportion of tagged ﬁsh in the second sample should be equal to the proportion of 400 ﬁsh out of the total population.

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Apply the idea

Therefore, there is estimated to be 4800 ﬁsh in the lake.

Reflect and check To check our answer, we can multiply the total population by the proportion of tagged fish to see if it gives us the 400 fish originally tagged.

Example 3 Maryke is researching the monthly spending habits of students on school lunches at a local high school. She found two secondary sources of data: • Source A: This data set provides the average monthly spending of students on lunch for the entire school year. • Source B: This data set provides raw data for the monthly spending of each student for each month of the school year, broken down by demographics like age, sex, and postal code. a If Maryke wants to ﬁnd out in which month the spending was the lowest and in which month the spending was the highest, for year 12s, which source should she refer to? Explain.

Create a strategy Decide which source provides data on individual months of the school year and where she can separate out the year 12s.

Apply the idea Maryke should refer to Source B. Firstly, she can separate out the year 12 data. Also, she can determine in which month the spending was the lowest and the highest because it provides the monthly spending of each student for every month of the school year, allowing her to calculate the average for each speciﬁc month.

Reﬂect and check We some times call data which is all grouped together and summarised aggregate data and raw data disaggregated data. 9.06 Samples and populations mathspace.co

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b If Source A reports that the average monthly spending is $50 and there are 10 months in the school year, can Maryke determine if there was a month where students spent more than $100, on average? Why or why not?

Create a strategy Decide whether the monthly average can be used to ﬁnd the spending habits from an individual month.

Apply the idea

Reﬂect and check

No, Maryke cannot determine if there was a month where students, on average, spent more than $100 just from the average monthly spending data of Source A. The overall average does not provide details of spending in individual months.

Many times, it is best to ﬁnd secondary sources that report raw data values rather than data that has already been summarised. If you need to use the data for a different reason than the researcher who collected the data, the information in their report may not be useful for your statistical study.

Idea summary • In statistics, data can be gathered from every member of a group, known as a population, or from a subset of the population, known as a sample. • Gathering data from a sample instead of the entire population is often more practical and less expensive, however, the sample needs to be representative of the population for the data to be accurate. • A representative sample has the same demographic diversity as the population, relevant characteristics to the study, and random selection. • The size of the sample is crucial; a sample that’s too large may be too costly or time-consuming, while a sample that’s too small may not represent the population and is not useful to make conclusions.

9.06 Practice questions Understanding 1

494

Explorer 1−5

Adventurer 1−5

Trailblazer 1−4, 6

True or false: a

In a sample survey, information is obtained from the entire population.

A census of the whole population is always possible.

Using a sample can save time and money for the researcher.

A sample should have the same diversity as the population.

A convenience sample is a good sampling method to use.

The population and sample means will always be exactly the same.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

If the population has 42 000 members, would this sample size be large enough? a

4000

40 000

A study is being done on physical activity in a small town. Would this characteristic be important to consider when selecting the sample? a

Favourite colour

Gender

Age

Favourite sport

Preferred method of transportation

Name

Identify the population and sample in each scenario: a

The mayor of Anytown wants to determine the opinion of his residents on the new mall planned to be built. He surveys 500 residents at random.

A bookstore owner wants to know what genres of books are most popular amongst her customers. She decides to pull up the receipts for 20 purchases out of the 650 sales made over the weekend.

Antony wants to know what teams the students at his school support. He ﬁnds that 30 students support the Sydney Swans, and 55 students support the Melbourne Demons.

For a statistical survey, the population is deemed to be all the students that attend the local secondary school. Are these samples of that population? a

All of the seniors

The ﬁrst 30 students to arrive at school that day

All of the students that were in the library at lunch time

All of the teachers

Some of the teachers

For a statistical survey, the population is deemed to be people aged under 18 who live in Sydney, Australia. Are these samples of that population? a

40 professors chosen from University of Sydney

Every person who lives in Bondi

All elementary students who are visiting Taronga Zoo in a single day

Students enrolled in the on-campus preschools or high schools in Parramatta

100 children chosen from a Sydney junior rugby league team

9.06 Samples and populations mathspace.co

495

Fluency 7

7, 8, 9bc, 10, 11ab, 13, 15−17

7, 8, 9cd, 11cd, 13−17

Match each sample histogram to the population it was taken from. i

iii

10 9 8 7 6 5 4 3 2 1 0

9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

10 9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

iv 10

0 10 20 30 40 50 60 70 80

9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

9 8 7 6 5 4 3 2 1 0

496

7, 8, 9ab, 10, 11ab, 12, 15−17

0 10 20 30 40 50 60 70 80

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

10 9 8 7 6 5 4 3 2 1 0 10 9 8 7 6 5 4 3 2 1 0

0 10 20 30 40 50 60 70 80

Ex 1

A sample of 50 people is drawn from a population. In this sample, the youngest person is 18 years old and the oldest person is 64 years old. Are these populations that the sample could have been chosen from? a

People in a grocery store

People who got tickets for speeding

People at a drop-in centre for seniors

Employees at a grocery store

For each population, give an example of a biased sample of the population. a

The population is all the students that attend the local high school.

The population is all people aged under 18, who live in Sydney.

The population is all people in a city who play in any organised sporting competition.

The population is all people who own a pet dog.

Sean wants to know what year 10s think of their English class, so he polls 70 random year 10s. Is the sample chosen biased or fair?

A school principal wants to estimate the number of students who ride a bicycle to school. Is each sample biased?

All students who are in the school band.

Eight students in the hallway.

Ten students from each year, chosen at random.

130 students during the lunch periods.

At a certain chocolate factory, 30% of products contain nuts. 400 chocolates are tested to check if they meet the required quality for sale. Of those tested, 61% contained nuts. a

What is the population?

What proportion of the population contains nuts?

What proportion of the sample contains nuts?

Are the 400 chocolates that were tested a representative sample?

At a popular orchard, 65% of apples are Gala apples. 500 apples are picked for a tastetesting event. Of those picked, 310 were Gala apples. a

What is the population?

What proportion of the sample is of the “Gala” variety?

Are the 500 apples that were picked a representative sample?

9.06 Samples and populations mathspace.co

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15 Ex 2

Lachlan asks 120 year 12 students at his school how much time they spend on homework per night. 78 year 12 students say they do more than 3 hours. At a meeting of the student council, Lachlan reports “65% of students at this school do too much homework.” For each statement, decide if it explains why Lachlan’s report is misleading. a

The survey does not represent the population of the school.

The question should have been multiple choice.

The question was biased.

The sample size was too small.

The conclusion does not align with the survey question.

An ecologist collects 250 of a particular bee species in a meadow, marks them, and then releases them. In a follow-up collection, the ecologist gathers 200 bees, and 82 of them are found to be marked. Estimate the total bee population in the meadow. A television station wants to estimate the number of viewers it had for a new show. They know that the country’s population is 13 620 000. When they randomly selected 5000 people and asked them if they had watched the show, they found that 400 of them had. Estimate the number of people in the entire population who watched the show.

Ex 3

Yakov is studying the performance of students in two different classes over a term. He found two secondary sources of data:

• Source A: This data set provides the individual marks of every student from both classes. • Source B: This data set provides the average final mark of students from both classes. If Yakov wants to know how many students from one class scored above 90%, which source should he use?

Which source should Yakov use if he wants to calculate the median mark for only one class?

Source B indicates that the average mark for one class is 85%. Can Yakov determine how many students scored exactly 85% using Source B?

Reasoning 18

18, 19, 21, 22

19, 21, 22

20, 21, 23

Pablo has concluded that about 36% of people would drive an electric car based on his survey results: Would drive an electric car Would not drive an electric car

58 104

Jenny has concluded that about 44% of people would drive an electric car based on her survey results: Would drive an electric car Would not drive an electric car

224 285

Which claim do you think is more valid? Explain your reasoning.

498

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

For a statistical survey, the population is deemed to be all people who own a pet dog. Ariel decides to ﬁnd her sample by talking to every person at her local dog park. Describe the advantages and disadvantages of her sampling method.

Explain why these samples are biased:

Hannah is surveying customers at a shopping precinct. She wants to know which stores customers shop at the most. She walks around an entertainment store and chooses 30 customers from the store for the survey.

A TV station wants to know what the most popular type of music is, so they ask listeners to contact them and vote for their favourite type of music.

The community health nurse wants to survey the students in a school about their eating habits. At lunchtime, she stands by a vending machine and surveys every student who purchases something from the machine.

A political polling company calls 1000 people at home between 4:00 p.m. and 7:00 p.m. on week nights to ﬁnd out who they are most likely to vote for in an upcoming election. They publish their numbers based on the responses of only the 410 people who answered their call. a

How could they have used random sampling to choose the 1000 people to call?

Explain why the company’s published results will not be accurate.

A researcher captures 300 dingoes, tags them, then releases them. The next day, they capture 750 dingoes, of which 90 have tags. The day after that, they capture 100 dingoes, of which 6 have tags. a

Compare the sample proportions from the second and third day of research.

Explain why the sample proportions are different.

Explain which sample statistic you would use.

An environmentalist is interested in tracking the population of koalas in Australia. She begins by tagging 480 koalas. In a follow-up collection, she catches 820 koalas, of which 62 are tagged. In a final round of collection, she catches 820 koalas, of which 58 have tags. a

Explain why the sample proportions are different.

Explain which sample statistic they would use.

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Problem-solving 25, 26 24

24−26

24−27

You are writing an essay on Australia’s role in World War II. While researching, you come across two sources of information: • Source A: A school textbook published in 1990 detailing Australia’s military engagements, key figures, and notable battles during the war. • Source B: A documentary released in 2015 that focuses on the personal stories of Australian soldiers, interviews with war historians, and unseen archival footage.

If you find conflicting information between the two sources, how might you determine which one is more accurate?

Why is it important to use multiple secondary sources when trying to get a comprehensive understanding of a historical event?

Marine biologists want to determine if a local species of ﬁsh is growing to a smaller size than it used to. They collect and measure 150 of the 450 ﬁsh known to exist in the area. a

Did they catch a large enough proportion of the species to be able to determine if the ﬁsh are growing to a smaller size?

What additional information would make us more conﬁdent that the sample is representative of the population?

For a statistical survey, the population is considered to be all of the athletes at an Australian Football League (AFL) game. a

Describe two different samples that could be taken at the game to represent this population.

Describe two samples that could be taken at this game that do not represent this population. Explain why these samples aren’t appropriate.

Farma, a newly established pharmaceutical company in Australia, has recently conducted clinical trials for a new medication. The trials involved 1000 participants across Sydney. According to the data they released, it appears that 10% of the trial participants experienced mild side effects such as headaches and fatigue. Farma plans to roll out this medication across the country, making it available to the 5 million residents of Sydney first. Consider the potential bias, accuracy, and statistical validity in your analysis.

500

Based on the trial data, estimate the number of Sydney residents who might experience side effects if the medication were given to the entire target population.

Discuss the ethical considerations of releasing this prediction to the public. How might the way the data is reported influence public opinion about the medication?

Do you think the sample size used in the initial trial is representative of the entire target population of Sydney? Why or why not?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

9 Chapter review 1

For the surveys, should the data be grouped or ungrouped when constructing a frequency table? a

A survey conducted of 3500 people, asking them how tall they are in centimetres.

A survey conducted of 1000 people, asking them how many books they read in the past year.

The scores for a recent mathematics test are displayed in the stem-and-leaf plot. The scores are out of 110.

Stem

Leaf

How many students took the test?

5 7 7 9 9

What test score occurred most frequently?

0 3 5 7

How many marks separate the highest score and the lowest score?

0 3 3 3 3 5 8

0 2 2 4 5 7 7

0 2 5 7 9

0 Key: 10|0 = 100

The histogram displays amount of time that Aaliyah spent watching television on her streaming platform over different days. Minutes Frequency 0 ≤ Minutes ≤ 10 10 < Minutes ≤ 20 20 < Minutes ≤ 30 30 < Minutes ≤ 40 40 < Minutes ≤ 50 50 < Minutes ≤ 60 60 < Minutes ≤ 70 70 < Minutes ≤ 80

Number of days

Time spent watching television 15

0 10 20 30 40 50 60 70 80 Time spent (in minutes)

Complete the frequency table based on the histogram.

Aaliyah’s subscription offers 1450 minutes of free streaming time, then charges $0.50 per minute for any additional time. How much was she charged for watching these episodes?

Chapter 9 review mathspace.co

501

A musical has a large cast of students with various ages. The column graph shows the ages of all the students in the cast. 25

Ages of students in a musical

Frequency

20 15 10 5 0

10 11 12 13 14 15 16 17 18 Age

How many students are in the cast?

What age is most common?

Find the range of the students’ ages.

Find the interquartile range.

These data sets each contain 10 movie ratings out of 10: Set A: 2, 3, 4, 5, 6, 7, 8, 9, 10, 10 Set B: 1, 2, 3, 4, 5, 9, 9, 9, 10, 10 Set C: 1, 1, 3, 4, 5, 7, 8, 8, 9, 10

Set B has the largest interquartile range. What does this tell you about the spread of ratings in this data set?

Based on the ﬁve-number summaries and the shapes of the data sets, identify which set is symmetric, which is skewed left, and which is skewed right.

In a game show, a contestant must complete 8 different mini games, and the time (in seconds) taken to complete each is recorded. One contestant’s times for the ﬁrst 7 games are: 47, 50, 52, 60, 66, 80, 82 Find their time in the 8th segment if the upper quartile of all 8 times is 75 seconds.

For the data set shown in the dot plot: a

Construct the ﬁve-number summary.

Draw a box plot.

Describe the shape of the box plot.

Stem

Leaf

5 8 8

0 2 5 5 6 8

0 1

2 3 3 5

Key: 1|2 = 12

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The box plot shows the number of books a group of students read in a year: a

Find and interpret the range.

What percentage of students read from 12 to 20 books?

The middle 50% of students read within how many books of each other?

In which quartile is the number of books read the most spread out?

The bottom 50% of students read within how many books of each other?

Number of Books

The box plots summarise the results from a botanical study. The treatment group received an experimental fertiliser to boost plant growth, and the control group received regular water. The boxplots show the number of weeks each group of plants took to reach maturity: Control group

12 15 18 21 24

Treatment group

12 15 18 21 24

Compare the medians of the control group and the treatment group.

Compare the ranges of the control group and the treatment group.

Compare the interquartile ranges of the control group and the treatment group.

Does the fertiliser have a positive effect on plant growth speed? Explain your answer.

For the cumulative frequency histogram: a

Find the total number of scores recorded.

Find the number of times a score between 150 and 200 occurred.

Find the percentage of scores that were 150 or more. Round your answer to one decimal place.

300 Cumulative frequency

250 200 150 100 50 0

50 100 150 200 250 Scores

Chapter 9 review mathspace.co

503

The number of attendees at various workshops during a tech conference was noted. The list represents the number of attendees in each workshop: 24, 23, 23, 22, 20, 21, 21, 25, 25, 25, 25, 25, 25, 22, 22, 25, 22, 22, 21, 21

Complete the table.

In how many workshops were there at least 24 attendees?

In how many workshops were there fewer than 23 attendees?

Approximately half of the data values recorded are greater than what number of attendees?

Number of attendees

Use the cumulative frequency histogram to estimate: a

The median

The lower quartile

The upper quartile

Number of workshops (f )

20 21 22 23 24 25

100 90 80 70 60 50 40 30 20 10

Score 44

504

Cumulative frequency (cf )

The residents of two apartment buildings were asked the number of indoor plants they own. The frequency of various responses are presented in the dot plots: a

Which building has more plants?

By how many plants does the mode of each building differ?

Which building has more variability in the number of plants?

Does either set have an outlier?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2 3 4 5 Building X

2 3 4 5 Building Y

Match the box plot shown to the correct histogram.

Score

7 8 9 10 11 12 13 14 15 16 17 18

D Frequency

9 8 7 6 5 4 3 2 1 0

10 9 8 7 6 5 4 3 2 1

8 9 10 11 12 13 14 15 16 17 18 Score

10 Frequency

9 8 7 6 5 4 3 2 1

Frequency

8 9 10 11 12 13 14 15 16 17 18 Score

10 9 8 7 6 5 4 3 2 1 0

8 9 10 11 12 13 14 15 16 17 18 Score

Two athletes compete in long jump events and train together over a season. They compete in 10 competitions and their ﬁnal jump (in metres) for each competition is shown: • Antoine: 6.8, 7.4, 6.9, 7.8, 7.3, 7.5, 7.1, 6.5, 7.2, 7.6 • Benji: 7.0, 7.2, 6.8, 6.7, 7.9, 6.0, 8.1, 7.5, 8.0, 7.7 a

Complete the table of statistics:

Which athlete jumps more consistently?

When considering Benji’s average jump, is it reasonable to remove the outlier before calculating the mean? Explain your answer.

Would removing the outlier from Benji’s data set change your answer to part (b)? Explain your answer.

Minimum Q1 Median Q3 Maximum Mean (1 d.p.) Range Interquartile range

Antoine 6.5 6.9 7.3 7.5 7.8

Benji

7.4

Chapter 9 review mathspace.co

505

506

A fruit farm has 52 000 cherries of different varieties. During a day of harvest, 800 cherries are randomly picked. Of those picked, 468 were Bing cherries. a

What is the population?

What proportion of the sample is of the “Bing” variety?

Do the 800 cherries picked represent a random sample?

Estimate the number of cherries in the entire population that are Bing cherries.

A gym manager wants to estimate the number of members who attend yoga classes. Is each sample biased? a

All members who participate in the morning spin class.

Twelve members near the water reﬁll station.

Every 10th member that enters the gym.

Entomologists are tracking the habits of a local beetle species. At the beginning of their research project, they collect and tag 600 beetles. The next week, they collect 350 beetles, of which 98 had tags. The week after, they collect 200 beetles, of which 42 had tags. a

Why are the sample proportions different?

Explain which sample statistics you would use to estimate the population.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The discipline of statistics aspires to look beyond individual details to draw larger conclusions. –David Spiegelhalter

Big ideas • Different representations of data highlight different characteristics of the data. • Many sets of bivariate data can be modelled using familiar functions. • Analysing data patterns and trends enables making informed estimations and predictions about values within or beyond the known data range. • Effective data representation requires the use of accurate and unbiased techniques to provide an accurate and unbiased understanding of the information being conveyed.

Curriculum Outcomes VC2M10AP02 investigate reports of studies in digital media and elsewhere for information on their planning and implementation VC2M10ST02 construct scatterplots and consider a line of good fit; comment on the association between the 2 numerical variables in terms of strength, direction and linearity VC2M10ST03 construct two-way tables and discuss possible relationship between categorical variables VC2M10ST04 analyse claims, inferences and conclusions of statistical reports in the media and other places, by linking claims to displays, statistics and representative data, including ethical considerations and identification of potential sources of bias VC2M10ST05 plan and conduct statistical investigations of situations that involve bivariate data, including where the independent variable is time; evaluate and report findings with consideration of limitations of any inferences VC2M10AST03 use digital tools to investigate bivariate numerical data sets; where appropriate use a straight line to describe the relationship allowing for variation, make predictions based on this straight line and discuss limitations

10 Bivariate statistics Chapter outline 10.01 Two-way tables 10.02 Associations between categorical variables 10.03 Bivariate numerical data 10.04 Associations between numerical variables 10.05 Line of good fit 10.06 Interpolation and extrapolation 10.07 Misrepresentations of data Investigation: Misrepresentations in the media Investigation: Conduct a statistical investigation 10.08 Optional: Correlation coefficient Chapter 10 review

510 522 541 550 565 576 590 605 611 615

There’s a correlation between the number of honeybees and crop production!

10.01 Two-way tables After this lesson, you will be able to… • construct a two-way table using the given data. • identify and explain joint and marginal frequencies in a two-way table. • apply basic calculations to find percentages and totals in a two-way table. • interpret data presented in a two-way table.

Two-way tables Exploration Students were asked whether they prefer high-top sneakers or low-top sneakers, and they were also asked whether they prefer to shop online or shop in a store. Some of the results from the survey are: • The survey included 75 students. • 30 students said they prefer high-top sneakers. • 35 students said they prefer to shop online. • 7 students said they prefer to shop in a store and high-top sneakers. • 33 students said they prefer low-top sneakers and shopping in a store. 1. How many students prefer low-top sneakers? 2. How many students prefer low-top sneakers and shopping online? 3. Is there an easier way to display the data in order to answer the ﬁrst two questions?

A two-way table can be used to summarise two categorical variables or attributes. The cells in the table represent the intersections of the categories, displaying the frequency count for each combination of categories. This table contains information about the preference of art medium, and age. If we read across each row, we can tell how many people surveyed prefer drawing, painting or graphics and how many of those are children or adults. If we read down each column, we can tell how many children or adults surveyed prefer drawing, painting or graphics.

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Child

Adult

Total

Drawing

Painting

Graphics

Total

100

Joint frequencies

Add across

Add down

Marginal frequencies

Each joint frequency gives us speciﬁc combinations of conditions. Meanwhile, each marginal frequency gives us overall totals. Both types of information are important in understanding the data in a two-way table and performing calculations to draw conclusions.

Example 1 Some students in Ms. Rochelle’s class were polled on how many siblings they have. The results are provided in the table: 1st period 2nd period 3rd period

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 7 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 7 Period 1

Period 2

Period 3

Total

Less than two siblings Two or more siblings Total a Fill in the two-way table showing how many students have less than two siblings, and how many have two or more siblings.

Create a strategy Identify the joint frequencies of the number of students from each period who have less than two siblings or two or more siblings.

Apply the idea Here’s the complete two-way table: Less than two siblings Two or more siblings Total

Period 1 10 10 20

Period 2 11 9 20

Period 3 13 7 20

Total 34 26 60

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b How many of the students in Period 1 have two or more siblings?

Create a strategy

Apply the idea

Look at the ‘Period 1’ column and the row for ‘Two or more siblings’ in the table.

Number of students = 10

c What proportion of the students have less than two siblings?

Create a strategy

Apply the idea

Divide the marginal frequency of the number of students with less than two siblings by the total number of students polled.

Example 2 A university conducted a survey asking students to rate the effectiveness of online classes and traditional in-person classes. Students were divided into two categories: undergraduate and postgraduate students. The data from the survey is shown in the two-way tables. Undergraduate students

Online classes In-person classes

Not effective 27 32

Slightly effective 11 12

Somewhat effective 2 5

Effective 13 6

Very effective 31 29

Total 84 84

Postgraduate students

Online classes In-person classes

Not effective 9 3

Slightly effective 14 0

Somewhat effective 4 8

Effective 3 13

Very effective 5 11

Total 35 35

a Calculate the percentage of undergraduate students who found classes to be Effective or Very effective for online classes and for traditional in-person classes. Round your answers to one decimal place.

Create a strategy Add the joint frequencies for undergraduate students who rated classes as Effective or Very effective separately for both online and in-person classes, then divide each total by the respective class’s total number of undergraduate students.

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Apply the idea Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round

For undergraduate students, the percentages for Effective or Very Effective ratings are 52.4% for online classes and 41.7% for traditional in-person classes. b Calculate the percentage of postgraduate students who found classes to be Effective or Very effective for online classes and for traditional in-person classes. Round your answers to one decimal place.

Create a strategy Add the joint frequencies for postgraduate students who rated classes as Effective or Very effective separately for both online and in-person classes, then divide each total by the respective class’s total number of postgraduate students.

Apply the idea Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round Write as a fraction of the total

Add the numerator

Convert fraction to percentage and round

For postgraduate students, the percentages for Effective or Very effective ratings are 22.9% for online classes and 68.6% for traditional in-person classes.

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c What does this data suggest about the perception of online classes among different student levels? A

Undergraduate students slightly prefer online classes to in-person classes, while postgraduate students strongly prefer in-person classes.

Both student groups greatly prefer online classes to in-person classes.

Undergraduate students have about equal preferences, while postgraduate students strongly prefer online classes to in-person classes.

Undergraduate students strongly prefer online classes to in-person classes, while postgraduate students have about equal preferences.

Create a strategy Look at the percentages that we calculated in parts (a) and (b).

Apply the idea Undergraduate students have a slightly higher percentage favouring online classes (52.4%) over in- person classes (41.7%). On the other hand, postgraduate students have a signiﬁcantly higher percentage preferring in-person classes (68.6%) to online classes (22.9%), so the correct answer is option A.

Idea summary Types of frequencies to consider in a two-way table are: • Joint frequency: How often a combination of two conditions occur • Marginal frequency: How often a particular category of data occurs (or the total row and column data) Both types of information are important in understanding the data in a two-way table and performing calculations to draw conclusions.

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10.01 Practice questions Understanding 1

Explorer

Adventurer

1−3

2−4

This table gives information on the treatment plans prescribed to cats with feline diabetes. Depending on the severity of the case, cats are either prescribed a special diet or insulin injections. Venetian Pet Clinic Mohawk Animal Clinic Outback Animal Care Centre Total

Trailblazer

Special diet 14 11 15 40

Insulin injections 14 9 16 39

Total 28 20 31 79

How many cats from Venetian Pet Clinic were prescribed insulin injections?

How many cats from Outback Animal Care Centre were prescribed insulin injections?

Which animal clinic prescribed a special diet less frequently than insulin injections?

This two-way frequency table displays the number of deaf-blind children in each state and their age group. Victoria New South Wales Queensland Total

0–5 year-olds 63 52 56 171

6–17 year-olds 388 143 193 724

Total 451 195 249 895

For each of the groups:

Classify as either a joint frequency or a marginal frequency.

Find the value.

The number of deaf-blind children in Victoria.

The number of deaf-blind 0–5 year-olds.

The number of deaf-blind 6–17 year-olds in New South Wales.

The number of deaf-blind children in Victoria who are more than 5 years old.

This table shows how much homework students do each night in different year levels. Use the information in the table to determine whether the statements are true or false. 1.5 hours or more Less than 1.5 hours Total

Year 8 students 7 18 25

Year 9 students 28 6 34

Total 35 24 59

Most students in general do homework for less than 1.5 hours per night.

Most Year 8 students do homework for 1.5 hours or more per night.

There were more students surveyed who do homework for 1.5 hours or more per night than there were who do less than 1.5 hours per night.

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This table shows the preferred flavour of milkshake for children and adults. Is each statement true or false based on the table? Child Chocolate Vanilla Strawberry Total

16 39 98

Total 93 50 230

43 children prefer chocolate.

A total of 50 adults were surveyed.

There were more adults surveyed than children.

87 people prefer vanilla.

230 people were surveyed.

Fluency 5

Adult 50 71 11

5−9

5, 7, 9, 10

5, 7, 10, 11

Use the given information to complete each two-way frequency table. a

This table gives information about the number of sales of each type of shoe at Foot Swag, a shoe store in a shopping centre. Running shoes

Basketball slides

Work boots

43 85

13 36

29 48

Teenagers Adults Total b

Seventy-one students were asked if they prefer chocolate (C) or vanilla (V ) ice cream. Each of the students was also asked if they prefer sprinkles (S) or hot fudge (F ) as an ice cream topping. The results are displayed in the tree diagram: 12 C

30 71

41 71 V

Total 84 85 169

30 18 30

29 41

12 41

Sprinkles

Hot fudge

Total

Chocolate Vanilla Total

Laura asked some people in her community what their preference was between beef, chicken, and ﬁsh.

20 respondents said they preferred beef, 8 of those were children. 15 children said they preferred chicken, and 15 adults said they preferred ﬁsh. 35 children were surveryed, and there were 73 respondents overall. Beef

Chicken

Fish

Total

Children Adults Total 516

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Ex 1

Ex 2

Special diet 14 11 15 40

What does 28 mean in this context?

What does 11 mean in this context?

Insulin injections 14 9 16 39

Total 28 20 31 79

Sophia asked 38 children and 54 adults in her community about their dietary preferences. The options given were: vegetarian, vegan or neither. 40 respondents said they were vegetarian, 15 of those were children. 18 children said they were vegan, and 22 adults said they were neither vegetarian nor vegan. a

Construct a two-way table based on the results of Sophia’s survey.

What is the proportion of respondents that are vegetarian?

What is the proportion of adults that are vegan?

Mr. Tobit asked the students on football teams to pick their favourite subject. He displayed the results in a two-way table. a

How many defence players did not pick Maths as their favourite subject?

How many players picked Music?

How many offence players are on the football teams?

At a local high school, a survey is conducted to gauge student satisfaction with canteen food. The responses are measured on a five-point Likert scale, with 1 being “very unsatisfied” and 5 being “very satisfied”. The survey also takes into account whether the student is a junior student or senior student.

Maths Music Science English

Offence 11 8 13 15

Defence 12 18 11 10

Rating 1 Rating 2 Rating 3 Rating 4 Rating 5

Junior students 23 35 50 70 22

Senior students 30 20 45 40 15

What percentage of juniors were very satisﬁed with the canteen food? What about the seniors?

Determine which rating is most common among junior students. What about senior students?

Comment on any patterns or discrepancies you observe between the two groups of students.

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This table shows the number of trains arriving either on time or late at a particular station: a

How many trains were late on Friday?

How many trains passed through the station on Wednesday?

How many trains were on time for the entire week?

What fraction of trains were on time for the whole week?

Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Arriving on time 23 12 22 30 11 15 14

Arriving late 38 25 29 27 14 11 10

At a local university, students were asked for their favourite subject at high school and what they had chosen to study at university. The results are shown in a table. Maths Science Music Art Total

Maths favourite 76 64 64 6 210

Science favourite 20 46 11 19 96

Music favourite 61 53 67 38 219

Art favourite 43 59 59 74 235

Total 200 222 201 137 760

Based on this data:

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What percentage of students surveyed said their favourite subject was Maths at high school? Round your answer to two decimal places.

What percentage of students are studying either Music or Arts at university? Round your answer to two decimal places.

What fraction of students had Music as their favourite subject in high school but decided to study a different ﬁeld at university?

What percentage of students have chosen to study in the same ﬁeld as their favourite subject from high school? Round your answer to two decimal places.

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13, 14

A scientist recorded some data on the diet of two types of possums observed on Phillip Island: • 310 possums were observed. • 45 brushtail possums were observed to eat a diet of fruit. • 180 possums observed were honey possums. • 120 possums were observed to eat insects. • Of the 110 possums that were observed to eat nuts, 75 were honey possums. a

Complete the table shown based on the given information about possums: Brushtail possum

Honey possum

Total

Fruit Insects Nuts Total

Compare the total number of brushtail possums to the number of honey possums.

Compare the number of possums who were observed eating fruit to the total number of possums eating insects and nuts.

Compare the number of brushtail possums observed eating nuts to the number of honey possums observed eating insects.

Determine if each statement is true or false. Justify your response. a

The marginal frequencies for the columns must have the same sum as the marginal frequencies for the rows.

Each of the joint frequencies in a single row can be equal to the marginal frequency of that row.

A poll was given where adults and teenagers were asked if they like pineapple on their pizza. The possible responses were “yes”, “no” and “sometimes”. The results of the poll show that: • 60 people responded to the poll. • 18 adults responded “yes”. • 24 of the people who responded were adults. • The number of people who responded “yes” was equal to the number of people who responded “no” plus the number of people who responded “sometimes”. Find and explain the error made in each statement: a

Since 18 adults responded yes, there were 60 − 18 = 42 adults who responded either no or sometimes.

There were 18 adults who said yes and 24 of the people were adults so the total of adults was 42.

The number of people who responded “yes” was equal to the number of people who responded “no” plus the number of people who responded “sometimes”. So we can take the total number of people, 60, and divide by 3 to ﬁnd the number of people who said each response.

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Problem-solving 15, 16 15

16, 17

16−18

Yuri surveyed a group of people about the type of employment they had. He recorded the data in the graph. 40 35

Number of People

30 25 Men

20 15

Women

10 5 0

None

Part-time Casual Type of Employment

Full-time

Complete the two-way table with the information from the graph. No job

Casual

Part-time

Full-time

Men Women

How many people are unemployed?

What percentage of men are employed full-time? Round your answer to one decimal place.

What fraction of people employed casually are male?

If 111 people from the group are randomly selected, how many would you expect to be employed part-time?

In a biology study, the appearance of an animal carrying a certain gene is recorded. Assume that all animals must carry the MC1R gene or the Agouti gene but not both.

Light fur Dark fur a

520

Agouti gene 37 x

If the proportion of all the animals recorded carrying the MC1R gene and having light fur is

MC1R gene 36 28

, ﬁnd the value of x.

What does the value of x mean in this context?

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Lucinda sets out to see how many of her classmates use the term “soccer” or “football” to describe her favourite sport. Lucinda plans to survey her classmates from her ﬁrst three periods, and creates the following table for organising the data. Football

Soccer

Total

Period 1 Period 2 Period 3 Total Complete the frequency table so that it meets the conditions: • None of her class periods have the same number of students. • Overall, most students use the term “soccer”. • One of Lucinda’s class periods is equally divided between “soccer” and “football”. • The number of students who prefer “football” in one class period was equal to the number of students who prefer “soccer” in a different class period. 18

At a high school with 5000 students, a survey was conducted to investigate the participation of junior and senior students in school sports activities. Out of a sample of 100 students, a two-way table was produced. Always Often Sometimes Rarely Never Total

Junior students 12 9 10 7 11 49

Senior students 8 7 11 12 13 51

Total 20 16 21 19 24 100

Estimate the number of senior students in the entire school population who would say they “Never” participate in school sports.

On a particularly hot week, the school canteen had a sale on ice blocks. The sports coach believes that the sale might influence sports participation due to students preferring to sit out with an ice block instead of playing. How might you design a followup survey or data collection method to investigate this hypothesis using two-way tables?

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10.02 Associations between categorical variables After this lesson, you will be able to… • calculate and interpret row and column percentages from given data in a two-way table. • analyse data using row and column percentage tables to determine associations. • use 100% stacked column graphs to visually represent and analyse data. • describe the association between variables, if any, and provide reasoning based on the data. • differentiate between an association and causation between variables.

Row and column percentages Two-way tables display relationships between two categorical data sets. Categories label the table’s top and left side, while frequencies ﬁll the table. Row and column totals are usually included. This two-way table displays the statistics of the passengers and crew who sailed on the Titanic on its fateful maiden voyage in 1912: Survived Died Total

First class 202 123 325

Second class 118 167 285

Third class 178 528 706

Crew 212 696 908

Total 710 1514 2224

Although it is interesting to note that 202 First Class passengers survived, it is far more useful to know the percentage break up of survivors from each class. Row percentage tables and column percentage tables give us more useful information than the raw data alone. To know the proportion of survivors from each class, we need to calculate row percentages. To ﬁnd the row percentage, divide the value in the table by the row total and multiply by 100%. The row percentages are shown in this table with working shown in some of the cells: First class

Second class

Survived Died

11%

Third class

Crew

Total

25%

30%

100%

35%

46%

100%

This percentage frequency table now gives us more useful information than the raw data. The ﬁrst row gives us the percentage breakdown of survivors by class. We can now easily read that 46% of the people who died were crew members, whereas only 8% of the people who died were in ﬁrst class.

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We can also calculate the percentage in each class type that survived or died. To do this, we calculate column percentages. To ﬁnd the column percentage, divide the value in the table by the column total and multiply by 100%. The column percentages are shown in this table with working shown in some of the cells: First class

Second class

Third class

Crew

25%

23%

59%

75%

77%

100%

Survived Died Total

100%

The first column gives us the percentage breakdown of survivals and deaths in first class. From the raw data, we can see that a similar number of first class (202) and third class (178) passengers survived, but this can be misleading. The percentage frequency table shows us that 62% of first class passengers survived, whereas only 25% of third class passengers survived.

Example 1 Maria surveyed a group of people according to their age about the type of job they had. She recorded the data in this graph: 40

Number of people

35 30 25 Under 50

Over 50

15 10 5 0

None

Casual Part-time Type of employment

Full-time

a Complete the two-way table displaying the row percentages, rounding your answers to one decimal place. Under 50 Over 50

None 17.6% 22.2%

Casual 27.8%

Part-time 29.4%

Full-time 41.2%

Total 100% 100%

Create a strategy Create a table for the data. Then for each missing value, divide the value in the table by the row total and multiply by 100%.

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Apply the idea From the graph, we can make the following two-way table: Under 50 Over 50 Total

None 15 20 35

Casual 10 25 35

Part-time 25 15 40

Full-time 35 30 65

Total 85 90

From the table, we can see that 10 of the casual employees are under 50, and the row total for under 50 is 85. Find the row percentage

Evaluate and round

From the table, we can see that 15 of the part-time employees are over 50, and the row total for over 50 is 90. Find the row percentage

Evaluate and round

From the table, we can see that there were 30 full-time employees that are over 50. Find the row percentage

Evaluate and round

The row percentage table is: Under 50 Over 50

None 17.6% 22.2%

Casual 11.8% 27.8%

Part-time 29.4% 16.7%

Full-time 41.2% 33.3%

Total 100% 100%

Reflect and check For casual employees that are over 50, we also could have subtracted the other values in the row from 100. Casual over 50 Percent = 100 − 17.6 − 29.4 – 41.2 = 11.8

b Complete the two-way table displaying the column percentages, rounding your answers to one decimal place. None Under 50 Over 50 Total

57.1% 100%

Casual 28.6% 100%

Part-time 62.5% 37.5% 100%

Full-time

100%

Create a strategy Use the table from (a) for each missing value, divide the value in the table by the column total and multiply by 100%.

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Apply the idea Under 50 Over 50 Total

None 15 20 35

Casual 10 25 35

Part-time 25 15 40

Full-time 35 30 65

Total 85 90

From the table, we can see that there were 15 under 50 in the None category, and the None column total is 35. Find the column percentage Evaluate and round From the table, we can see that there were 35 full-time under 50, and the full-time column total is 65. Find the column percentage Evaluate and round From the table, we can see that there were 25 casual employees that are over 50, and the casual column total is 35. Find the column percentage Evaluate and round From the table, we can see that there were 30 full-time employees that are over 50, and the fulltime column total is 65. Find the column percentage Evaluate and round The column percentage table is shown: Under 50 Over 50 Total

None 42.9% 57.1% 100%

Casual 28.6% 71.4% 100%

Part-time 62.5% 37.5% 100%

Full-time 53.8% 46.2% 100%

Reflect and check Notice that, when we have row percentage tables or column percentage tables, the values only sum to 100% in one of the margins. In this case, we would not get 100% when adding across the rows.

Idea summary To find the row percentage for a particular value, divide the value in the table by the row total and multiply by 100%. To find the column percentage for a particular value, divide the value in the table by the column total and multiply by 100%.

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Associations between categorical variables To find if there is an association between the variables in the Titanic table, we can ask the question, “Is the survival rate dependent on the class of the passenger?” In order to find this, we must first identify the independent variable. This is also known as the explanatory variable, which is the factor that is manipulated or controlled in an experiment or study to test its effects on the dependent variable. It explains, influences, or predicts changes in the response variable. In this situation, it is the class of passenger. Which percentage frequency table should we use? When to use a row percentage table Independent variable forms row headings

When to use a column percentage table Independent variable forms column headings

How do we tell if there is an association? For a row percentage table Look at values in the columns

For a column percentage table Look at values across the rows

The independent variable forms the heading of each column, therefore the column percentage frequency table will best indicate any patterns. Survived Died Total

First class 62% 38% 100%

Second class 41% 59% 100%

Third class 25% 75% 100%

Crew 23% 77% 100%

When we read across the first row in the column percentage frequency table and look at the numbers, we can see a clear difference in the percentages of passengers that lived or died in each class. This suggests there is an association between the class of passenger and the rate of survival. If the values are similar (±5%), we say there is no clear association. How do we describe the association? 1. State if there appears to be an apparent association or not. In this example, there appears to be an association between the variables. 2. Describe the association. In this example, we could suggest class of passenger affects the survival rate of the passengers. 3. Provide reasoning. We could say that based on the data, the higher the class of passenger, the more likely a passenger was to survive. Two-way tables show potential links between variables, with percentages making associations clear. Subtler links need deeper statistical study. Note: ‘association’ means one variable’s change seems to impact the other, not necessarily cause the change.

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Example 2 Glen surveyed all the students in year 12 at his school and summarised the results in a table: Height ≥ 170 cm Height < 170 cm Total

Play netball 33 13 46

Do not play netball 72 30 102

Total 105 43 148

a Which variable is the independent variable? A

Play netball

Height

Create a strategy

Apply the idea

Choose a variable which, when changed, would affect the other variable.

The decision of a student to play or not may depend on their actual height, whereas their height won’t change depending on their decision to play netball or not. So the height is the independent variable, option B.

b To examine if there is an association between height and playing netball, should Glen use a column or row percentage frequency table?

Create a strategy

Apply the idea

In a percentage frequency table, we want to sum We found in part (a) that the height is the the percentages of the independent variable. independent variable, which are the rows. To sum the percentages of heights, we should use a row frequency table.

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c Complete the row percentage frequency table for this data. Round your answers to the nearest percentage. Height ≥ 170 cm Height < 170 cm

Create a strategy

Play netball ⬚% ⬚%

Do not play netball ⬚% ⬚%

Total ⬚% ⬚%

Divide each number of students by the total number of students in that row and multiply the answer by 100%.

Apply the idea Here is the original table with the row totals: Height ≥ 170 cm Height < 170 cm

Play netball 33 13

Do not play netball 72 30

Total 105 43

Here is the table with the calculations for the row percentages: Play netball

Do not play netball

Total

Height ≥ 170 cm

100%

Height < 170 cm

100%

d Looking at the columns of the completed table, does there appear to be an association between height and playing netball? A

No, there does not appear to be any association as the numbers are similar.

es, there appears to be an association as the numbers are quite different. It seems that Y taller people like to play netball.

Create a strategy In a row percentage table, if the values are similar, then we say there is no clear association.

Apply the idea The completed row percentage table is shown: Height ≥ 170 cm Height < 170 cm

Play netball 31% 30%

Do not play netball 69% 70%

Total 100% 100%

The numbers in each column are quite similar with just 1% difference in values. Therefore, we can say that there does not appear to be any association between height and playing netball, option A.

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Idea summary When analysing associations between categorical variables, consider the type of percentage frequency table and how to identify and describe associations: • Which percentage frequency table should we use? When to use a row percentage table Independent variable forms row headings

When to use a column percentage table Independent variable forms column headings

• How do we tell if there is an association? For a row percentage table Look at values in the columns

For a column percentage table Look at values across the rows

If the values are similar, then we can say that there is no clear association. • How do we describe the association? 1. State if there is or is not an association apparent. For example: There appears to be an association between the variables. 2. Describe the association. For example: The class of passenger affects the survival rate of the passengers. 3. Provide reasoning. For example: Based on the data, the higher the class of passenger, the more likely the passenger were to survive.

100% stacked column graphs Association between variables can often be seen more clearly in a stacked column graph. Shown is a stacked column graph for the data from the Titanic table earlier. When we look at each column, we can see that the proportion of death to survival in each column is different. This indicates that there is an association between the variables. Death vs. survival: Titanic 1912 100% 90% 80%

Died, 38 Died, 59

70% 60%

Died, 75

Died, 77

Survived, 25

Survived, 23

Third class

Crew

50% 40% 30%

Survived, 62 Survived, 41

20% 10% 0% First class

Second class

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To draw a stacked column graph: • Label the horizontal axis with the independent variables. • Label the vertical axis with percentages from 0% to 100%. • Draw a column for each independent variable that reaches the height of 100% on the vertical axis. • To divide each column into the percentages as shown in the frequency table, start from the bottom of the column, count to the ﬁrst percentage and draw a horizontal line to mark it off, then count up to the second percentage from the horizontal line and then mark off again, until all sections are complete. • Write the label and percentage in each section of the columns which indicates the dependent variables displayed or provide a key.

Example 3 A group of year 12 students surveyed their class and recorded the hair colour and eye colour for each student. The results are displayed in the 100% stacked column chart shown: Hair colour for various eye colours 100% 80% Blonde

60%

Red Brown

40%

Black

20% 0% Blue

Green

Brown

Hazel

Eye colour

a What is the independent variable for this chart? A

Eye colour

Hair colour

Create a strategy In a stacked column graph, the independent variable is placed along the horizontal axis.

Apply the idea The stacked column graph has the eye colour placed along the horizontal axis. So it is the independent variable, option A.

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b Does the chart suggest an association between eye colour and hair colour? A

Yes, as the corresponding segments are similar in size.

Yes, as the corresponding segments are of different sizes.

No, as the corresponding segments are similar in size.

No, as the corresponding segments are of different sizes.

Create a strategy Check whether the proportion of the segments in each column are similar sizes.

Apply the idea The stacked column graph segments differ in size, which suggests an association between eye colour and hair colour, option B.

c Can we say that having blue eyes causes a high chance of having blonde hair? A

Yes. The data shows that students with blue eyes are more likely to have blonde hair.

No. There appears to be an association, but we cannot say that one causes the other.

Create a strategy Consider the difference between association and causation.

Apply the idea An association does not mean one variable causes the other variable to change, but that a change in one variable appears to affect the other. The chance of having blonde hair can be caused by many factors. Having blue eyes may affect the chance of having blonde hair, but we cannot say it is a cause. The correct answer is option B.

Idea summary When analysing a stacked column graph, we can determine if there is an association between the categorical variables by comparing the proportions of the segments in each column: • If the segments vary in size, we may say there is an association between the variables. • If the segments are similar in size, then there is no clear association between the variables. Association differs from causation. An association does not mean one variable causes the other variable to change, but that a change in one variable appears to affect the other.

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10.02 Practice questions Understanding 1

Explorer

Adventurer

1−4

Trailblazer

1−4

This table gives information on the treatment plans prescribed to dogs with canine diabetes. Depending on the severity of the case, dogs are either prescribed a special diet or insulin injections. Parkside Veterinary Clinic Riverbend Animal Hospital

Special diet 23 16

Insulin injections 15 12

Total 38 28

This table gives the row percentages on the treatment plans prescribed based on the previous table. Parkside Veterinary Clinic Riverbend Animal Hospital

Special diet 61% 57%

Insulin injections 39% 43%

Total 100% 100%

What percentage of dogs from the Riverbend Animal Hospital were prescribed with insulin injections?

Will the percentages remain the same when calculating the column percentages?

Is there an association between the pet clinic and the type of prescriptions for dogs with canine diabetes?

What type of percentage frequency table should be used when the explanatory variable forms the column headings?

When constructing a stacked column graph, is the explanatory variable labelled on the horizontal axis or the vertical axis?

Is there an association between household composition and distribution of money based on the data in the stacked column graph? Household expenditure by domestic situation 100% 90% 80% 70% 60% 50% 40% 30% 20% 10%

Housing 532

Single

Married

Transportation

Food

Married with kids Insurance

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Single with kids

Entertainment

Other

Fluency 5

5−8, 11, 12

Adults 30 45 75

Children 40 55 95

Total 70 100 170

A teacher recorded the hair type and colour of their students. The results are shown in a column graph. 10 9 8 Number of students

Ex 1

6−14

In a survey, people were asked about their favourite subjects. The results are shown in the table. Calculate the row percentages for each subject rounded to two decimal places. Math Science Total

5−9, 11, 12, 14

7 6

Straight

Curly

4 3 2 1 0

Brown

Blonde

Black

Complete the two-way table displaying the row percentages rounded to one decimal place. Straight Curly

Red

Red 5.9% 10%

Brown

Blonde 5.9%

Black 41.2%

40%

Total 100% 100%

Complete the two-way table displaying the column percentages rounded to one decimal place. Red Straight Curly Total

50% 100%

Brown 66.7% 100%

Blonde 25% 75% 100%

Black

100%

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A survey was conducted about people’s favourite movie genre. The results are shown in a column graph. 50 45

Number of people

40 35 30

Teens

Children

20 15 10 5 0

Comedy 30% 30%

Teens Children Total

Ex 2

534

Action

Drama 45%

Horror 5%

40%

Total 100% 100%

50% 100%

Action 33.3% 100%

Drama 81.8% 18.2% 100%

Horror

100%

Isaac surveyed all of the students in Year 12 at his school and summarised the results in a table. Height ≥ 170 cm Height < 170 cm

Horror

Complete the two-way table displaying the column percentages rounded to one decimal place. Comedy

Action Drama Movie genre

Complete the two-way table displaying the row percentages rounded to one decimal place. Teens Children

Comedy

Play tennis 62 60

Do not play tennis 138 140

Which variable is the explanatory variable?

Create a row percentage frequency table for this data. Round the values to the nearest percentage.

Members of a gym who are either currently pregnant or postpartum were asked what kind of training they do. Each of them only did one kind of training. The table shows the results.

Currently pregnant Currently postpartum

Cardio

Weight

Which variable is the explanatory variable?

Create a row percentage frequency table for this data. Round the values to the nearest percentage.

Does there appear to be an association between the type of training and the pregnancy status of gym members?

Does a person’s pregnancy status cause them to choose a certain type of training?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Some students were asked if they are left-handed or right-handed. The results are provided in the table: Not female Female Total

Left-handed 8 7 15

Total 55 72 127

Construct a row percentage two-way table. Round your answers to one decimal place.

Construct a column percentage two-way table. Round your answers to one decimal place.

The results of a survey of 279 people on their employment status and parenting status is shown in a row percentage table. People without kids People with kids

Ex 3

Right-handed 47 65 112

None 8% 10%

Part-time 17% 47%

Full-time 75% 43%

Total 100% 100%

Is employment status or parenting status the explanatory variable?

Does there appear to be an association between parenting status and employment status?

A stacked column chart shows the ﬁtness levels for various categories of smokers. Fitness of smokers 100% 80% 60% 40% 20% 0%

Very heavy (1-5)

Heavy (16-25)

Below average fitness

Moderate (6-15) Normal fitness

Light (1-5)

Non-smoker (0)

Above average fitness

What is the response variable for this chart?

Does there appear to be an association between ﬁtness level and category of smoker?

Can we say that being a non-smoker causes a high chance of an above average ﬁtness level?

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These tables and 100% stacked column graph shows the results of a survey on smoking: Introverts Extroverts

Smokers 37 51

Non-smokers 69 123

Introverts Extroverts

Smokers 35% 29%

Non-smokers 65% 71%

Total 100% 100%

Personality type of smokers 100

Percentage

80 60

Smokers Non-smokers

40 20 0

536

Extroverts Introverts Personality type

Is smoking status or personality type the explanatory variable?

Does the 100% stacked column graph indicate an association between the variables?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

170 people were surveyed about their music preference. The results have been recorded in a table. Music Preference Rock and Roll Classical Pop Rap Country and Western R and B Punk Total

Vegetarian 24 8 17 6 17 6 4 82

Non-vegetarian 19 15 17 2 24 9 2 88

Total 43 23 34 8 41 15 6 170

What is the explanatory variable in this data set?

Which of the 100% stacked column charts should be used to look for an association between the variables? A

Music preferences by dietary restriction 100% 80% 60%

Non-vegetarian Vegetarian

40% 20% 0%

Rock & Classical roll

Pop

Rap Country & R&B western

Punk

Music preference

Music preferences by dietary restriction 100% 80%

Punk R&B

60%

Country & western Rap Pop

40%

Classical Rock & roll

20% 0% Vegetarian

Non-vegetarian

Dietary restriction

Does this stacked column chart suggest that there is an association between music preference and dietary restriction? 10.02 Associations between categorical variables mathspace.co

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Reasoning

15, 16, 18

17, 19

What does it mean if the sections in each column of a stacked column graph are of similar size?

In Australia, 130 random people were surveyed, examining their carbon footprint and the city they lived in. The people were then categorised as living in either an urban or regional location, and whether they caused high carbon emissions or low carbon emissions. The results are displayed in a table. High emissions Low emissions Total a b c d

Urban 48 30 78

Mr. Tranor asked his class to pick their favourite subject. He displayed the results in a twoway table.

a b c d e f

Maths 13 20

Music 9 12

Science 11 18

English 11 10

How many left-handed children did not pick maths as their favourite subject? How many students picked music as their favourite subject? Is favourite subject or dominant hand the explanatory variable? To examine if there is an association between the variables, should we use a column or row percentage table? Create this frequency table, rounding your answers to the nearest percentage. Does there appear to be an association between a child’s dominant hand and their favourite subject? Explain your answer.

In a study of car accidents, data was found on the number of passengers in the car and whether or not the car rolled over. No. of passengers Roll over No roll over Total a b c d

538

Total 68 62 130

Is location or emission level the explanatory variable? To examine if there is an association between the variables, should we use a column or row percentage table? Construct this table for the data, rounding your answers to the nearest whole number. Does there appear to be an association between location and emission level? Explain your answer.

Right-handed children Left-handed children

Regional 20 32 52

<5 335 1622 1957

5—9 18 42 60

10—15 17 42 59

> 15 2 5 7

Total 372 1711 2083

Which variable is the explanatory variable? To examine if there is an association between the variables, should we use a column or row percentage table? Create this percentage frequency table for the data. Round your answers to one decimal place. Does there appear to be an association between passengers and rolling over? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

At a local university, students were asked what their favourite subject at high school was and what they have decided to major in at university. The results are shown in the following table. Favourite subject

Maths major Science major Music major Art major Total

Maths 66 51 38 12 167

Music 21 43 68 48 180

Art 42 25 37 76 180

Total 185 159 176 153 673

Which variable is the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table for this data. Round your answers to two decimal places.

Does there appear to be an association between favourite subject at school and university major? Explain your answer.

A lecturer commented that “having mathematics as their favourite subject at high school causes many students to take mathematics at university”. Is he correct? Explain your answer.

Problem-solving 20, 21 20

Science 56 40 33 17 146

20, 21

The local library constructed divided bar graphs showing the last six months of book loans by category. Assume the total number of loans was exactly the same for each month.

20−22

Jun Jul Aug

In which month were business books the most borrowed type of book?

Sep

In which month were children’s books the most popular?

Oct

Which variable is the response variable?

Nov

Do the divided bar graphs suggest that there is an association between the variables?

Fiction Design Environmental

Children’s Business

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A group of people were asked if they are employed and if they have a smartphone. The results are shown in the table: Owns a smartphone Does not own a smartphone Total

Employed 34 71 105

Unemployed 18 87 105

Total 52 158 210

Is employment status or smartphone ownership the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table rounding your answers to the nearest percentage.

Does there appear to be an association between employment and owning a smartphone? Explain your answer.

Does a person’s employment status cause them to own a smartphone? Explain your answer.

Create a 100% stacked column graph for your percentage frequency table.

Does the 100% stacked column graph indicate an association between the variables? Explain your answer.

118 people were surveyed about their age and how many hours they watch television each week. The results are shown in a table.

Less than 3 hours 4 to 8 hours 9 to 15 hours Over 15 hours Total

540

10 years and under 9 2 7 10 28

11 to 20 years 8 10 8 9 35

21 to 30 years 2 2 8 9 21

31 years or older 5 9 10 10 34

Total 24 23 33 38 118

Is age or time spent watching TV the explanatory variable?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table rounding your answers to two decimal places.

Create a 100% stacked column graph for this percentage frequency table.

Does the 100% stacked column graph indicate an association between the variables? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

10.03 Bivariate numerical data After this lesson, you will be able to… • identify what bivariate numerical data is. • identify the independent and dependent variables in a pair of bivariate data. • plot and interpret scatterplots to analyse bivariate data. • apply understanding of bivariate numerical data to solve real-world problems.

Bivariate numerical data Bivariate numerical data is the name for numerical data consisting of pairs of values. Each study participant will have two pieces of data collected. We use these pairs to find out whether there is a simple relation between variables.

Person 1 Person 2 Person 3 Person 4 Person 5

Age ( years) 30 40 50 60 70

Bone density (g/cm³) 1.35 1.28 1.22 1.10 0.97

For a study about bone health, a person’s bone density is measured against their age. Their age is the independent variable and can be any value. Their bone density is the dependent variable that is recorded against their age.

Each person’s age and bone density make a pair of values in the bivariate data set. A single data point in a bivariate data set is written in the form (x, y), with x being the independent variable and y being the dependent variable. We display bivariate data graphically by plotting the data points Bone density with the value of the independent variable on the horizontal Bone density ( g/cm3) axis and the value of the dependent variable on the vertical 1.4 axis. This is known as a scatterplot. We may then check whether age is related to bone density by analysing the relationship between the variables graphically. It’s important to note that the existence of a relationship between two variables in a scatterplot does not necessarily imply that one causes the other.

1.2 1

0.8 0.6 0.4 0.2

Age (years)

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Example 1 Consider the following variables: • Number of ice cream cones sold • Temperature (°C) a Which of the following statements makes sense? A

The temperature affects the number of ice cream cones sold.

The number of ice cream cones sold affects the temperature.

Create a strategy

Apply the idea

Determine which variable has an effect on the other.

When hot weather occurs, more ice cream cones will be sold. Less ice cream cones will be sold in cold weather. So, the correct answer is option A.

b Which is the dependent variable and which is the independent variable?

Create a strategy

Apply the idea

Recall that the independent variable is not affected by the other variable. In contrast, the dependent variable is affected or changed by the other variable.

Based on the previous part, we can see that the independent variable is the temperature and the dependent variable is the number of ice cream cones sold, since the temperature affects the number of ice creams sold.

Example 2 The prices of ten houses are graphed against the house’s Price (thousands) land area on the following graph: 1200 1000 800 600 400 200

Describe the relationship between land area and house price in the data.

Area (m2) 200 400 600 800 1000 1200

Create a strategy Describe what happens to the dependent variable ( price) as the independent variable (area) increases.

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Apply the idea The area increases from left to right. We can see from the graph that the price increases (rises) from left to right. So as the land area increases, the house price increases.

Example 3 Scientists were looking for a relationship between the number of hours of sleep we receive and the effect it has on our motor and process skills. Subjects were asked to sleep for different amounts of time, and they were all asked to undergo the same driving challenge in which their reaction time was measured. The table shows the results, which are to be presented as a scatterplot. Amount of sleep (hours) 9 6 4 10 3 7 2 5

Reaction time (seconds) 3 3.3 3.5 3 3.7 3.2 3.85 3.55

a Create a scatterplot for the observations in the table.

Create a strategy Plot the points on a Cartesian plane.

Apply the idea 3.9

Reaction time (seconds)

The scatterplot of the data is shown. Reaction time is dependent on how much sleep the person gets. So the values on the horizontal axis represent the number of hours of sleep and the values on the vertical axis represent the reaction time.

3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1

Sleep (hours) 1 2 3 4 5 6 7 8 9 10

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b According to the results, describe the relationship between amount of sleep and reaction time.

Create a strategy Describe what happens to the dependent variable (reaction time) as the independent variable (sleep) increases.

Apply the idea The sleep hours increase from left to right. We can see from the graph that the reaction time decreases ( falls) from left to right. So as the hours of sleep increase, the reaction time decreases.

Idea summary In statistics, we often study pairs of values to understand potential relationships between them. This is called bivariate numerical data. Each pair of values in bivariate data is known as a data point. These data points consist of an independent variable, which could be any value, and a dependent variable, which is the outcome we measure against the independent variable. To visualise these pairs of values, we plot the data on a scatterplot. By plotting these data points, we can observe any potential relationships between the variables.

10.03 Practice questions Understanding 1

544

Explorer 1, 2ac, 3, 4

Adventurer 1, 2bc, 3, 4

Trailblazer 1, 2cd, 3, 4

Is each an example of univariate data or bivariate data? a

A school collects data on the height of each student in two classes in order to compare the classes.

A scientist collects data on iron levels in soil and growth of a type of weed in order to investigate the relationship between them.

A school collects data on the shoe size of each student in the school.

A hockey club collects data on time travelled to get to training from members of three teams in order to compare them.

A political group collects data on the taxable income and addresses of the local electoral constituents in order to investigate if there is an association between them.

A psychologist collects data on the number of days in daycare and NAPLAN results in grade three in order to investigate the relationship between them.

John wishes to investigate whether there is an association between the amount of natural sunlight in a classroom and student’s exam results.

Emma wishes to investigate whether players in the A hockey team have higher average mathematics test results than players in the B hockey team.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Create a scatterplot for each set of data: a

x 1 3 5 7 y 2 7 11 15

9 17

x y

1 3

3 7

5 11

7 13

9 17

x 1 2 4 7 y 20 18 17 15

8 11

x y

1 7

3 3

5 11

9 15

12 19

Is there likely to be a relationship between each pair of variables? a

Study time and test score

Length of hair and number of pets

Typing speed and dancing ability

Driving time and petrol used

Identify whether the lines are increasing, decreasing, or neither. a

y 4

1 −4 −3 −2 −1 −1

x 1

−2

−3

−4

d 4

−4 −3 −2 −1 −1

−3

−4

6, 7, 10, 12, 14, 16

−2

Fluency

x 1

−4 −3 −2 −1 −1

6−8, 10−12, 14, 16

5, 7−11, 13−16

For each of the pairs of variables: i

What is the independent variable?

Number of dogs owned and number of destroyed shoes

Size of a pizza and cost of a pizza

Time spent studying and exam mark

What is the dependent variable?

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Ex 1

Consider the variables: • Ticket sales

• Revenue from a show

Which variable affects the other?

What are the dependent and independent variables?

A kite can rise to a height of 161 m when the wind is blowing at 6 kilometres per hour and to a height of 322 m when the wind is blowing at 12 kilometres per hour. a

What are the two variables?

From the two variables, state the dependent and independent variables.

Does one variable have a direct effect on the other? If yes, which variable affects the other? a

Temperature (°C), Distance from the equator (km)

Weekly income, Number of friends

Time of travel (minutes), Distance covered (m)

Age (years), Time spent watching television in a day

Time spent working, Wages earned

Time spent to ﬁnish a novel, Number of pages

Height (cm), Number of children

Number of pets, Time spent caring for pets

Time spent training, Athletic performance

The graph shows the height of a ball after it is dropped off the side of a building. a

Which variable is the dependent variable?

Which variable is the independent variable?

Height

Time

Ex 2

A cafe records the number of soups sold and the daily maximum temperature on a graph. What is the relationship between temperature and soups sold in the data?

Soups sold 60 48 36 24 12 4

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Temperature 16 20

A company records the height and annual salary ($’000) of ten employees on a graph. What is the relationship between height and salary in the data?

Annual salary 100 80 60 40 20 Height (cm) 40

Ex 3

120

160 200

The table presents data on the number of immigrants and the unemployment rate in Australia over a 10-year period: Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 a b c

Number of immigrants (thousands) 250 265 275 280 290 295 300 305 310 320

Unemployment rate (%) 5.2 5.0 4.8 4.6 4.5 4.3 4.1 4.0 3.9 3.7

What are the independent and dependent variables? Create a scatterplot for the set of data. According to the results, what is the relationship between the number of immigrants and the unemployment rate?

The table presents data on the global annual average carbon dioxide emissions and the average global temperature anomalies over a 10-year period: Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 a b c

CO2 emissions (million metric tonnes) 35 000 35 500 36 000 36 500 37 000 37 500 38 000 38 500 39 000 39 500

Global temperature anomaly (°C) 0.6 0.65 0.68 0.7 0.75 0.8 0.83 0.85 0.88 0.9

Which are the independent and dependent variables? Create a scatterplot for the set of data. Based on the scatterplot, how do the CO2 emissions relate to the global temperature anomalies? 10.03 Bivariate numerical data mathspace.co

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This table shows the number of apples eaten per week and the number of visits to the doctor per year by six students: Number of apples eaten per week Number of visits to the doctor per year

7 6

9 3

11 0

Create a scatterplot for the set of data. Vanessa wants to visit the doctor as few times as possible each year. According to the plot, should she eat 0 or 11 apples per week?

The height of the tide at a beach is recorded at nine different hours in one day: 13 −0.21

24 −0.05

7 0.92

2 0.55

4 0.84

What are the dependent and independent variables?

Create a scatterplot for the set of data.

15 −0.76

6 1.01

11 0.31

21 −0.68

This table shows the age and height of six children in a family: 3 90

4 105

9 130

11 145

13 160

16 165

Create a scatterplot for the set of data.

Bob is the second youngest child in the family. Use the plot from part (a) to ﬁnd out how tall he is.

Reasoning

17, 18

Six students were asked how many people there are in their family, and the number of trees planted in their garden. Their responses were plotted on a graph. Each dot represents one student.

548

6 11

Age ( years) Height (cm)

3 14

Time (hours) Height (metres)

0 17

One student, Jack, has 7 people in his family. How many trees are planted in his garden?

Do you think there is a connection between the number of people in a student’s family and the number of trees in their garden? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

17−19

32 Number of trees in the garden 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 Number of people in the family 3 4 5 6 7 8 9 10 11 12 13 14 15

This table shows the average time spent on a website (in minutes) and the average amount spent (in dollars) for a group of online shoppers in a particular month: Week 1 Week 2 Week 3 Week 4

Average time spent (minutes) 15 30 45 60

Average amount spent (dollars) 75 130 185 250

Create a scatterplot for the set of data.

A statement is made: “The more time a shopper spends on a website, the less they spend.” Critique this claim based on your scatterplot.

The table shows the number of books read per year by six students and the mark they received in their yearly maths exam. Number of books read per year Mark in yearly maths exam

5 50

15 65

20 70

30 80

40 90

50 100

Create a scatterplot for the set of data.

In this group of students, Valentina achieved the highest mark in the yearly maths exam. Use the plot from part (a) to ﬁnd out how many books she read throughout the year.

The teacher told the class: “Students who read more books also tend to score higher on maths exams.” According to the plot in part (a), is the teacher correct? Explain your answer.

Problem-solving 20, 21

20−22

Georgia has a high income but does not like her job. Which dot would represent her? Explain your answer.

3 2

Job Satisfaction

20−22

4 1

0 0

Income

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The scatterplot represents the social media usage (in minutes per day) and the respective grades of ﬁve students: A, B, C, D, and E.

Which student seems to spend the least amount of time on social media?

Which student has the lowest grades despite not spending the most time on social media?

Which student seems to manage their time best on social media while maintaining good grades?

Student survey Grades

B B A A A E E E C C D D Social media usage (minutes per day)

I dentify a pair of variables in everyday life that you believe may have a relationship. Explain what the variables are and why you think there may be a relationship.

Identify which variable is the independent variable and which is the dependent variable in your chosen pair from part (a). Explain your choice.

Identify a pair of variables in everyday life that you believe do not have a relationship. Explain what the variables are and why you think they do not have a relationship.

10.04 Associations between numerical variables After this lesson, you will be able to… • interpret scatterplots to describe the association between two numerical variables. • identify the form of a data set as linear or non-linear. • examine scatterplots to determine the form, direction and strength of data points. • recognise outliers in a data set. • distinguish between correlation and causation using real-world examples.

Associations between numerical variables We analyse how the independent variable (x) inﬂuences the dependent variable ( y) when determining associations between numerical bivariate variables. If an association is linear, then it can be referred to as a correlation. To identify an association between variables, we examine: • Form: The overall pattern or shape of the data points—linear or non-linear.

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• Direction (if linear): The trend of the data—positive (increasing y with increasing x, similar to a positive gradient) or negative (decreasing y with increasing x, similar to a negative gradient). • Strength (if linear): The extent to which points follow a straight line—strong (close to a line), moderate, weak, or no correlation. While data points might display non-linear forms like curves, we’ll focus on linear models. A perfect correlation means all points lie exactly on a straight line, and no correlation signifies a random scatter of points. Most plots fall between these extremes, showing weak, moderate, or strong correlation.

Interactive exploration

Discover this concept in action online

mathspace.co

Perfect positive correlation

Perfect negative correlation

Strong positive correlation

Strong negative correlation

Moderate positive correlation

Moderate negative correlation

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Weak positive correlation

Weak negative correlation

No correlation

We may also see outliers, which are data points drastically out of the pattern. Correlation means that the variables are related, and often does not mean that the changes in the dependent variable are caused by the independent variable. Causation means there is a cause and effect and must be shown with carefully designed studies. Two variables may be strongly correlated, but not necessarily causal for a few reasons. 1. Third (contributing) variable: A third variable is contributing to both x and y, so the relationship between x and y is not causal. • For example: There is a correlation between number of cats and number of street lights in a city, but cats don’t cause street lights. There is a third variable of population size. 2. Coincidence: It is purely coincidence that the two variables are related. • For example: There is a correlation between butter sales and number of female CEOs. 3. Unproven: In some cases, to be causation, it must be proven using proper and repeatable statistical studies. • For example: Cigarette smoking was not proven to cause cancer until the 1950-60s, though it was suspected and correlated before that.

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Example 1 From the scatterplot given, we can conclude that:

Vision

x 130 140 150 160 170 180 Height (cm)

A There is no direct relationship between someone’s vision and their height. B Those who are taller have better vision. C The worse someone’s vision, the taller they are.

Apply the idea Examine the scatterplot and determine whether there is a discernable relationship between height and vision. Based on the given scatterplot, there is no direct relationship between someone’s vision and their height. So, the correct answer is option A.

Example 2 The table shows the marks of 12 students in English and French. Student 1 2 3 4 5 6 7 8 9 10 11 12

Mark in English 85 71 57 60 79 76 71 91 50 49 66 92

Mark in French 89 71 56 62 86 76 77 86 90 47 67 92

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a Which of the following scatterplots correctly represents the data? 100

100

90 80 70 60 50

40 40 50 60 70 80 90 100

Mark in English

100 Mark in French

B Mark in French

Mark in French

90 80 70 60 50 40 40 50 60 70 80 90 100 Mark in English

Apply the idea Draw your own scatterplot and plot each given point to identify the correct graph. The correct answer is A.

b Is the correlation between students’ English and French marks positive or negative? A

Positive

Negative

Create a strategy

Apply the idea

Think of what happen to the students’ mark in French as their marks in English increase.

As the students’ marks in English increase, so too do their marks in French. So the answer is A: Positive

c Is the correlation between students’ English and French marks strong or weak? A

Strong

Weak

Apply the idea When the points on a scatterplot tend to follow a single line, the correlation is strong. When the points on a scatterplot are scattered greatly around a single line, the correlation is weak. So the answer is A: Strong

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d Which student’s scores represent an outlier? A

Student 10

Student 9

Student 12

None of them

Apply the idea Which point, if any, on the scatterplot is positioned away from the rest of the data? The answer is B: Student 9

Example 3 This table of values that show four excerpts from a database comparing the income per capita of a country and the child mortality rate of the country. If a scatterplot was created from the entire database, what relationship would you expect it to have? A Strongly positive

No relationship

Income per capita 1465 11 428 2621 32 468 C

Child mortality rate 67 16 35 9

Strongly negative

Create a strategy

Apply the idea

Look at how the mortality rate changes as the income increases.

As the income per capita increases, the child mortality rate decreases, so there is a negative correlation between the two sets of data. The correct answer is C.

Example 4 A study found a strong positive association between the temperature and the number of beach drownings. a Does this mean that the temperature causes people to drown? A

Yes

Apply the idea The statement that there’s a strong positive association between temperature and beach drownings does not mean temperature causes people to drown. This simply indicates a correlation, not causation. So, the correct answer is B: No.

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b Which of the following is the most likely explanation for the strong positive association? A

Contributing variables - there are other causal relationships and variables that come into play, and these may lead to an indirect positive association between the temperature and the number of beach drownings.

Coincidence - there are no other contributing factors or reasonable arguments to be made for the strong positive association between the temperature and the number of beach drownings.

Create a strategy Is there a common factor that has a direct relationship with both temperature and the number of beach drownings?

Apply the idea A plausible argument would be that the increase in temperature could lead to more people visiting the beach and therefore more chances of drowning incidents. There might be other variables at play, like stronger currents during warmer periods or increased alcohol consumption. So, the correct answer is A.

Idea summary To identify an association or a correlation between variables, we examine: • Form: The overall pattern or shape of the data points—linear or non-linear. • Direction (if linear): The trend of the data—positive (increasing y with increasing x, similar to a positive gradient) or negative (decreasing y with increasing x, similar to a negative gradient). • Strength (if linear): The extent to which points follow a straight line—strong (close to a line), moderate, weak, or no correlation. • Presence of outliers However, correlation doesn’t always imply causation. Apparent relationships between x and y don’t necessarily mean changes in one cause changes in the other. There could be an unobserved variable indirectly linking x and y.

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10.04 Practice questions Understanding 1

Explorer

Adventurer

1−3

Trailblazer

1−3

Four different classes with four different professors had the same final exam. The exam results and number of classes attended by each student is displayed for each class. Does each graph suggest that there is an association between number of classes attended and final exam mark? a

b 100 Final exam mark

Final exam mark

100

x 0

10 20 Number of class attended

Number of classes attended

100

100 Final exam mark

Final exam mark

x 0

Number of classes attended

Each point on the scatterplot shows the time (in weeks) Sumon spent training for a half marathon and the corresponding kilometres they were able to run: Using the scatterplot, are these statements true or false? a

The number of weeks that Sumon trained for the half marathon is the independent variable.

The y-coordinates of the points represent the time spent by Sumon training.

There is evidence to suggest that the longer Sumon trains, the further they can run.

Running distance (km) 20 15 10 5 No. of weeks 5

Training for a half marathon causes further running distance over time.

The correlation between the number of weeks training and the number of kilometres Sumon is able to run is positive. 10.04 Associations between numerical variables mathspace.co

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Complete each sentence with one of the words from this list. • association • causation • coincidence a b c d

The strength and direction of a scatterplot describe the ⬚. When two variables are related, we say there is an ⬚. To show ⬚, we must do a carefully designed study.

When correlation is not causation, some possible explanations are ⬚ and ⬚.

Fluency 4

• contributing (third) variables • correlation

4−6, 8, 9, 12−14, 16

4−8, 10, 12, 14, 16

4−8, 10, 11, 15, 16

Does each scatterplot show a linear or non-linear relationship? a

5 x

x 5

5 x

x 5

5 x 5

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Describe the strength and direction for each linear association. a

5 x 5

5 x

x 5

This scatterplot shows the relationship between air and sea temperature: a

Describe the relationship between air and sea temperature.

Describe the correlation between air and sea temperature.

35 Sea temperature

Ex 1

30 25

Air temperature

Scientists conducted a study where each person was asked to read a paragraph then recount as much information as they could remember. They found that the longer the paragraph, the less information each person could retain. If the length of the paragraph were plotted (on the horizontal axis) against the amount of information retained (on the vertical axis), would the correlation be negative or positive?

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Ex 2

This table shows the marks of 12 students in Maths and Sports: a

Construct a scatterplot for the students’ marks in Maths versus their marks in Sports.

Describe the correlation between students’ marks in Maths and marks in Sports.

Which student’s scores appear to represent an outlier?

560

Sports 44 74 52 70 67 60 73 86 84 93 57 92

20 41

25 44

30 39

35 34

40 30

45 25

50 22

55 18

60 19

Construct a scatterplot for this data.

Describe the correlation between a person’s age and the number of accidents.

Does this data set contain any outliers?

65 17

This table shows the number of hours of sleep a person gets per week associated with a sample of persons of different age groups. Age Hours (per week)

Maths 63 92 60 79 88 81 61 91 72 42 66 92

This table shows the number of traﬃc accidents associated with a sample of drivers of different age groups. Age Accidents

Student 1 2 3 4 5 6 7 8 9 10 11 12

20 42

25 39

30 45

35 37

40 38

45 41

50 36

55 34

60 33

Construct a scatterplot for this data.

What is the correlation between a person’s age and the hours of sleep per week?

Does this data set contain any outliers?

The market price of bananas varies throughout the year. Each month, a consumer group compared the average quantity of bananas supplied by each producer to the average market price ( per unit). The results are shown in a table. a

Construct a scatterplot using the data from the table.

Describe the correlation between the supply quantity and the market price of bananas.

According to this data, when will a supplier of bananas receive a higher price per banana?

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Supply (kg) 550 600 650 700 750 800 850 900 950 1000

65 30

Price (dollars) 16.25 15.75 15.75 14.75 14.50 14.00 13.75 12.75 12.50 12.00

This table lists the time taken to sprint 400 m by runners who all ran in different temperatures as part of a study. a

Construct a scatterplot to represent the data in the table.

How many runners were tested in the study?

Describe the correlation between temperature and sprint time for the data.

Which data point represents an outlier?

Time (sec) 60 67 48 69 65 49 57 53

A researcher is studying the relationship between the number of passers-by in an emergency, and the time taken (in seconds) before a passer-by helps a stranger during an emergency. The data is recorded in this table. Number of passers-by (n) Time until help is offered (t)

Ex 3

Temperature (°C) 5 2 10 8 1 7 6 4

1 8

2 19

3 26

4 37

5 51

6 65

Construct a scatterplot using the data from the table.

As more passers-by are present what happens to the time taken until help is offered?

Describe the correlation between the number of passers-by and the time until assistance is offered.

Does the scatterplot contain any outliers?

This table of values shows the time taken to finish a lap on a race track for ten average speeds. If a scatterplot was created from the entire database, would you expect the relationship to be strongly positive, strongly negative or that there was no relationship? Speed (x) Time ( y)

20 85

25 87

30 75

35 82

40 69

45 73

50 60

55 57

60 45

65 49

This table of values shows four excerpts from a database comparing the literacy rate of a country and the gross domestic product (GDP) per capita of the country. If a scatterplot was created from the entire database would you expect the relationship to be strongly positive, strongly negative or that there was no relationship? Literacy rate 85% 92% 78% 95%

GDP per capita $5 000 $10 000 $3 000 $15 000

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Ex 4

For each data example, is there likely a causal relationship between the variables? a

Number of times a coin lands on heads and the likelihood that it lands on heads on the next ﬂip

Amount of weight training a person does and their strength

Hair length and gender

Marital status and ﬁlm preference

Height and number of children

Number of pets and time spent caring for pets (hours)

Time spent to ﬁnish novel and number of pages in the novel

Time spent training and athletic performance

Reasoning

17, 18

18, 20

18−20

A student collects some data that shows there is a positive increase between height and the ability to play soccer. Does that mean if you are tall, you are better at soccer? Justify your answer.

As preparation for a Science test a group of 10 students was given a revision sheet containing 60 questions. The table shows the number of questions from the revision sheet successfully completed by each student and the mark, out of 100 of that student on the test.

562

Which variable is independent and which variable is dependent?

Construct a scatterplot of the data.

Describe the correlation between the two variables suggested by the scatterplot. Explain your reasoning.

Explain why the relationship is not perfectly linear.

Number of questions 11 14 36 60 57 42 20 27 50 59

For the scatterplot: a

Explain why the correlation between the variables is weak.

Determine whether the pairs of variables could be represented by the data set. i

arks in an English examination and distance M travelled from home to school

Cost of cars and cost of petrol

iii

istance travelled in a car and the cost of a D driver’s licence

Height and weight of students at school

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Test result 20 23 62 97 100 66 35 52 87 99

A researcher is studying the relationship between the average temperature of a city and the number of people migrating to that city. She collected data from 10 different cities and plotted the data on a scatterplot. Describe a possible pattern that might be observed in the scatterplot if there is a relationship between the average temperature and migration.

Problem-solving 21, 23

21−24

A chemical company is testing the effect of different chemicals on slowing the melting of snow. They are currently testing four different chemicals with the results showing how much ice is remaining after 3 hours at a certain temperature. A

B 100 Snow remaining (%)

Snow remaining (%)

100

x 0

Temperature outside (°C)

100

100 Snow remaining (%)

Snow remaining (%)

21−24

x 0

Temperature outside (°C)

A snow sculpture company wants a chemical that will keep snow from melting. a

Which chemical should they choose to ensure that at least half of the snow is still there after 3 hours?

If the company operates at a temperature of 3°C to 4°C, which chemical should they choose?

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After a Maths exam, Uther commented that he left out Question 10 because it was not relevant to the topics that were assessed, but that the other questions were ﬁne. The teacher felt that Question 10 and Question 7 were assessing the same skills, but decided to look at the results of students across all 8 classes. She plotted the percentage of students who had answered Question 10 correctly against the percentage of students who had answered Question 7 correctly. a

% that got Q10 correct 4

% that got Q7 correct

Determine whether the statements describe the data. i

There is no relationship between students’ performance in Q7 and Q10.

Students performed similarly in Q7 and Q10.

iii

Students who performed poorly in Q7 performed well in Q10.

Let’s assume that Uther performed exceptionally well in every other question. Which region should the teacher expect Uther’s data point to be in?

Determine whether each statement is true or false. Provide an example to support your claim. a

If event A has a causal relationship with event B, then there is a correlation between event A and event B.

If there is a correlation between event A and event B, then there is a causal relationship between event A and event B.

If event A is positively associated with event B, then there is a causal relationship between event A and event B.

Construct a scenario with data that suggests a weak positive relationship between two variables. Justify why the relationship may be weak.

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10.05 Line of good fit After this lesson, you will be able to… • draw a line of good fit for a given data set. • interpret the gradient and y-intercept of a line of good fit. • use the equation of the line of good fit to make predictions about one variable based on the other. • evaluate the correlation between two variables using a line of good fit.

Line of good fit Exploration This scatterplot shows the car value in thousands of dollars and the number of years since the car was purchased for 20 different cars. 1. Is there a correlation between the years since purchased and the value in thousands of dollars? 2. Which of the lines on the graph ﬁts the data best - dashed or solid?

Car value Value (in thousands of dollars) y 30 25 20 15 10 5

Time since purchase (years) x 1

A line of good fit is a straight line drawn by eye that best represents the data on a scatterplot. It is sometimes called a trend line. We can use lines of good fit to help us make predictions or conclusions about the data. To draw a line of good fit, we want: • a line that passes through the centre of the points. Unless the correlation is perfect, it cannot pass through all of the points. • approximately the same number of points above and below the line • to minimise the vertical distances from the points to the line

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Given a set of data relating two variables x and y, it may be possible to form a linear model. This model can then be used to understand the relationship between the variables.

A line of good ﬁt has an equation of the form y = mx + c.

Height (cm) 12

The m value shows the gradient, identifying whether the correlation 6 is positive or negative. 4 • If the gradient is positive, when the independent variable increases by 1 unit, the dependent variable increases by m units. 2 Weekly growth The correlation in this case is positive. 1 2 3 4 5 6 7 8 9 • If the gradient is negative, when the independent variable increases by 1 unit, the dependent variable decreases by m units. The correlation in this case is negative. The c value shows the vertical intercept (also known as the y-intercept): • When the independent variable is 0, the value of the dependent variable is c. When we interpret the vertical intercept at (0, c), we need to consider if it makes sense for the independent variable to be zero and the dependent variable to have a value of c.

Example 1 The scatterplot shows the data for two variables, x and y:

10 9 8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 9 10

Draw a line of good ﬁt for the data.

Create a strategy Draw a line that follows the trend of the points, and has the same number of points above and below the line.

Apply the idea 10 9 8 7 6 5 4 3 2 1

Here is an example of a line of good ﬁt that follows the trend of the data and has the same number of points above and below the line.

x 1 2 3 4 5 6 7 8 9 10

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Example 2 The number of hours spent watching TV each evening, h, is measured against the percentage results, m, achieved in the Economics exam. The following linear regression model is ﬁtted to the data: m = −10h + 97 a The vertical intercept indicates that when a student watched no TV, the mark in the exam, on average, was ⬚%.

Create a strategy

The vertical intercept occurs when x = 0. In this case, the variable h is used instead of x. The value h = 0 represents a student that watched 0 hours of TV.

Apply the idea When a student watched no TV, h = 0. m = −10h + 97

Given equation

m = −10(0) + 97

Set h = 0

m = 97

Evaluate

So, the vertical intercept indicates that when a student watched no TV, the mark in the exam, on average, was 97%.

b Does the interpretation in the previous part make sense in this context? A

es, it is reasonable that a student who spends no time watching TV could get an exam Y mark of 97%.

o, it does not make sense because the exam mark is not within the valid range for the N dependent variable.

Create a strategy

Apply the idea

In this case, the independent variable is “hours watching TV”.

It is possible to watch no TV and earn a mark of 97, so the answer is option A.

Is watching no TV possible in the context? Is earning a mark of 97 possible in this context?

c State the gradient of the line.

Create a strategy

Apply the idea

When an equation is of the form y = mx + c, the gradient of the line is m, the number in front of x. In this case, h is used instead of x, so the gradient will be the number in front of h.

The number in front of the independent variable (h) is −10, so the gradient is −10.

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d If a student increases the amount of TV they watch by 3.5 hours, how much can we expect their Economics exam mark to drop by?

Create a strategy The gradient indicates how much their exam mark will decrease by for each additional hour they spend watching TV. The gradient of −10 indicates that their exam mark will decrease by 10% for each additional hour they spend watching TV.

Apply the idea Exam mark drop = 10% × 3.5 = 35%

Multiply the decrease percentage by hours Evaluate

Idea summary To draw a line of good ﬁt, we want to minimise the vertical distances from the points to the line. This will roughly create a line that passes through the centre of the points. The number of points above and below the line should be equal. A line of good ﬁt has an equation of the form y = mx + c. The m value shows the gradient, and whether the correlation is positive or negative. • If the gradient is positive, when the independent variable increases by 1 unit, the dependent variable increases by m units. The correlation in this case is positive. • If the gradient is negative, when the independent variable increases by 1 unit, the dependent variable decreases by m units. The correlation in this case is negative. The c value shows the vertical intercept (also known as the y-intercept): • When the independent variable is 0, the value of the dependent variable is c.

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10.05 Practice questions Understanding 1

Explorer

Adventurer

1−4

Trailblazer

1−4

The equation for the line of good ﬁt is given by P = −4t + 116, where t is time. Would the correlation be increasing, decreasing, or neither?

Choose the line of good fit for this scatterplot. 9

8 7

Line 2

Line 3

6 5 4

Line 1

3 2 1 1

2 3 4 5 6

8 9

Does there appear to be a linear relationship between the variables? a

5 x 5

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Looking at the scatterplot: y

45 40 35 30 25 20 15 10 5

x 1

a b

Is the relationship between the x and y-variables positive or negative? Which could be the equation for the line of good ﬁt? A

y = −4x − 4

y = 44 + 4x

y = −4x + 44

y = 4x − 4

Fluency Ex 1

5, 6a–c, 7, 9ac, 10−12

5, 6c–e, 8, 9bd, 10−12

This scatterplot shows the data for two variables, x and y. Draw a line of good ﬁt for the data. 19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7 8 9 10 11 1213 14 1516 17 1819

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5, 6d–f, 8−12

For each scatterplot: i

Draw a line of good fit by hand.

Write the equation of the line of good fit.

5 x 5

x 5

5 x 5

2 x 12

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This scatterplot graphs data for the number of people in a room and the room temperature. Draw a line of good ﬁt for this scatterplot.

38 36 34 32 30 28 26 24 22 20

Number of people

Room temperature 10 20 30 40 50 60 70 80 90

This scatterplot graphs data for the number of copies of a particular book sold at various prices. Draw a line of good ﬁt for this scatterplot.

190 180 170 160 150 140 130 120 110 100 90

Copies sold

Price 18 20 22 24 26 28 30 32 34 36

Ex 2

For each line of good fit: i

What is the gradient of the line?

Describe the correlation of the data set as positive or negative.

iii

Describe the change in y as x increases by 1 unit.

What is the value of the y-intercept?

y = 4.62x + 7.58

y = −7.68x + 5.49

y = −5.69x + 7.84

y = 3.16x − 2.72

The amount of money households spend on dining out each week, D, is measured against their weekly income, I. This linear model D = 0.3I + 27 is fit to the data.

572

Explain the meaning of the vertical intercept.

Does the interpretation in the previous part make sense in this context?

State the gradient of the line.

If the weekly income of a family increases by $200, by how much can we expect their spending on dining out to increase?

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This table shows the price of various used cars. Age ( years) Price (thousands of dollars)

1 16

2 13

0 21.99

Draw an approximate line of good ﬁt.

Find the y-intercept of the line of good ﬁt.

What does the y-intercept represent?

At what age would a car be worth $0?

5 10

7 8.6

4 12.5

3 11

4 11

8 4.5

2 14.5

Price (thousands of dollars) 24 20 16 12 8 4 Age (years) 4

Scientists record the number of aphids in areas with different numbers of ladybirds and graph the data in the scatterplot shown.

A 3200

A = −3.82L + 3865. 21 represents the line of good fit. a

How much does the average aphid population change by with each extra ladybird? Round your answer to the nearest aphid. Find the average aphid population of a region with no ladybirds. Round your answer to the nearest aphid.

2400 1600 800 L 200 400 600 800 1000

Reasoning 13

13, 14

The heights (in centimetres) and the weights (in kilograms) of 9 primary school children is shown on the scatterplot:

13−15

Weight (kg) 60 50

State the y-value of the y-intercept.

Describe the meaning of the y-intercept in context.

Does the interpretation in the previous part make sense in this context? Explain your answer.

(0, 27.7)

20 10 Height (cm) 20 40 60 80 100 120 140

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Concern over student use of the social media app LookAtMe leads to a study of student grades in Maths versus minutes spent using the app. The results are shown in the table: Minutes, M Grade, P %

292 26

153 63

354 13

253 37

11 97

42 89

195 51

7 98

162 59

254 36

Predict the grade of a student who spends no time on the LookAtMe app. Use a scatterplot and its line of good ﬁt to justify your response.

Use your model to explain and interpret the relationship between minutes spent on the LookAtMe app and Maths grades.

Lorena and Frasier each draw a line of good fit. a

Frasier said he noticed that the there was a point that was far away, so he moved his line closer to it. Does Frasier’s line meet the requirements for a line of good fit? Explain.

Frasier

7 6 5 4

Lorena said she noticed that the there was a point that was far away, but decided to ignore it. Does Lorena’s line meet the requirements for a line of good fit? Explain.

Lorena

3 2 1

x 2 4 6 8 10 12 14 16 18

Problem-solving 16−18 16

16−19

This scatterplot graphs data for the number of ice blocks sold at a shop on days with different temperatures. a

Sketch the line of good ﬁt for this data.

Use your line of good ﬁt to estimate the number of ice blocks that will be sold on a:

31°C day

42°C day

Does the number of ice blocks sold increase or decrease as the temperature increases?

17−19

Ice blocks 140 120 100 80 60 40 20 Temperature (°C) 14 16 18 20 22 24 26 28 30 32 34 36 38 40

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The distance of several locations from the equator and their temperature on a particular day is measured and their values are presented on the scatterplot. a

Which equation could represent the line relating distance (x) and temperature ( y)? A

y = −0.005x + 49

y = −0.005x – 49

y = 0.005x + 49

y = 0.005x − 49

20 Distance (1000 km)

Estimate the distance from the equator, x, if the temperature is 30.59°C.

For the scatterplot shown: a

Find the equation of the line of good ﬁt.

Use the line of good ﬁt to approximate the value of y for x = 6.9.

Beekeepers and scientists study a phenomenon known as colony collapse disorder (CCD), where the majority of worker bees in a hive disappear, leaving behind the queen and immature bees. The percentage of beehive losses that can be attributed to CCD each year, since 2005, is shown in a table. a b

Temperature (°C)

Construct a scatterplot and find a line of good fit for this data.

11 10 9 8 7 6 5 4 3 2 1

Year 2005 2006 2007 2008 2009 2010

4 5 6

8 9

x 1

2 3

Hives lost to CCD, (H)% 18 13 23 26 28 30

0 1 2 3 4 5

More data has been collected, and the results are listed: • In 2011, 30% of hives were lost to CCD. • In 2012, 33% of hives were lost to CCD. • In 2013, 36% of hives were lost to CCD. • In 2014, 40% of hives were lost to CCD. How would this new data affect the line of good fit?

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10.06 Interpolation and extrapolation After this lesson, you will be able to… • understand and differentiate between interpolation and extrapolation. • predict values using both the equation of a line and a graph of a line of good fit. • evaluate the reliability of predictions. • apply their understanding of interpolation and extrapolation to solve real-world problems.

Interpolation and extrapolation We can draw lines of good fit by hand, but by using technology, we can find a regression line which is the best line that summarises the data set. The least-squares regression line is a line of fit that is calculated using technology.

Interactive exploration

Discover this concept in action online

mathspace.co

Once we have the equation of a line of fit, we can use it to make predictions. We can: • Substitute the x-value into the equation of the line to calculate the predicted y-value. • Evaluate whether the prediction is reliable. Predictions can also be made using the graph of the line of good fit. We go up from the x-axis to the line of fit and then across to the y-axis to find the predicted value. S

Ice Cream Sales

400

300

200

100

T 10

576

20 Temperature (°C)

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These terms describe the range in which we make predictions: • Interpolation: Prediction within the range of x-values in the data • Extrapolation: Prediction outside the range of x-values in the data y

Interpolations

Extrapolations

The reliability of predictions depends on the correlation strength, whether the data is interpolated or extrapolated, and the number of points in the data set. • Strong correlation + interpolation = reliable prediction • Moderate/weak correlation + interpolation = less reliable prediction • Extrapolation = generally unreliable prediction • More data points increase reliability

Example 1 The equation y = −8.71x + 6.79 is a line of good fit for a bivariate data set. When x = 3.49 predict the value of y.

Create a strategy Substitute the value of x into the equation for the line of good ﬁt.

Apply the idea y = −8.71x + 6.79

Write the equation

y = −8.71 × 3.49 + 6.79

Substitute x = 3.49

y = −23.6079

Evaluate

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Example 2 Chirping crickets can be an excellent indication on how hot or cool it is outside. Different species of crickets have different chirping rates, but for a particular species the following data was recorded. Temperature (°C) Number of chirps per minute

14 77

17 115

21 150

24 176

220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

a According to the graph, what is the temperature when the crickets make 140 chirps each minute?

Create a strategy If we move from the frequency of chirps to the temperature, we find the x-value for a given y-value. Start with y = 140 and find the x-value.

Apply the idea We can draw a line from our y-value, 140, horizontally to the line of the relationship, then another line vertically down to the x-axis. What is the x-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 20°C.

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b According to the graph, how many chirps per minute will the crickets make if the temperature is 27°C?

Create a strategy When x = 27, find the value of y.

Apply the idea We can draw a line from our x-value, 27, vertically to the line of the relationship, then another line horizontally left to the y-axis. What is the y-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 210 chirps per minute.

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c According to the graph, how many chirps are the crickets making each minute if the temperature is 19°C?

Create a strategy When x = 19, find the value of y.

Apply the idea We can draw a line from our x-value, 19, vertically to the line of the relationship, then another line horizontally left to the y-axis. What is the y-value we end up at? 220

Chirps per minute

200 180 160 140 120 100 80 60 40 20

Temperature °C 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

The answer is 130 chirps per minute.

d Which prediction is more reliable, part (b) or (c)?

Create a strategy Remember that for reliability of predictions: • Strong correlation + interpolation = reliable prediction • Moderate/weak correlation + interpolation = less reliable prediction • Extrapolation = generally unreliable prediction • More data points increase reliability

Apply the idea On the initial scatterplot, the minimum number of chirps per minute is 77 and the maximum number of chirps per minute is 176. This means that part (b) uses extrapolation and part (c) uses interpolation. Part (c) is more reliable than part (b).

Reflect and check There are only four points, so none of the predictions are incredibly reliable. However, the correlation is strong, so that improves the reliability somewhat.

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Idea summary Predictions can be made either using a graph of the line of good ﬁt or substituting into the equation of the least-squares regression line. The reliability of predictions depends on correlation strength, interpolation or extrapolation, and the number of data points. Interpolation: Prediction within the range of x-values in the data Extrapolation: Prediction outside the range of x-values in the data Reliability of predictions: • • • •

Strong correlation + interpolation = reliable prediction Moderate/weak correlation + interpolation = less reliable prediction Extrapolation = generally unreliable prediction More data points increase reliability

10.06 Practice questions Understanding 1

Explorer 1, 2, 3a

Adventurer 1, 2, 3b

Trailblazer 1, 2, 3c

Is each statement true or false? a

A line of good ﬁt can only be used for interpolation, not extrapolation.

Extrapolation using a least-squares regression line is generally more reliable than interpolation.

Interpolation can lead to inaccurate predictions if the relationship changes beyond the range of known data points.

Scatterplots may not accurately represent the relationship between variables if there are outliers, non-linear relationships, or if the data is too sparse.

It is not important to consider data validity and limitations when using interpolation or extrapolation.

Considering data validity and limitations helps ensure that the predictions made are accurate and reliable, and that they do not lead to incorrect conclusions or decisions based on the data.

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Does each graph show interpolation or extrapolation? a

90 80 70 60 50 40 30 20 10

80 70 60 50 40 30 20 10

10 20 30 40 50 60 70 80 90

Find the y-value for the given least-squares regression line and x-value. a

y = 0.45x + 7.1, at x = 3

y = 1.75x - 3.2, at x = -2.4

Fluency 4

4a, 5, 6, 8, 9, 11, 13

y = -2x - 7, at x = 5

4b, 5, 6, 8, 9, 11, 13

4c, 5−7, 10, 12

Are these predictions an extrapolation or an interpolation? a

A prediction for the y-value when x = 5 is made from the data set: x y

4 0

7 2

8 4

11 7

12 6

13 4

17 8

18 8

19 11

20 8

A prediction for the y-value when x = 33 is made from the data set: x y

582

10 20 30 40 50 60 70 80 90

37 72

54 53

58 26

59 21

43 73

55 47

60 12

38 102

64 10

35 112

A prediction for the x-value when y = 95.69 is made from the data set: x

94.4

96.4

94.4

97.8

94.9

95.9

94.4

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 Ex 1

A line of good fit for this data set is y = 1.26x − 57.01: When x = 3.49, predict the value of y. Is this an example of interpolation or extrapolation? Round your answer to two decimal places. x y

93 51.2

57 25.4

86 38.9

97 58.6

78 38.2

96 60.8

68 26.3

69 28.5

54 5.4

92 92

Scientists conducted a study to see people’s reaction times after having different amounts of sleep. The results are recorded in the table: Number of hours of sleep (x)

1.1

1.5

2.1

2.5

3.5

Reaction time in seconds ( y)

4.66

4.1

4.66

3.7

3.6

3.4

The data has been graphed along with a line of good fit found by eye. a b

Predict the reaction time for someone who has slept for 5 hours. Predict the number of hours someone sleeps if they have a reaction time of 4 seconds.

Reaction time 5 4 3 2 1 1

Number of hours of sleep 4 5 2 3

The number of ﬁsh in a river is measured over a ﬁve year period: Time in years (t) Number of fish (F )

0 1903

1 1994

2 1995

The data has been graphed along with a line of good fit found by eye. a

Predict the number of years until there are no ﬁsh left in the river.

Predict the number of ﬁsh remaining in the river after 7 years.

According to the line of good ﬁt, how many years are there until there are 900 ﬁsh left in the river?

3 1602 2000 1800 1600 1400 1200 1000 800 600 400 200

4 1695

5 1311

t 2 4 6 8 10 12 14 16 18 20

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Ex 2

The number of people at the beach can be affected by the temperature. After observing a particular beach, the data was recorded in a table. Temperature (°C) Number of people at the beach

15 80

18 120

22 180

25 230

Sketch a line of good fit by eye.

Approximately, how many people are at the beach when the temperature is 20°C?

200

Approximately, what is the temperature when there are 200 people at the beach?

150

Estimate how many people are at the beach when the temperature is 19°C?

100

c d e

225

People

175 125 75

Is the prediction for 30°C or 19°C more reliable?

50 25

Temperature °C 12 14 16 18 20 22 24 26 28

One litre of gas is raised to various temperatures and its pressure is measured is shown in a table. Temperature (K) Pressure (Pa)

300 2400

302 2416

304 2434

308 2462

310 2478

312 2496

314 2512

316 2526

318 2546

The data has been graphed along with a line of good ﬁt:

584

Use the line of good ﬁt to predict the pressure when the temperature is 306 K.

Is the prediction in part (a) an example of interpolation or extrapolation?

Is the prediction in part (a) reliable?

Is it reasonable to use the line of good fit to predict pressure within each of these range of temperatures? i

300 ≤ Temp ≤ 320

300 ≤ Temp ≤ 600

iii

0 ≤ Temp ≤ 320

280 ≤ Temp ≤ 340

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2580 2560 2540 2520 2500 2480 2460 2440 2420 2400

Pressure (Pa)

Temperature (K) 296 300 304 308 312 316

An ice cream shop records the number of ice creams sold and the maximum temperature of each day for two weeks: Temp (°C) Sales

35 70

29 45

34 71

33 67

25 30

25 28

34 65

34 75

The data has been graphed along with a line of good ﬁt:

Predict the number of ice creams sold if the maximum temperature for a day is 38°C.

Predict the maximum temperature if 10 ice creams were sold.

Is it reasonable to use the line of good ﬁt to predict ice cream sales within each of these range of temperatures? i

0 ≤ Temp ≤ 40

20 ≤ Temp ≤ 40

iii

20 ≤ Temp ≤ 40

0 ≤ Temp ≤ 60

32 60

100 90 80 70 60 50 40 30 20 10

35 74

33 63

28 35

33 62

32 65

Sales

Temp (°C) 22 24 26 28 30 32 34 36 38 40

A plane’s altitude (A) is measured at several times (t) during its descent. Time (t seconds) Altitude (A metres)

0 9000

200 7815

400 7092

The data has been graphed along with a line of good fit drawn by eye. a

Predict the altitude of the plane 100 seconds into the descent.

Predict the altitude of the plane 500 seconds into the descent.

For how many seconds has the plane been descending when it is at an altitude of 7500 metres?

How many seconds did the plane take to descend to the ground?

1700 593 A

9000 8000 7000 6000 5000 4000 3000 2000 1000 t 200 400 600 800 1000 1200 1400 1600

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A car company looked at the relationship between how much it had spent on advertising (A) and the amount of sales (S) each month over several months. The data has been plotted on the graph and a line of good ﬁt drawn. a

Two points on the line are (3200, 300) and (5600, 450). Find the gradient of the line of good ﬁt. The line of good ﬁt can be written in the form S = mA + c, where m is the gradient, c is the vertical intercept, S is sales in thousands of dollars, and A is the amount spent advertising costs.

S 800 700 600 500 400 300 200 100 A 1000 2000 3000 4000 5000 6000 7000 8000

Find the value of c, the vertical intercept of the line. c

Use the line of good ﬁt to estimate the number of sales next month if $4800 is to be spent on advertising.

The depth a diver, x, has descended below the surface of the water is plotted against her lung capacity, y.

y 105

Does the line of good ﬁt have a positive or negative gradient?

100

Find the gradient of the line.

Find the equation of the line of good ﬁt.

Use the line of good ﬁt to estimate the lung capacity, y, at a depth of 4 metres.

85 80 x 1 2 3 4 5 6 7 8 9 10

Reasoning 14

15−17

16, 17

Several cars underwent a brake test and their age, x, in years was measured against their stopping distance, y, in metres. The scatterplot shows the results and a line of good ﬁt of y = 3x + 30 that approximates the positive correlation.

14, 16, 17 y 70 60 50

Use the line of good ﬁt to estimate the stopping distance of a car that is 7.5 years old.

Is the predicted value in part (a) reliable? Explain your answer.

x 1 2 3 4 5 6 7 8 9 10 11 12

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The land size of a sample of properties and their selling price were graphed on the scatterplot: Another property that is not represented in the scatterplot is sold for $670 000 and measured 3400 square metres in size. Is this consistent with the relationship shown in the graph? Explain your answer.

Price (100 000s of dollars)

12 10 8 6 4 2

x 00 00 00 00 00 00 00 10 20 30 40 50 60 70 Size (square metres)

Are these predictions examples of interpolations or extrapolations? Justify your answer. a

Using the provided table that shows a car’s fuel eﬃciency at various speeds, a prediction for the fuel eﬃciency at 70 km/h is made from the data set.

40 50 60 80 90 100

Fuel efficiency (L/100 km) 6.5 5.8 5.2 4.9 5.4 6.1

Year 2015 2016 2017 2018 2019

Revenue ($ million) 10 12 15 18 22

Speed (km/h)

Using the table of a company’s annual revenue over the past 5 years, a prediction for the next year is made from the data set.

Using this scatterplot that shows a student’s test scores at different hours of study, a prediction for the test score at 15 hours of study is made from the data set.

Test score

80 70 60 50 40 30 20 10

Hours of study 2 4 6 8 10 12 14 16 18

Discuss the limitations of using interpolation and extrapolation for making predictions in the context of climate change.

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Problem-solving 19−22 18

x 1000

2000

3000

5000

6000

7000

8000

9000

Find the equation of the line of good ﬁt shown.

If your distance from the equator is 4000 km, estimate the temperature, y, using the equation in the previous part.

Is the predicted value in part (b) reliable? Explain your answer.

A line of good ﬁt has been drawn to approximate the relationship between sea temperature, T in °C, and the area of healthy coral, A in hectares, in a particular location. Two particular points, (2, 700) and (23, 175), lie on the line. a

Find the equation of the line of good ﬁt.

Using the line of good ﬁt, ﬁnd the area A of coral expected when the sea temperature is 26°C. Is the predicted value in part (b) reliable? Explain your answer.

A 800 700 600 500 400 300 200 100

T 5

5 3.1

6 2.9

The table shows data on the number of kilograms of litter collected each week in a national park x weeks after the park managers started an antilittering campaign: Weeks (x) Kilograms of litter collected ( y)

588

4000

18, 20, 22−24

The distance in kilometres, x, of several locations from the equator and their temperature, y in °C, is measured and graphed on a scatterplot. 50 45 40 35 30 25 20 15 10 5

19, 21−23

1 3.9

2 3.7

3 3.5

4 3.1

Construct a scatterplot for this data.

Sketch a line of good ﬁt on your scatterplot.

Find the equation for your line of good ﬁt.

Use your equation of the line of good ﬁt to ﬁnd the number of kilograms of litter collected 8 weeks after the start of the anti-littering campaign.

Is the predicted value in part (d) reliable? Explain your answer.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7 2.7

The ﬂeet manager for the Australian Automotive C Association wants to estimate how car 550 maintenance costs, C (in hundreds of dollars), 500 are related to the distance, K (in thousands of kilometres), driven each year. 450 The data collected and the least-squares regression line are shown on a scatterplot.

400

The equation of the least-squares regression line for the graph is C = 9.9K + 146.2

300

Interpret what the coeﬃcient of K represents in this context.

Explain what the constant term in the equation represents in this context.

Is this interpretation of the constant term reasonable given the context? Explain your answer.

350

250 200 150 100 50 K 5

Use the least-squares equation C = 9.9K + 146.2 to predict the annual maintenance cost for a car that is driven 36 000 km per year.

Is this predicted value reliable? Explain your answer.

Is there any reason to believe that there is a causal relationship between distance driven and maintenance costs? Explain your answer.

The height of a particular projectile y in metres was measured at different times t in seconds and the curve of good ﬁt was drawn. a

Using the curve of good ﬁt, what is the predicted height of the projectile after 1 second?

Using the curve of good ﬁt, what is the predicted height of the projectile after 4 seconds?

The height at a given time has the following relationship y = −5t2 + 100. Which of these statements about interpolation and extrapolation are true?

y 100 80 60 40 20 t 1

xtrapolation is less reliable since we should use a E line of good ﬁt instead.

xtrapolation is less reliable since we make assumptions about the relationship E outside of the observed data.

Draw a scatterplot with a line of good fit which would not be suitable for interpolation. Explain.

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589

Based on the given scatterplot and line of good fit, are either interpolation or extrapolation reliable? Explain.

8 7 6 5 4 3 2 1

x 1

2 3 4 5 6

8 9

10.07 Misrepresentations of data After this lesson, you will be able to… • understand what data misrepresentation is and its implications. • explore common strategies of data manipulation and identify them in realworld examples. • develop skills to critically evaluate data displays and graphs for potential misrepresentation.

Data manipulation strategies Statistics are often used in the media to support facts, strengthen arguments or provide insights. However, they can also be manipulated to inﬂuence opinions. It’s vital to interpret statistics with caution as they can be skewed intentionally or unintentionally. Common ways of data manipulation include: • Cherry-picking data: selectively using favourable results • Strategic choice of measures of centre and spread: measures of centre and measures of spread may give different impressions if outliers are present • Biased surveys: question phrasing and interviewer influence can affect responses • Sampling technique: poor methods can introduce bias; small or non-random samples are typical culprits • Over-generalisation: asserting conclusions for a larger population based on a nonrepresentative group • Overstating significance of findings: unclear sample sizes or error margins can mislead • Correlation and causation: mistaking observed relationships for causality • Comparing apples and oranges: meaninglessly comparing dissimilar data sets • Misinterpreting statistics or probability: common in media and legal contexts Misrepresentation may be accidental or deliberate, and critical evaluation of ﬁgures is essential in our information-rich society. 590

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Example 1 A company that sells a weight-loss supplement and their website says that “this product led to an average weight loss of 5 kilograms over a month”. This is based on the results of two studies of 10 people whose weight loss/gain is shown in this table. The results of the second study were excluded as they said there were too many outliers. Study 1 Study 2

+2 +2

−4 +1

−6 −1

−8 −3

−4 −2

−5 −5

−6 −2

−7 +2

−6 +3

−6 +1

Which option(s) explains why this is misleading? A Cherry-picking data

Comparing apples and oranges

C Biased survey question

Poor sampling technique

Create a strategy Decide which one most accurately describes the scenario: • Cherry-picking data: selectively using favourable results • Comparing apples and oranges: meaninglessly comparing dissimilar data sets • Biased surveys: question phrasing attempts to inﬂuence responses • Poor sampling technique: poor methods can introduce bias; small or non-random samples are typical culprits

Apply the idea

Reﬂect and check

The company only chose to use a study that showed favourable results. They failed to reference other studies that could provide a more complete and accurate picture of the supplement’s effectiveness.

Using the process of elimination: • The answer is not B because the study is relevant to the claim. • The answer is not C because they are not using a survey to collect data.

Both studies only had 10 participants which is too small to draw any conclusions from. The answer is A: Cherry-picking data and D: Poor sampling technique.

Idea summary Common ways that data may be manipulated during reporting: • Cherry-picking data • Strategic choice of measures of centre and spread • Biased surveys • Sampling technique

• • • • •

Over-generalisation Overstating significance of findings Correlation and causation Comparing apples and oranges Misinterpreting statistics or probability

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Misleading graphs Charts are often used in the media to show comparisons and describe trends.

Exploration

78 77 76 75 74 73 72 71 70 69 68

In favour of bill X

80 70 Percent in favour

Percent in favour

Graph 1 and Graph 2 show which political party is in favour of a particular bill.

60 50 40 30 20 10

Liberal

Labor

Other

Liberal

Labor

Political party

Graph 1

Graph 2

Other

1. Are they showing the same or different data? Are they conveying the same message? Explain. 2. What do you notice that is different about the graphs? Explain. 3. When would each graph be useful for a media source to show to viewers? Why or why not?

Sometimes, graphs are also used to misrepresent information and promote a biased view. Some of the ways in which a chart can be misleading include: Misleading axes or scale: • Biased labelling • Missing a scale on one of the axes • Using a scale that isn’t uniform • Having an inappropriate scale for the given data (making variations seem less important or more important than they are) • Using inappropriate scaling on chart elements like bars or sectors • Using a logarithmic scale without specifying Incorrect display creation • Using unconventional methods or using a display incorrectly • Only displaying a select portion of the data • Using the wrong type of chart With any set of data, it is important to analyse the data within the correct context.

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Example 2

Women in cabinet around the world

12 10

Tony Abbott’s new cabinet

8 6 4

Uganda

Indonesia

New Zealand

Liberia

Zimbabwe

Afghanistan

Egypt

1 Australia

Malaysia

China

United States

India

Cuba

Rwanda

2 Canada

Authorised by G. Wright, Australian Labor Party 5/9 Sydney Ave, Barton ACT 2600

The Australian Labour Party released this graph after Tony Abbott was elected as Prime Minister:

Source: https://alp.org.au/our-history/labor-governments-and-women/

Which of the following comments apply? A This graph is misleading because the scale on the vertical axis is not uniform. B This graph is misleading because it assumes that there are always an equal number of male and female cabinet members to choose from. C This graph is misleading because it assumes all cabinet sizes are the same. D The graph is misleading because there are no European countries included.

Create a strategy Check if each statement matches what the graph shows.

Apply the idea The vertical axis has a uniform value of 2, so option A is incorrect. There is no assumption on male cabinet members since their data is not shown, so option B is incorrect. The cabinet sizes of each country are not the same, so the amount of women in Tony Abbott’s cabinet may be proportionally more comparable to the amount of women in other countries’ cabinets. Option C is correct. The title is “Women in cabinet around the world” and there are no mentioned European countries, so option D is correct.

Reflect and check Instead of having the total number on the vertical axis, a proportion would be more appropriate. If a country has 10 members in the parliament and one is female, then that is more representative than a country that has 60 members in the parliament and only three are female.

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Example 3 Which feature of this graph is misleading? Risk of dying from COVID-19 versus normal annual risk

50% 10% 1% 0.5% 0.1% 0.05% 0.01% 0.005% 0.001%

Normal male death rate Normal female death rate Death rate with COVID-19

10 20 30 40 50 60 70 80 90 100 Age

Create a strategy

Apply the idea

Many of the issues relate to the scale and type The y-axis does not use a linear scale. Instead, of graph. Analyse the axes to see what might be a logarithmic scale was used but it was not misleading. specified.

Reflect and check Even if you specify that a logarithmic scale was used, like the image shown, it can still mislead someone who does not know how to interpret a logarithmic scale. It would be best to provide an additional explanation with the graph that explains how the data changes. Risk of dying from COVID-19 versus normal annual risk 10% 1%

Normal male death rate Normal female death rate Death rate with COVID-19

0.1% 0.01% 0

10 20 30 40 50 60 70 80 90 100 Age

*Log scale used to see differences in rates at younger ages

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Idea summary Some of the ways in which the axes and scale can be misleading include: • • • •

Biased labelling Missing a scale on one of the axes Using a scale that isn’t uniform Having an inappropriate scale for the given data (making variations seem less important or more important than they are) • Using inappropriate scaling on chart elements like bars or sectors • Using a logarithmic scale without specifying Some of the ways in which a display can be misleading include: • Using unconventional methods or using a display incorrectly • Only displaying a select portion of the data • Using the wrong type of chart

10.07 Practice questions Understanding 1

Explorer 1−3

Adventurer

300 people were asked if they would support any of the three candidates for an election. a

What should the percentages in a conventional pie chart add up to?

What do the percentages in this pie chart add up to? Supports Huckabee Is this an appropriate data display?

Trailblazer

1−4

1−4 Presidential run 2012

Supports Palin 70% 63% 60% Supports Romney

A survey is being conducted to make a conclusion about mobile phone use in Australia. Which samples would likely be representative of the population? a

All the people who respond to a survey sent out on a popular social media platform.

All researchers at all universities in Australia.

Adding a question to the census.

Every tenth person going into a shopping centre in a randomly selected suburb or area for each town or city.

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Determine whether each statement, that describes how this graph might misrepresent the statistics, is true or false. a

The wrong graph type was used.

The graph doesn’t start at zero.

The scale on the vertical axis is not uniform.

Line graphs are often helpful to show trends over time.

Line graph 16 8 4 2 1 0

Fluency 5

The data in the table is based on a base 10 logarithmic scale. It shows the number of influenza cases at the beginning of a flu season. Convert the data to a normal scale. Round each answer to the nearest whole number. Date October 15 October 18 October 21 October 24 October 27 October 30

Ex 1

Cases on log10 scale 1.5 2.3 2.24 2.53 2.4 2.6

5, 6, 9−11, 13, 14, 16

5, 6, 8, 10, 11, 13, 14, 16

5, 7, 8, 11, 12, 14−16

State the type of data manipulation described in each example. a

There is a strong correlation between temperature and ice cream consumption, therefore rising temperatures cause higher ice cream consumption.

75% of a high school class surveyed said they like sport, therefore all high school students like sport.

Asking an air traﬃc controller about his stress levels and comparing that to the response of a four-year-old child.

A politician cites a single year’s crime statistics to argue that crime has dramatically decreased under their 4-year tenure.

A newspaper states: “Oil prices have risen by 33% over the last two years”. The article states that in 2018, the oil prices increased by 10% and in 2019, the oil price increased by 23%. Are these statements true or false?

596

The headline is true and accurate.

We need to know the prices to determine whether this statement is accurate.

An increase of 10% followed by an increase of 23% does not give an increase of 33% overall.

A graph is required for more complete analysis.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

In NSW, data collected on assaults occurring in hospitals showed that from 2006 to 2011, the number of assaults increased, from 2011 to 2014, the number of assaults remained constant and in 2015, there was a drop in the number of assaults. Based on this, a newspaper reported “Assaults in hospitals have risen dramatically from 2006 to 2015”. Are these statements accurate?

The reporter has not included the data since 2011 accurately in their representation.

The reporter has represented the trend correctly.

The reporter has misinterpreted the data, as the number of assaults has decreased from 2006 to 2015.

The reporter has misrepresented the ﬁndings by not reporting a drop in assaults in 2015.

A new affordable housing initiative requires 10% of units in new buildings to be $250 000 or less, so the units available for purchase are priced at: $250 000, $250 000, $250 000, $600 000, $605 000, $620 000, $630 000, $640 500, $645 000, $650 000, $660 000, $662 500, $668 000, $675 000, $680 000, $695 000, $700 000, $710 000, $715 000, $725 000, $740 000, $744 000, $750 000, $755 000, $759 000, $760 500, $765 000, $770 000, $772 000, $775 000 The marketer for the building says “Most units cost $250 000.”

Do the prices of the units meet the affordable housing initiative requirement?

Is the marketer’s statement accurate?

Which data manipulation strategy is the marketer using?

The marketer decides to change his statement to “Units starting at $250 000.” Is this new statement accurate?

A survey sent to dentists and hygienists asked for their preference on brands of toothpaste. The participants were allowed to choose more than one brand if they wished. The results of the survey are shown in the table. Percentage

Brand A 65%

Brand B 75%

Brand C 72%

Brand D 79%

Brand B advertises the claim “Most dentists recommend Brand B.”

Which data manipulation strategy is the toothpaste company using?

Which statement describes why the claim is misleading? A

The statement is inaccurate.

The survey was conducted by the company that owns toothpaste Brand B.

The statement implies that dentists recommend Brand B over the other brands.

The advertisement does not include the hygienists that were surveyed.

A statement on the box of an anti-wrinkle face cream reads: Noticeably firmer, younger feeling skin in just 4 weeks. Below the statement in small print, it says: *Reported by 25 women. Why is this claim misleading? 10.07 Misrepresentations of data mathspace.co

597

A restaurant is encouraging its staff to promote a new menu item, mushroom risotto. The servers are encouraged to ask each customer this question to improve sales: “Do you want a nutritious risotto for lunch or the standard sandwich?” a

Which manipulation strategy are the servers using?

Suggest a way for the servers to promote the new item menu that is less manipulative.

What feature is misleading about the line graph? 60 000

Turnover million

50 000 40 000 30 000 20 000 10 000 0 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 Year

The ﬁgure shows the number of tourists from each country to the United States:

CN AU

Brazil: 613 347

France: 930 265

Germany: 1 263 344

Japan: 2 169 716

UK X

Canada: 14 043 658 Mexico: 4 295 124 United Kingdom: 2 905 909

Top 20 tourist-generating countries to United States of America

South Korea: 560 405 Italy: 558 594

India: 447 079

Colombia: 288 439

Australia: 526 441

Netherlands: 416 870

Argentina: 271 737

China: 488 484

Venezuela: 344 802

Switzerland: 262 595

Spain: 451 530

Ireland: 296 771

Israel: 237 818

What is the issue with the length of the line segments representing each country?

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Ex 2

Consider the graph:

Average house price ($)

Massive increase in house prices this year 82 000 81 000 80 000 79 000

2008

Year

2009

What two features of this graph might cause it to misrepresent the statistics? 15

When evaluating new drugs, health officials assess the drug’s potential for addiction, among other effects. One aspect of this evaluation involves studying how the drug levels change in the blood over time. Drugs that quickly peak and then decrease rapidly in the blood could be at a higher risk for abuse. Note: while drugs that maintain relatively stable blood levels might have a reduced risk of causing peaks and valleys associated with abuse patterns, this alone does not ensure that the drug won’t be addictive. The line graph shows the amount of a drug that remains in the blood over time. Medication in blood over time Medication concentration in blood (ng/mL)

Ex 3

100 40 mg 80 mg single dose 10 mg q12h steady-state

0 1 2 3 4 5 6 7 8 9 10 11 12 Hours from dosing

The manufacturer claims this drug has a very slow decrease, so it won’t be addictive. This is not an accurate claim. Which aspects of the graph allows for this misleading claim?

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The pie chart shows the number of nuclear weapons possessed by each country. Estimated nuclear stockpiles, 2005 Russia U.S. 7,200 10,315 deployed 8,800 inactive Russia has about 8,800 warheads in reserve or awaiting dismantling

In addition to known nuclear powers, North Korea says it has nuclear weapons. Iran could be next. Est. warheads China

410

India

75-110

France

350

Pakistan

150-110

Britain

200

N. Korea

7-15?

Israel

100-170

Source: Time Magazine a

Considering only the size of the slices (i.e. ignoring the numbers), which country has the largest nuclear stockpile?

Looking only at the numbers listed for each country, which country has the largest nuclear stockpile?

Has the pie chart been drawn accurately?

Reasoning 17

17, 18, 20

17, 19, 20

19−21

In the polls for an upcoming election, various polls are analysed to see who is likely to win the election. Most polls put Kodos ahead of Kang by between 10% and 12%. A few polls have Kang ahead of Kodos by a few percentage points. Kang tells his followers that they are sure to win the election based on the latest polls. Explain why this statement misrepresents the statistics.

600

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Consider the line graphs of the same set of sales ﬁgures: 9000

Sales ($)

8000 7000 6000 5000 4000 3000 2000 1000

Month 1

Describe how the difference in scale between the two graphs might be misleading. 9000

Sales ($)

8000 7000 6000 5000 4000 3000 2000 1000 1

Month 5

Consider the given column graph: Quarterly operating deficit 55 50 $ (million)

45 40 35 0

Sept - 94 Dec - 94 Mar - 95

Why is this a particularly poor example of a graph with broken axes?

Explain why it is not generally good practice to use a vertical scale with a broken axis for column graphs.

Describe a graphical display that would be better suited to this situation.

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The same information is displayed in two different ways. 15

Average global temperatures Degrees (Celsius)

Average global temperatures Degrees (Celsius) 20 18 16 14 12

10 8 6 4 13 1920

2 Years

2020

1920

Years

2020

Which graph could be used to support the claim that the global temperature is increasing significantly?

Which graph could be used to support the claim that the global temperature is not increasing significantly?

At the United Nation’s climate summit, it was stated that limiting the planet’s temperature to a 1.5 °C overall increase by 2100 could avoid catastrophic events. How does this claim help you understand what is considered a “significant increase” in temperature?

Read the hypothetical article: Your mobile phone is ageing you - Research indicates that smartphones are causing premature skin ageing in over half of all users. ‘Smooth as Silk’, an anti-ageing skin product company, does a survey which reveals that more than 50% of frequent smartphone users display early signs of ageing such as ‘tech-neck’ (wrinkles from device use). The study also identifies ‘screen face’ ( pigmentation from screen’s blue light) among significant users. Dr. Louise Fairweather’s survey involving more than 2000 participants suggests blue light exposure accelerates ageing and recommends reducing screen time or using blue light skin products for skin protection.

602

Was the survey’s sample selection well done or not?

Based on the company doing the study, do these results surprise you?

Do you find the article’s headline fitting? Why or why not?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Problem-solving 22, 23 22

22, 23

23, 24

Compare the two graphs and explain which graph is misleading and why. Graph 1

Graph 2 100 Days of freezing temps

Days of freezing temps

85 84 83 82 81 2017

80 60 40 20

2018

2017

2018

This graph was shown on an Argentinian news service with the comment that ‘Argentina’s testing rate for Covid-19 is almost as high as that of the United States’. COVID-19 testing by country

14 100 330

258

Brazil

Argentina

15 700

22 300

7000

USA

Italy

Germany

Norway

Number of COVID-19 tests per million people

Source: Reddit May 2020 a

Explain why the graph and comment are misleading.

Redraw the graph correctly.

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Consider the shown line graph: The Main NAEP shows dramatic gains in math. From 1990 to 2000, fourth grade math scores increased fifteen points - equivalent to about 1.2 years of learning

Scale score

240 238 236 234 232 230 228 226 224 222 220 218 216 214 212 210 1990

1992

1996

2000

Source: Brookings report on American education a

The labels on the horizontal axis are not evenly spaced. Explain why this may be misleading.

Is the scale on vertical axis reasonable or misleading? Justify your reasoning.

Did you know?

Bivariate statistics help us unlock hidden connections! By analysing two variables together, it reveals patterns—like how study time relates to exam scores or how exercise impacts happiness. This approach allows us to see if one factor influences another or if they simply change in tandem. Businesses use it to understand customer behaviour, like how weather affects product sales, while scientists rely on it to study everything from environmental factors to health trends. Bivariate statistics are a powerful tool for spotting relationships, making predictions, and turning complex data into meaningful insights that can guide decisions in countless fields. 604

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

INVESTIGATION

Misrepresentations in the media Investigate the misuse of statistics in media and its impacts.

Objectives • To recognise and identify the common tactics used to mislead with statistics in media • To analyse statistical data and visual representations and evaluate their truthfulness

Materials •

Device with internet access

•

Pen or pencil

•

Paper

Statistics in the media When we read an article, or the government releases a statement or a friend is convincing you of an argument, we often use statistics to support our argument. To be even more convincing, we might use a visual display like a pie chart or a column graph to show the information. When someone is making an argument or trying to make their point, they may be making deliberate choices to make the data look a certain way in order to prove their point. Sometimes this is done accidentally and other times it is done to deliberately mislead. There are many ways someone can twist data to make a point: • Cherry-picking data - selecting the results that are favourable. • Biased surveys and sampling - the phrasing of survey questions and the interviewer can often influence the responses given, as well as how they have selected the people being surveyed. • Over-generalisation - when asserting a conclusion from a particular group to a wider population when the initial group is not representative of the larger population. • Overstating significance of findings - sometimes it is not clearly stated how large the sample was or how large the margin of error is. If the error margin is large, then the true result may be significantly different from the stated figure • Correlation and causation - if a relationship is observed between two variables A and B, it is tempting to jump to the conclusion that A causes B. However, it may be that B causes A, the two may in part cause each other, there may be a third factor that causes both or the relationship may not exist at all (and the observation was made purely by chance). • Bad comparisons - sometimes two data sets cannot be meaningfully compared against each other. For example, if native speakers of English are compared against non-native speakers

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in an English reading comprehension task, we cannot conclude that the native speakers are “more intelligent” when they inevitably do better. • Misinterpreting statistics or probability - probabilities, in particular, are commonly misunderstood and misrepresented in the media. For example, stories of lottery winners are commonplace in society, which serves to obscure the astronomically small chance (typically less than 1 in 300 million or more) of winning if you bought a ticket. In this activity, you will navigate through your favourite news or entertainment website and try to find examples of the misuse of statistics. These may involve tactics such as cherry-picking data, biased surveys and sampling, over-generalisation, overstating the significance of findings, false correlation and causation, bad comparisons, or misinterpreting statistics or probability. Procedure: • Open a news or entertainment website. • Browse through articles and try to identify misuse of statistics. • Write down the examples you find, noting the tactics being used. • What tactics were used in the articles you found? • How do these tactics affect the interpretation of the data?

Investigate In light of the articles and instances of statistical misuse you’ve examined, consider the following: 1.

Which misuse tactic did you find the most? Why do you think this tactic is so commonly used?

How could the misleading statistics you found influence the public’s perception of the subject matter?

Juvenile crime rates (%)

Let’s take a look at this chart and its description: • We can use two pieces of data that seem to be changing together, like movie attendance and juvenile crime in a city, and start making up stories. • But in truth, we don’t know if one is being caused by the other. It’s possible that the increased crime rate led to the movie cinemas shutting down. It is even more likely that a third factor, like increasing unemployment, might explain both trends. Alternatively, these two things might be totally unrelated.

Movie attendance (thousands)

Misleading graphs

0.15

0.1

0.05 2010

2014

2018

Year “Children are committing crimes

If you look through enough data, you’ll find something to because movies aren’t interesting tell a story about, and using statistics and charts can make anymore.” the story you tell seem even more convincing. But being convincing with statistics is not the same as discovering the truth with statistics, and we need to be careful about what conclusions we draw and what we consider as a fact when someone tells a story.

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Let’s try another one:

1.5 Petrol price ($/L)

Analysing data isn’t just about how you present the data you look at, but also which data points within a data set that you choose.

1.25 1 0.75

2010 2012 2014 2016

Year “Petrol prices still getting cheaper.”

1.5 Petrol price ($/L)

According to the chart, the price of petrol does seem to be steadily decreasing. But if we zoom out over the last 20 years, another story emerges - they seem to be increasing in general.

1 0.5

1998 2002 2006 2010 2014 Year Petrol prices over a longer time period

Investigate Analyse real-world examples of misleading graphs using your understanding of accurate and deceptive data representation.

Look at the following images and their titles. What are these graphs trying to convince you to believe? The first is showing us how much more China likes cats than Brazil. The second is showing us that Dante easily beat Miguel in the school election. Are either of these statements true? What feature of the graphs is being used to try to convince us of these things?

Number of votes

Convincing visuals

Number of pet cats (millions)

Look through some newspapers or online news websites and find a graph that is misleading. Explain what is misleading in the way it’s presented and comment on why the author may have presented the graph that way.

40 20 0

Brazil China Dante destroys Miguel in election

154 152 150

Miguel Dante

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Let’s look at the graph of the election result. How much larger is the vote difference? Only 3 votes.

155

Very close school election 150

Number of votes

154 153

100

152 151 M ig ue Da l nt e

150

50 0 M ig ue Da l nt e

Which of these graphs more accurately tells the story of the school election?

Dante destroys Miguel in election Number of votes

Dante has definitely won the election, but did he “destroy” Miguel? The y-axis in the original graph starts at 150. If we resize the graph the start at 0, then the graph might tell a different story.

Let’s look at the graph about China and Brazil and the number of pet cats. The raw data demonstrates that there are around twice as many cats in China than Brazil, which is why the cat on the right has been displayed as being twice as tall. However, by using a 2D image of a cat instead of a column graph, then the amount of space the picture of the second cat takes up is actually four times bigger. In the image below, you can see that when columns are used, the difference doesn’t seem as dramatic.

20 0

Brazil

China

40 20 0

Brazil China

Number of pet cats (per 100 000 people)

Number of pet cats (millions)

Another consideration is to look at the data not just as a raw number of cats, but as the number of cats per capita. Because China has a much larger population, when we adjust the graph to show the number of cats per 100 000 people, another story emerges. 12 8 4 0

Brazil China

Two charts can use the same data and be used to generate graphs that display different focuses.

Investigate Utilise your skills in manipulating data presentation to construct a narrative. Find a graph used in a magazine, newspaper or online news site. Remake the graph, using the same data to convince readers of a different point of view.

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Check this article Read this caption carefully and examine the chart that accompanies it. What problems can you find? What questions might you want to ask the people who wrote it?

Height (cm)

200

190

180 Basketball

AFL

Soccer

“Recent findings have found that approximately half of all soccer players are below the average height of people in A-league soccer competition. Parents that wish to have their children compete in basketball competitively should not allow their children to play soccer.”

Create your own The table below shows the average marks for students in Year 8 in their final year maths test, their height, their sex, the number of DVD sales in Australia that year, and whether the principal of the school was over 175 cm tall.

Year 2013 2014 2015 2016 2017 2018

Sex Female Male Female Male Female Male Female Male Female Male Female Male

Average Average maths mark maths mark (split by sex) (combined) 68 73 78 72 74 76 81 75 69 83 77 71 84 80 76 80 81 82

Average height 159 165 158 164 159 163 159 167 160 168 159 163

Average height (combined)

DVD sales (millions of dollars)

162

2834

161

14.9

161

11.4

163

9.9

164

8.4

161

7.3

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Prompts You do not have to use all the data, just choose the bits that help your story. Look at just two or three columns and see if there is a certain story you can tell by using the data in the right (or wrong) way. Here are some possible headlines you could make if you get stuck: • Better test results in boys getting taller • Less time watching DVDs results in higher test scores • Taller principals motivate students to grow taller

Careful Note that this exercise asks us to use the deceitful tactics that people often employ to mislead the public. These are bad practices, and we do not want to encourage you to use them. However, everyone needs to understand the kinds of tricks that people in the media use every day to grab attention - or worse. If you understand how the tactics work, you’ll be better prepared to see through them and avoid being misled.

Investigate Assume the role of a digital news writer and leverage your understanding of data manipulation to create an engaging but potentially misleading narrative. Imagine you are a writer for an online news site that needs more views. You are asked to write an article using this data that will grab people’s attention - even if it isn’t statistically relevant, or is completely misleading. Using the tactics found above, write a misleading headline and display the data in a misleading chart to make your headline work. Write the first paragraph of your news story to go with it. In this investigation, we dived deep into the world of statistical misrepresentation in the media. We’ve seen how data can be manipulated or misinterpreted to create misleading narratives and have practised recognising these tactics.

Discussion

610

What steps can you take to evaluate the reliability of statistics presented in the media?

How does understanding these tactics help in forming your own opinion about different topics?

How can you apply this understanding in your daily life while consuming media?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

INVESTIGATION

Conduct a statistical investigation Investigate the process of conducting a statistical investigation.

Objectives • To describe the four parts of a statistical investigation • To conduct a statistical investigation to explore an association in bivariate data

Materials •

Device with internet access

•

Pen

•

Paper

Identify a problem Statistical investigation process The statistical investigation process is a process that begins with the need to solve a real-world problem and aims to reflect the way statisticians work. It is a cyclic process that involves several stages: Identify a problem: clarify the problem and formulate questions that can be answered with data.

Collect data

Collect data: design and implement a plan to collect or obtain appropriate data. Analyse data: select and apply appropriate techniques to analyse the data.

Identify a problem

Analyse data

Interpret and communicate the results: interpret the results of analysis in a way that relates to the original question. Interpret and communicate

For example, if a scientist is using the statistical investigation process to investigate a possible relationship between litres of soft drink consumed per week and blood sugar levels for a set of people (bivariate data), then the steps may look something like this:

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Identify a problem • Does there appear to be an association? • Do people who drink more soft drink tend to have higher blood sugar levels? • Who is the population of interest? Collect data Collect bivariate data (litres of liquids consumed and blood sugar levels) for a sufficient number of people in the population. The source of the data should be recorded so that someone reading the final report could verify the data themselves. A statistician must consider whether they will survey an entire population of interest (census) or a representative group from within the population (sample). The process of selecting a sample must be as unbiased as possible to keep the data as representative as possible. Analyse data This is where we “do the maths” by applying graphical or numerical techniques to analyse the data. The scientist might construct a percentage two-way frequency table or a scatterplot, determine the strength of the relationship by determining the correlation co-efficient, or find an equation of a line or curve that best describes the apparent relationship. Interpret and communicate the results Comment on whether the analysis indicates that there is an association between the variables. The interpretation of the results should be related to the original question and communicated in a systematic and concise manner.

Investigate In this investigation, you will work through the statistical investigation process.

612

Describe a problem you would like to solve or a question you would like to answer. Choose a problem that involves bivariate, numerical data that you will collect and analyse over a period of time.

Clarify the problem by writing it as a statistical question.

Describe the population of interest. This will be people or objects that you will gather data on to answer your statistical question.

Do you already have a prediction as to what the outcome of you investigation will be? How might this lead to bias?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Collect data After clarifying the problem, collect the numerical data from the population of interest. Describe, explain and justify the methods that used to obtain data. If you are using primary data, you should explain: • the choice of census or sample • how you selected your sample and how it is representative of the population • the methods you used to collect the data (e.g. questionnaire, measurement, counting) • any problems that you encountered in collecting the data If you have used secondary data, you should state the source of the data and provide relevant information on how it was collected. Once the data is collected, you will need to present the data in tables, graphs or lists; preferably using familiar mathematical formats.

Investigate Answer these questions about the statistical question you chose in Activity 1. 1.

What type of numerical data do you need to collect to answer your statistical question?

Will you use a census or a sample? Why?

If you are using a sample, how will you select the sample? Is your selection method random? Is it representative of the population?

Describe how you will collect the data.

Take some time now to collect your data. Did you have any problems while collecting the data? If so, describe them.

How will you organise numerical the data? Why did you choose to organise it that way?

Analyse data To analyse the data collected, we must perform the mathematical calculations we have learned in this course. Perform the mathematical analysis • Describe, explain and justify the mathematical process used • Define the required variable and constant parameters • Calculations should be presented systematically and explained using mathematical language • Clearly state the final results of the analysis The analysis section contains the mathematical calculations along with an explanation and justification of the interpretations leading to the conclusions that we have drawn.

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Investigate Answer these questions about the statistical question you chose in Activity 1. 1.

What types of calculations are required for you to answer your statistical question?

For your data, which variable is the independent variable and which one is the dependent variable?

Does there appear to be a correlation between the variables? If so, is it positive or negative?

Interpret and communicate The final step of the investigation is an interpretation of the mathematical and statistical results in the context of the investigation. It should be a concise statement of the most important information and must not introduce any new information. A good conclusion should concisely: • restate the question and describe the population of interest • summarise the mathematical processes used to analyse the data and the results of the analysis • state any conclusions, in the context of the original question • describe important limitations • state any assumptions made in the process

Investigate Answer these questions about the statistical question you chose and the results of your investigation in the previous activities. 1.

Describe the results of the analysis in context.

State any assumptions you made when collecting and analysing the data or when interpreting the results.

Describe important limitations of your investigation. For example, what can’t we conclude from the data?

Did the result of your investigation support your original hypothesis?

Discussion

614

What was your favourite part about the statistical investigation?

Did you face any challenges while completing the investigation? If so, describe them and discuss how you addressed the challenges.

In your own words, briefly describe the four parts of a statistical investigation.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

10 Chapter review 1

Emrik conducted a survey in his neighbourhood about the type of car people preferred. The options given were: Truck, Sedan, or SUV. 36 respondents said they preferred trucks, of which 20 were under 30 years old. Among 22 respondents who preferred sedans, 14 people were older than 30 years old while among 28 respondents who preferred SUVs, 18 people were under 30 years old.

Construct a two-way table based on the results of Emrik’s survey.

What is the proportion of respondents that prefer trucks?

What is the proportion of people older than 30 that prefer sedans?

At a university with 32 000 students, a survey was conducted, asking students what they were majoring in and what team or club they are currently a member of. The results are shown in the table. Physical therapy Education Software engineering Graphic design Total

Sports team 90 58 60 10 218

Book club 18 62 13 15 108

Dance club 55 60 47 35 197

Art club 27 45 38 47 157

Total 190 225 158 107 680

How many students are members of either the dance or art club?

What fraction of students studying software engineering are members of a team or club other than the dance club?

What percentage of students surveyed are majoring in physical therapy? Round your answer to two decimal places.

Estimate the number of students in the entire university population who are majoring in physical therapy.

The results of a survey of 379 people on their transportation preference based on their living area is shown in the row percentage table: People living in urban areas People living in suburban areas

Public transport 25% 5%

Bike 20% 15%

Private car 55% 80%

Total 100% 100%

Which variable is the explanatory variable, living area or transportation preference?

Does there appear to be an association between living area and transportation preference?

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The 100% stacked column graph show the results of Beverage preferences a survey on beverage preferences: 100% a Which is the explanatory variable, beverage 75% preference or preference on time of day? b

Does the 100% stacked column graph indicate an association between the variables?

25% 0%

616

Morning Evening person person

210 people were surveyed about their education level and how many hours they read each week. The results are shown in the table:

Less than 2 hours 3 to 6 hours 7 to 12 hours Over 12 hours Total

Tea Coffee

50%

High school or less 25 20 7 0 52

Some university 20 20 12 3 55

Bachelor’s degree 3 10 25 8 46

Graduate degree 2 12 31 12 57

Total 50 62 75 23 210

Which variable is the explanatory variable, education level or time spent reading?

To examine if there is an association between the variables, should we use a column or row percentage table?

Create this percentage frequency table. Round your answers to two decimal places.

Does there appear to be an association between the variables?

The local music store constructed divided bar graphs showing the last six months of instrument sales by category (there are ﬁve categories: S P W K B Jan String, Percussion, Wind, Keyboard, and Brass). Assume that S P W K B Feb the total number of sales was consistent for each month. a

In which month were bass instruments the most purchased type of instrument?

In which month were string instruments the highest selling type of instrument?

Which variable is the response variable?

Does the divided bar graph suggest that there is an association between the variables?

Mar

Apr

May

Jun

Percussion Wind String Keyboard Brass

Does one variable have a direct effect on the other? If yes, identify the independent and dependent variables. a

Time spent studying a subject, Number of chapters covered

Shoe size, Number of siblings

Duration of sunlight exposure, Plant growth

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

The table shows the average number of days of exercise and the average number of sick days of students over a six month period: Number of days of exercise 2 7 12 15 20 24

Month 1 Month 2 Month 3 Month 4 Month 5 Month 6

Number of sick days 7 5 4 3 1 0

Create a scatterplot for the set of data.

A statement is made: “As the number of days a student exercises increases, the number of days they are sick decreases.” Is this claim correct? Explain your answer.

Describe the strength and direction for each linear association. a

5 x 5

x 5

The table represents the monthly sales ﬁgures of a new software product in its ﬁrst year in the market. a

Construct a scatterplot for this data.

Describe the correlation between the month and the sales ﬁgures.

Does this data set contain any outliers?

Does the data show that the month of the year causes an increase in sales? Explain your answer.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Sales (in thousands) 15 18 20 21 22 22 23 22 23 24 24 26

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For each scatterplot: i

Draw a line of good ﬁt by hand.

Write the equation of the line of good ﬁt.

iii

Describe the correlation of the data set as positive or negative.

ii b

16 14

10 8

6 4

x 5

x 2 4 6 8 10 12 14 16 18

The height above the ground (in centimetres) of a radish sprout over time is shown on the scatterplot: Height (cm) 25 20 15 10 5 Age (days) 5

20 25

Describe the meaning of the y-intercept in context.

Does the interpretation in the previous part make sense in this context? Explain your answer.

After how many days will the sprout be 16 cm tall?

A line of good ﬁt for this data set is y = 2.5x + 20.5 x y a

618

2 25.5

2.7 28.1

3 28.4

3.5 29.2

Predict the value of y when x = 5.3

4 30.0

4.2 30.6 b

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 32.45

6 35.5

6.7 38.0

Is this interpolation or extrapolation?

A line of good ﬁt has been drawn to approximate the relationship between the number of days Charlee cooks at home, C and the amount of money she spends at restaurants in a month, M in dollars. A line of good ﬁt, M = −28.6C + 970, approximates the correlation. a

Using the line of good ﬁt, ﬁnd the amount of money Charlee would spend if she only cooks at home 1 day of the month.

Is the predicted value in part (a) reliable? Explain your answer.

Predict the number of days Charlee would cook at home if she spends $250 at restaurants in a month.

Is the predicted value in part (c) reliable? Explain your answer.

900 800 700 600 500 400 300 200 100

M (in dollars)

C (in days) 4

12 16 20 24 28

State the type of data manipulation described in each example: a

Chaysten’s car has broken down exactly 3 times, and all 3 times were during rainy days. He claims that the rain caused his car to break down.

68% of university students surveyed said they would not sign up for a drama class. The university removed the drama class from the next term’s schedule, stating the reason as “No student interest”.

A company received the ratings shown during their opening day of business. 5.0

4.5

4.8

4.0

0.0

4.5

5.0

4.4

4.7

5.0

4.4

The next day, they posted a sign in their window that said “Customer rating: 5 stars”. 16

400 people were surveyed on new models of smartphones, and which ones they preferred. Participants were allowed to choose more than one smartphone. a

The company marketing GammaDevice claims “Most people prefer GammaDevice”. Which data manipulation strategy are they using?

Why are the sizes of the sectors misleading?

Describe a graphical display that would be more appropriate for this data.

Prefers BetaPhone 70%

Prefers GammaDevice

70%

67%

Prefers AlphaModel

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Big ideas Networks provide insights into connections and relationships between elements, with properties enabling analysis and understanding.

Curriculum Outcomes VC2M10SP02 interpret networks and network diagrams used to represent relationships in practical situations and describe connectedness

Networks Chapter outline 11.01 Introduction to networks 11.02 Network representations 11.03 Planar graphs 11.04 Walks, paths, and trails 11.05 Optional - Traversable networks Chapter 11 review

622 636 648 658 674

Delivery companies can save millions of litres of fuel annually by optimising delivery routes!

11.01 Introduction to networks After this lesson, you will be able to… • identify vertices and edges in different types of graphs. • distinguish between directed and undirected graphs. • understand and calculate the degree of a vertex. • recognise and explain loops, multiple edges and isolated vertices. • apply knowledge of graphs to real-world scenarios. • interpret weighted graphs and calculate the weight of a network.

Graphs and networks What do roads, human ancestry, currency exchange, planning a wedding, and the internet have in common? They can all be represented with a network diagram also called a graph.

Belconnen

Pink

y Po

4 Cambray

Arda

nd Fernam

Mordan Range 3

Dolby

Eastfarthing

The roads and towns in this map form a graph. Daniela

Bruno

Pierre

Lina

Michelle Jason

This social network illustrates the relationships between individuals.

622

ke La l l e

en ’s Gre Tiller

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Gilgandra

Network diagrams are graphs that represent systems in the real-world. The terms graph and network are interchangeable. Networks are made up of a set of points, called vertices, that are connected by a set of lines called edges. A vertex (or node) is the fundamental building block of a graph and represents a single object or idea. It is drawn as a dot or circle. The plural of vertex is vertices. An edge is a line segment that begins and ends at a vertex. An edge represents a relationship between the objects or ideas that it links together, and the kind of relationship depends on the context. Edge Vertex

Vertex

On the left is a single vertex. On the right are two vertices with one edge between them. A graph is a collection of vertices with edges drawn between them. Edges do not have to be straight. The following graphs are a collection of vertices and edges. 7

8 9 4

Let’s take a moment and notice a few things about the graphs: • Some edges have direction (they look like arrows). • Sometimes a vertex is connected to itself by an edge. • Sometimes vertices are connected to another vertex by more than one edge. • The edges can have a numerical value. The following networks have labelled vertices. The labels can be something specific such as the name of a person or city or the labels can be more abstract like capital letters. B

Athens

Chronos

Rhea Coeus

Phoebe

A Florence C

Zeus Berlin

Leto

Amsterdam Paris Artemis

D Palermo Oxford

An edge that starts and ends at the same vertex is called a loop.

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When there are two or more edges that start at the same vertex and finish at the same vertex, we refer to these as multiple edges. E

This graph has a loop at vertex A and multiple edges between vertices D and F.

F B

A valid network has: • At least one vertex • All edges start at one vertex and end at one vertex Valid

Invalid B

A C

A D

D B

This is an invalid network because the edges here do not end at another vertex.

A connected network is one where all the vertices are connected (either directly or indirectly). Connected

Not connected A

E C

C E

624

D D

The vertex which is not connected is an “isolated vertex”.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 1 For this graph: A

a Which vertex is isolated?

Apply the idea Choose the vertex that is not connected to the other vertices. Vertex B is isolated, it is not connected to any other vertices by any edges.

b How many edges are there in the network?

Apply the idea There are 7 edges.

c How many vertices are there?

Apply the idea There are 5 vertices.

d Which vertex is the loop connected to?

Apply the idea Choose the vertex that has an edge that starts and ends at itself. The loop is connected to vertex A.

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Idea summary A graph is a diagram consisting of a set of vertices connected by edges. When graphs are applied to practical situations they are often referred to as networks. A vertex (or node) is the fundamental building block of a graph and represents a single object or idea. It is drawn as a dot or circle. An edge represents a relationship between the objects or ideas, and the kind of relationship depends on the context. A loop is an edge that starts and ends at the same vertex. Multiple edges are two or more edges that start at the same vertex and ﬁnish at the same vertex. An isolated vertex is a vertex that has no edges connected to it, not even a loop.

Directed and weighted graphs Edges can represent a one-way or two-way connection. • Directed edges (also called arcs) are drawn as arrows and represent a one-way connection. • Undirected edges are represented by lines and represent a two-way connection. Shark France

Spain

eats borders Fish

The ﬁrst graph has a directed edge because sharks eat ﬁsh and not the other way around - it is a one-way relationship. The second graph has an undirected edge because France and Spain share a border - it is a two-way relationship.

If there are directed edges in the graph, we call it a directed graph (or digraph). If there are no directed edges, we call it an undirected graph. A graph that is shown with edges that are a mix of arrows and lines is still a directed graph, where the lines represent a two-way connection. “Mixed” graphs like this are not common. It makes things easier if either all of the edges are lines, or all of the edges are arrows. We can redraw such a graph by replacing two-way edges with a pair of directed edges.

1.10 1.10

1.34 1.10

1.23 1.43 0.96

A graph with numerical values on its edges, called weights, is referred to as a weighted graph. The weight of the graph is the sum of all the values. The values written on the edges here, could represent distances or cost.

Some graphs can be both weighted and directed.

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Example 2 Consider this weighted and directed network: 6 13 14

a What is the weight of the edge from S to Q?

Apply the idea The weight is 7.

b What is the weight of the entire network?

Apply the idea Network weight = 14 + 10 + 13 + 9 + 6 + 4 + 7 Add the weights = 63

Evaluate

Idea summary Networks can be represented by directed or undirected graphs and weighted graphs. In a directed graph (or digraph) the edges indicate a one-way relationship and are represented by arrows. An undirected edge is drawn as a line segment that begins and ends at a vertex, indicating a two-way relationship. In a weighted graph, edges are assigned a number. Weighted graphs may be directed or undirected. The weight of a graph is the sum of the weights of all the edges.

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Degree of a vertex The degree of a vertex is the number of edges that either go into or out of a vertex. Loops are counted twice. This number tells us how many ways we can move from that vertex to another (or even the same) vertex within the graph. A useful trick is to make a ring close around the vertex. Count the number of times that edges cross the ring at a vertex, and we will have the degree.

Degree 3

Degree 5

Degree 3

Degree 6 Degree 5

Degree 4

A vertex is referred to as odd, even or isolated depending on the degree of a vertex being odd, even or zero respectively. If the degree of a vertex is zero it is an isolated vertex.

Example 3 Consider this network:

A B

a What is the degree of vertex C?

Apply the idea Vertex C is only connected by one edge. The degree of vertex C is 1.

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b What is the degree of vertex D?

Apply the idea Vertex D is connected to three edges. The degree of vertex D is 3.

c What is the degree of vertex B?

Apply the idea Vertex B is connected to three edges in which one edge is a loop that is counted twice. The degree of vertex B is 4.

d Which vertices are odd?

Apply the idea Vertex A is connected to two edges, so it has a degree of 2, which is not an odd number. Based on our answers from parts (a) to (c), vertices C and D have a degree of an odd number.

Idea summary The degree of a vertex is the number of edges that connect to the vertex. Loops are counted twice. For an odd vertex, the degree is an odd number. For an even vertex, the degree is an even number. For an isolated vertex, the degree is zero.

11.01 Practice questions Understanding 1

Explorer 1, 2, 3a–d, 4a–d

Adventurer 1, 2, 3a–d, 4c–f

Trailblazer 1, 2, 3c–h, 4e–h

Are these statements true or false in relation to networks? a

There is always at least one vertex.

There is always at least one edge.

Each vertex must be connected to every other vertex by edges.

An edge can start and end at the same vertex.

Edges always start and end at vertices.

An edge can connect three vertices together.

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How many edges are connected to vertex A?

List the vertices connected to vertex E.

Name the vertex that is directly connected to all other vertices.

Name the vertex that is not directly connected to vertex A.

Are these networks valid or non-valid? a

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How many edges and vertices does each network have? a

Fluency Ex 1

5−7, 8ab, 9a–e, 10a, 11, 12ab

5−7, 8bc, 9c–g, 10a, 11, 12ab

How many vertices and edges does the network have?

Is the network directed or undirected?

5−7, 8cd, 9d–h, 10b, 11, 12cd

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How many vertices are there?

How many edges are there?

Which vertex has a loop?

Between which two vertices are there multiple edges?

Is the network directed or undirected?

What is the degree of vertex S?

How many vertices and edges does the network have?

How many edges need to be added for the isolated vertex to be connected?

Do these networks have multiple edges? a

Are these networks directed or undirected? a

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Ex 2

Find the weight of the edge connecting Q and S.

Calculate the weight of the entire network.

7 2 9 4 5

Find the weight of the edge from S to Q.

Calculate the weight of the entire network.

4 11 16

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Ex 3

634

How many loops does each network contain? a

For each network: i

Identify the degree for each vertex.

State if the network is connected.

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Reasoning

13, 14

Jonah says the network has 6 vertices and 8 edges because there are six circles and eight line segments. What mistake has he made and what is the correct answer?

Is it possible to draw one edge so that every vertex is even in this network? If so, what edge would need to be added? If not, explain why not.

Problem-solving 15, 16 15

15−17

15−18

Would a directed or undirected network be appropriate for these situations? a

The countries that border each other

The results of an elimination-style sports tournament

The animal food chain

Your parents and their ancestors

Ways to get from one classroom to another at school

How parts of the body are connected

How many edges begin at vertex Z?

What is the weight of edge BC?

What is the difference in total weighting if you travel from A to B to C to A in a clockwise direction compared to anticlockwise?

41 6

12 9 53

How many edges end at vertex T?

Calculate the total weight of all edges that end at vertex V.

Is it possible to travel along the edges from vertex W to P?

11 5

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Show that the sum of the degrees of the vertices for the network is equal to double the number of edges. Why does this rule always apply for undirected networks?

11.02 Network representations After this lesson, you will be able to… • identify the properties of simple and complete graphs. • represent relationships through network diagrams. • differentiate between directed and undirected networks. • apply knowledge of network representation to solve problems.

Simple and complete graphs A simple graph is a network where: • the edges have no direction • there are no multiple edges • there are no loops a

This is a simple graph.

This is not a simple graph.

The second image is not a simple graph because it has multiple edges between vertices b and d, and a loop at vertex b.

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A complete graph is a simple graph where each pair of vertices, in the network, are connected by an edge. These graphs show complete graphs with n nodes for Kn.

Example 1 Select the two simple networks: A

Create a strategy

Apply the idea

Use the fact that a network is simple if: • It has no loops • Each vertex is connected to another vertex by at most one edge • The edges have no direction

Options B and D are not simple networks due to being directed. Option E is not a simple network due to having a loop. Option F is not a simple network due to having a pair of multiple edges. Options A and C are simple networks.

Idea summary • A simple graph is a network whose edges have no direction, with no multiple edges or loops. • A complete graph is a simple graph where each node (vertex) is conected to every other node (vertex) directly, by an edge.

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Network representations When drawing a network for a given context, we may need to be intentional about how we represent the network graph.

Interactive exploration Discover this concept in action online

mathspace.co

Networks can be drawn in different ways, but still represent the same information. For example, the given two networks are the same, but drawn differently. When representing a network graph in a different way, we want to ensure that the vertices and the connections between them are the same.

Example 2 Rhoda maps out the electrical wiring for a unit she is building.

Electrical Box Ceiling Light Light Fan

Which of these network graphs could be used to represent the electrical wiring? A

Create a strategy The ceiling fan should have a degree of 5 and the electrical box should have degree of 4. We can start from there and then check the vertices that follow from those higher degree vertices.

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Apply the idea

Reflect and check

Option A has a vertex with degree 8, so this is not correct.

We can label the vertices and highlight the regular lights to check.

Option B has two vertices with five edges between them, so this is not correct.

Ceiling light

Option C is correct.

Ceiling light

Electrical box

Ceiling light

Ceiling fan

Idea summary A network can be drawn in many different ways and represent the same information. What is important is that the vertices and the connections between them are the same.

Reading network diagrams There are many contexts where network graphs can be used such as travel networks, wired or wireless networks in offices, eco-systems, and chemical structures. Let’s explore some situations that can be represented by networks.

Example 3 A beach volleyball competition team can have 3 players on the court at any time. The team has 5 players at the tournament. The given network diagram shows the combination of players that have already played together in their first three games.

Lucy Ryan Beth Jimmy Ellie

a What do the vertices represent in this network diagram?

Apply the idea The vertices represent the five different players on the team.

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b What does an edge between two vertices represent?

Apply the idea An edge between two players represents that the players in those vertices have played together in one of the first three games.

c Is this a simple graph?

Apply the idea A simple graph has no direction on the edges, no multiple edges, and no loops. Yes, this is a simple graph.

d There is one more game left in the tournament. Which three players should the coach choose so that every team member has played with every other team member at least once? A

Beth

Ellie

Jimmy

Lucy

Ryan

Apply the idea The graph should show a line connecting every player’s name to satisfy the condition that every team member has played with every other team member at least once.

Lucy Ryan Beth Jimmy

Ryan and Beth, as well as Jimmy and Beth, are not connected by a line. This means that Beth, Jimmy, and Ryan have not yet played together.

Ellie

So the coach should choose Beth, Jimmy, and Ryan. The correct answers are options A, B, and D.

Reflect and check After this last game, if we draw the new edges we will get a complete graph. Lucy Ryan Beth Jimmy Ellie

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Example 4 The given network diagram has 3 vertices that represent three rides at an amusement park: Carousel (C), Pirate Ship (P) and Haunted House (H). The weighted edges represent the amount of time, in minutes, it takes to travel from one ride to the next. The loops represent the time it takes to queue up for each ride.

3 C 5

4 11

a How long does it take to travel from the Carousel to the Pirate Ship?

Create a strategy

Apply the idea

Carousel is represented by vertex C and Pirate ship is represented by vertex P. We are looking for the weight of the edge between them.

The weight of the edge between C and P is 5, so it takes 5 minutes to travel from the Carousel to the Pirate Ship.

b How long does it take to queue up at the Pirate Ship?

Create a strategy

Apply the idea

The Pirate ship is represented by vertex P. We are looking for the weight of the loop at this vertex.

The weight of the loop at P is 13, so it takes 13 minutes to queue up at the Pirate Ship.

c Sam got in line for the Haunted house at 10 a.m. It takes 7 minutes to complete the tour of the Haunted House. He had his turn in the Haunted House, then travelled to the Pirate Ship. What time was it when he was able to board the Pirate Ship?

Create a strategy We can add the weights of the edges for each step of his journey.

Apply the idea We can list out each step: • Waiting in line at Haunted House: 11 minutes • Toured the Haunted House: 7 minutes • Travel from Haunted House to Pirate Ship: 6 minutes • Waiting in line at the Pirate Ship: 13 minutes Adding up all the step: 11 + 7 + 6 + 13 = 37 minutes It was 10:37 a.m. when he boarded the Pirate Ship.

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Idea summary Vertices can represent objects or places, edges can represent paths or connections, and weights can represent numerical data that connects the vertices like times or distances.

11.02 Practice questions Understanding Ex 1

Explorer 1, 2, 3ab, 4, 5

Adventurer 1, 2, 3ab, 4, 5

Trailblazer 1, 2, 3b–d, 5

Describe these networks as simple or not simple: a

Describe this graph as complete or not complete. Explain your answer.

For each network:

What do the edges represent?

What do the vertices represent?

iii

Find the number of vertices.

Find the number of edges.

Is the graph simple?

Is the graph complete?

This network represents the internet fibre optics cables connecting several cities: City B City G City D City A City F City E City C

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This network represents an electrical circuit that includes a globe, resistors and multiple switches. Switch

Battery

Resistor

Switch

Resistor

Globe

Switch

This network displays newspaper routes between several houses: House B House F

House C House A House E House D

This network represents the roads between several towns: Town B

Town F

Town A Town E

Town D Town C Town G

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Ex 2

Considering the map, draw a graph that represents the major roads connecting towns.

Young

Temora

A41

Boorowa

B94 Harden B94

Cootamundra B85 A41

B81

M31

Yass

Junee

A25

A41

Ex 3

Lake Burrinjuck

Select the network graph which represents the same information as the given network. A

Fluency

6, 8, 10, 11ab, 12

6, 8, 10−13

644

A group of 5 students is playing 3-on-3 basketball against another group. After one team gets to 21 points or 10 minutes have passed, they change who is playing, and start a new game. This network graph shows who has played together after three games.

6, 7, 9, 11, 13

Lisa James Belinda

Have James and Jenny played together? Explain why or why not?

They are going to play one more game before going home. Which players should play so that every team member has played with every other team member at least once?

A group of journalists have been asked to work in pairs to collaborate on an article. This network graph represents the journalists who have worked together. a

Have Xanthe and Tina worked together? Explain how you know.

Name the journalists who still need to collaborate with someone else so that each journalist will have collaborated with every other journalist at least once.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Jenny Hannah

Derek Xanthe Neil Tina Xavier

Ex 4

Over a year students work in pairs to complete poster projects. A group of 4 friends have already completed four projects together as shown with the network graph. a

Who has Aaron worked with?

Name the pairs of students that should work together on the next poster project so that all 4 friends have worked with each other.

The buses that run between these four towns are represented with this network graph. a

Which town can you not get to by bus?

Which town can you not leave by bus?

Describe a directed edge that could be added so you could get from any town to any other town by bus.

When converting between two currencies a bank charges a fee as a percentage. Let (A) represent Australian Dollars, (E) represent Euros, (N) represent New Zealand Dollars, (U) represent US Dollars, and (Y) represent Yen. The value on an edge represents the percentage fee that is charged to convert between those currencies. a

Which conversion has the highest fee percentage?

Is this graph a complete graph?

Why are there no loops?

Aaron

Rochelle

Mario

Bianca

Teribithia

Nod

Narnia

U 4

A 5

4 6

The given network diagram has three vertices that represent three rides at an amusement park: Roller coaster (R), Bumper cars (BC) and Big dipper (D). The weighted edges represent the amount of time, in minutes, it takes to travel from one ride to the next. The loops represent the time it takes to queue up for each ride. a

How long does it take to travel from the Roller coaster to the 7 Big dipper?

How long is the wait for the Bumper cars?

Each ride is two minutes long. If you started at the Roller coaster, then did the Bumper cars, then the Big dipper, how long would it take from the time you got in line for the Roller coaster to the time you got off the Big dipper?

8 BC 3 5

R 4

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The given graph represents the number of likes that four friends gave to each other’s social media posts over a week. a

How many likes did Fan receive in total?

How many likes did Zoya give in total?

Describe the type of network graph in two ways.

A 11

4 39

2 23

12 32

41 9 53

For each pair of networks, could one network be redrawn as the other with different labels for the vertices? a

A B

X W

X Y

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Reasoning

14, 17a

14, 15

14−17

If a network is not simple, can it be redrawn in a different way so that it is simple? Explain.

Can you draw a simple graph that has one vertex and one edge?

How many simple graphs can you draw with three vertices?

The given network represent social media connections on a particular app: a

Is the app one where two people are friends or is it one where people follow and get followed? Explain.

Would it be possible for this graph to become a non-simple graph as more people connect on this app?

Would it be possible for this graph to become a non-simple graph using a different app? Explain.

Problem-solving 18, 19, 20ab 18

Brad Sharon Adam Joanne Edward

18, 19, 20b

18, 19, 20bc

The two given network graphs represent A the same information. If vertex a maps to vertex 1, write out a mapping for the other seven vertices B

Consider the image of a train network showing the lines and stations: Expo Centre

Willamette River

Pioneer Square

Airport

Rose Quarter

Gateway

Hillsboro Gresham Beaverton

PSU Clackmas Town Centre

Draw a graph that represents the train network.

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Draw a directed graph to represent each of these food chain descriptions. Draw the arrows pointing from the animal that is eaten to the animal that eats it. a

Within a water system: • Tadpoles, water beetles and snails all survive by eating algae. • Small fish eat the tadpoles, whilst frogs survive on water beetles and snails. • The kingfisher, a skillful fishing bird, survives on eating small fish and frogs.

Rabbits and squirrels both eat plants. Foxes and hawks eat both rabbits and squirrels.

The killer whale depends on tuna as a primary food source. The tuna feeds on ﬁsh called mackerel. For mackerel to survive, they depend on microscopic organisms collectively known as zooplankton.

11.03 Planar graphs After this lesson, you will be able to… • identify a planar graph. • calculate the number of faces in a planar graph. • apply Euler’s formula to confirm if a graph is planar. • analyse the relationship between vertices, faces and edges in a planar graph.

Planar graphs You may have noticed that sometimes graphs are drawn with edges crossing each other. If it is possible to move the vertices into a configuration so that no edges cross each other, it is called a planar representation. If the graph can be drawn in this way, then it is called a planar graph. Note that the term planar means that the graph can be drawn in a planar representation, but it may not look like planar at first.

The top graph is a planar graph, and the three below it are of its equivalent planar representations.

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It isn’t always obvious at first glance whether a graph is planar. Sometimes you have to move the edges or vertices around in a few different ways until the edges do not cross anymore. Edges are like strings that can be dragged and stretched around. A

B C

Crossings

E D

B C

No more Crossings E

Planar?

Planar!

A non-planar graph is a graph where no matter how you draw it, there will always be edges that cross. For example, these two graphs are non-planar:

Once a planar graph is organised into a planar representation, we can define a face (or region) as the area of the plane bounded by edges. The part of the plane outside the graph is also a face. In all three planar representations of the graph from earlier, each face has been given a different colour below.

From the 4 colours, we can see that the graph has 4 faces. The octahedral graph below (which represents a 3-dimensional octahedron solid) has 8 faces.

Non-simple graphs that are planar also have faces, though some of their faces are bounded by only one or two edges:

The graph on the left has 6 faces, with 4 of them bounded by only 1 edge (where there are loops). The graph on the right has 5 faces, with 2 of them bounded by only 2 edges. If a graph is not planar, then there are no faces to define - the crossing of edges makes it impossible. 11.03 Planar graphs mathspace.co

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Interactive exploration Discover this concept in action online

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Example 1 Consider the graph shown.

a Match the graph with its equivalent planar representation. A

Create a strategy Redraw the edges that are crossing each other on the given graph. Make sure that the vertices that were connected by an edge on the original graph are still connected on the planar representation.

Apply the idea All the graphs among the options do not contain any intersecting edges. So they are all planar graphs. Only the graph in option C has the same edges, HD and BF, as in the original graph. So the correct answer is option C.

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b How many faces ( f ) does this representation have?

Create a strategy Count the number of regions the plane is divided into, including the part of the plane outside the graph.

Apply the idea There are 4 regions in the graph: 2 inside the rectangle, 1 between the rectangle and the edge BF, and 1 on the outside of the graph. So f = 4.

Idea summary If it is possible to move the vertices or edges into a configuration so that no edges cross each other, it is called a planar representation, and the graph is called a planar graph. A face (or region) is the area of the plane bounded by edges when organised into a planar representation. The part of the plane outside the graph is also a face. If a graph is not planar, then there are no faces to define - the crossing of edges makes it impossible.

Euler’s formula The Swiss mathematician Leonhard Euler (pronounced “OIL-er”) was a pioneer in the mathematics of networks in the 18th century. He noticed something interesting about networks that are connected and planar, which is that the number of vertices (v), the number of edges (e), and the number of faces ( f ), satisfy the formula: v+f−e=2 We call this formula Euler’s formula. Consider this planar network. It has 6 vertices, 9 edges, and 5 faces.

To show that Euler’s formula holds true for this graph, we can evaluate the left-hand side of the formula and show this is equal to 2: v+f−e=6+5−9 =2

Substitute v, f, and e Evaluate

So Euler’s formula holds true for this graph. So it is planar.

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Example 2 Consider the graph shown.

a How many vertices (v) does the graph have?

Create a strategy

Apply the idea

Count the number of vertices (circles).

Based on the given graph, there are 5 vertices so v = 5.

b How many edges (e) does this graph have?

Create a strategy

Apply the idea

Use the fact that an edge connects any two given vertices.

Based on the given graph, there are 6 edges that connect two given vertices so e = 6.

c Match the graph with its equivalent planar representation. A

Create a strategy Use the fact that the planar representation cannot have any edges crossing over one another.

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Apply the idea Among the choices, only option D represents the equivalent planar representation of the given graph as the edges connecting vertices B and D are redrawn so that all connected vertices are still connected in the planar representation. d How many faces ( f ) does this representation have?

Create a strategy Use Euler’s formula: v − e + f = 2 where v is the number of vertices, e is the number of edges, and f is the number of faces.

Apply the idea v−e+f=2

Write the formula

5−6+f=2

Substitute v = 5 and e = 6

−1 + f = 2 f=3

Make f the subject Evaluate

Example 3 A connected planar graph has 11 edges, and 8 faces. Solve for v, the number of vertices the planar graph has.

Create a strategy Use Euler’s formula: v − e + f = 2 where v is the number of vertices, e is the number of edges, and f is the number of faces.

Apply the idea v−e+f=2

Write the formula

v − 11 + 8 = 2

Substitute e = 11 and f = 8

v−3=2 v=5

Make v the subject Evaluate

Idea summary A connected planar graph satisﬁes Euler’s formula:

v+f−e=2 v f e

is the number of vertices is the number of faces is the number of edges

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11.03 Practice questions Understanding 1

Adventurer

1−3

Trailblazer

1−3

How many faces does this planar graph have?

How many edges?

How many vertices?

Calculate v + f − e.

Does it satisfy Euler’s rule for planar graphs, v + f − e = 2?

Move the vertex with degree 2 underneath the other vertices and redraw the graph to clearly show that it is planar.

Solve for the missing number of faces f, edges e, or vertices v: a

5+4−e=2

Fluency 4

654

9 + f − 14 = 2

4ab, 5a, 6a, 7ab, 8a, 9ab, 10a–c

v+6−8=2

4ab, 5a, 6b, 7bc, 8a, 9bc, 10b–d

4bc, 5bc, 6b, 7cd, 8b, 9cd, 10c–f

Draw an equivalent planar representation for each graph: a

Ex 1

Explorer

For each graph: i

Draw an equivalent planar representation.

Determine the number of faces f for the representation.

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For each of the pair of graphs: i

Select which graph is the connected planar graph.

Find the total number of edges, E.

iii

Calculate D, the sum of the degrees.

Write an algebraic equation relating the sum of degrees (D) and the number of edges (E).

Graph A:

Graph B:

Graph A:

Graph B:

For each graph, determine: i

The number of of vertices

The number of faces

iii

The number of edges

If the graph is planar

11.03 Planar graphs mathspace.co

655

Ex 2

Ex 3

For each graph: i

Find v, the number of vertices.

Find e, the number of edges.

iii

Draw an equivalent planar representation of the graph.

Find f, the number of faces of the planar graph.

Evaluate v + f − e for the graph.

Is it possible for a connected planar graph to have these parameters? a

v = 5, f = 6, and e = 9

v = 5, f = 5, and e = 10

v = 8, f = 6, and e = 13

v = 12, f = 10, and e = 20

For a connected planar graph with v vertices, e edges and f faces, ﬁnd: a

f given v = 7 and e = 14

e given v = 8 and f = 2

v given f = 3 and e = 12

e given v = 4 and f = 5

v given f = 8 and e = 15

f given v = 6 and e = 7

Reasoning 11

Graph A Graph B

Graph C Graph D

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11, 12

Complete the table: Graph A

Graph B

Graph C

Graph D

Vertices (v) Faces ( f ) Edges (e) v+f−e Planar? (Y/N) No. of vertices with odd degree b 12

What do you notice about the number of vertices with odd degree? Explain why.

Show that Euler’s rule also holds true for 3D shapes such as this cube.

Problem-solving 13, 14, 16a 13

13, 15, 16b, 17

A connected planar graph has 3 edges and 3 vertex. a

Find the number of faces, f.

Draw a planar graph with the given parameters.

A connected planar graph has 3 vertices and 3 faces. a

Find the number of edges, e.

Draw a planar graph with the given parameters.

A connected planar graph has 3 edges and 3 faces. a

Find the number of vertices, v.

Draw a planar graph with the given parameters.

For these non-planar graphs, ﬁnd the minimum number of edges that can be removed to make them planar: a

13, 14, 16

A connected planar graph has 4 vertices of degree 2, 2, 3, and 3. a

Find the number of edges, e.

Find the number of faces, f.

11.03 Planar graphs mathspace.co

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11.04 Walks, paths, and trails After this lesson, you will be able to… • identify what constitutes a walk in a graph. • differentiate between open and closed walks. • define and identify a path, trail, circuit, and cycle in a graph. • determine the shortest path between two vertices in a network. • apply this knowledge to solve problems related to walks, paths and cycles in graphs.

Walks A walk is the most general deﬁnition of how we can move around a graph. If we start at a vertex and then move along an edge to a new vertex, we are starting a walk. If we have a directed graph, we move in the direction of the arrows. We can continue the walk by selecting another edge leaving the second vertex, then a third edge leaving the third vertex, and so on. In a simple graph, we can represent a walk by a sequence of vertices.

Start A

Here is a simple, undirected graph, with the arrows representing a walk around the graph.

B D

We can record this walk as the sequence A, B, A, E, F, D, E.

End F

E G

If a walk starts and ends at the same vertex then it is closed, and if it starts and ends at different vertices then it is open. The walk marked on the first graph is an open walk.

Start A End E

658

This graph shows a different walk on the same graph; this time a closed walk.

This walk has sequence A, B, A, D, F, C, G, D, E, A. F

The walk is a closed walk because it starts and ends at the same vertex. We can also tell that it is a closed walk by looking at the sequence (without seeing the diagram) because the sequence starts and ends with the same letter.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Example 1 Select the valid walk in this network: A E, D, C, B, A B A, B, E, C, D C C, D, A, E, D D B, C, D, B, A

Create a strategy Choose the sequence where the network always has edges connecting one vertex to the next.

Apply the idea The sequence E, D, C, B, A in option A is a valid walk because there are edges from each vertex to the next in the sequence. In option B, E comes after B, but the vertices B and E are not connected, so it is not a valid walk. In option C, E comes after A, but the vertices A and E are not connected, so it is not a valid walk. In option D, B comes after D, but the vertices B and D are not connected, so it is not a valid walk. So the correct answer is option A.

Example 2 State a closed walk for the graph, starting from H with 3 edges.

Create a strategy

Apply the idea

Find a walk that starts and ﬁnishes at H with 3 edges.

A closed walk for the graph, starting from H with 3 edges is: H, X, Y, H Another closed walk is the same path but in the other direction: H, Y, X, H

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Idea summary A walk is the most general deﬁnition of how we can move around a graph. If we start at a vertex and then move along an edge to a new vertex, we are starting a walk. If we have a directed graph, we move in the direction of the arrows. If a walk starts and ends at the same vertex then it is closed, and if it starts and ends at different vertices then it is open. We record a walk by listing the vertices in order that we pass them.

Trails and circuits If a walk does not use the same edge more than once, then it is a trail, it can be an open or closed trail. Closed trails are also called circuits. Here is one of each on the same graph:

Start

End End

Start

The blue walk A, D, F, E, D, G, C, F on the left is an open trail, since no edge is used more than once, and it is open. The red walk G, C, F, E, D, F, A, D, G on the right is a closed trail (circuit), since no edge is used more than once, and it is closed. • A trail is a walk where no edge is repeated. • An open trail starts and ends on different vertices. • A closed trail (circuit) starts and ends on the same vertex. • All trails are walks and all circuits are closed walks.

Example 3 Which of the following are circuits in the network shown? A C, B, A, E, D, C B B, A, E, C, D, B C B, C, A, E, D, C D D, C, B, A, E, D

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Create a strategy Choose the trail that starts and ends on the same vertex and has no repeat edges.

Apply the idea Option A starts and ends on C. The edges covered are CB, BA, AE, ED, and DC. So no edges are repeated. So it is a circuit. Option B starts and ends on B. The edges covered are BA, AE, EC, CD, and DB. So no edges are repeated. So it is a circuit. Option C is not a circuit because it does not start and end on the same vertex. Option D starts and ends on D. The edges covered are DC, CB, BA, AE, and ED. So no edges are repeated. So it is a circuit. So the correct answers are options A, B, and D.

Example 4 State the trail with the fewest number of edges, that is a closed trail and starts at H.

Create a strategy Find the shortest walk that starts and ﬁnishes at H and passes each edge only once.

Apply the idea The shortest possible closed trail that starts at H has 4 edges: H, D, F, G, H We could also go in the other direction: H, G, F, D, H

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Example 5 Starting with vertices P, R, T, state the longest possible closed trail for the graph.

Create a strategy

Apply the idea

Find the longest trail that starts at P, goes to R, then to T, and ﬁnishes at P, without repeating any edges.

Using trial and error, we can ﬁnd the following path: Closed trail = P, R, T, S, Q, U, P

Idea summary • • • •

A trail is a walk where no edge is repeated. An open trail starts and ends on different vertices. A closed trail (circuit) starts and ends on the same vertex. All trails are walks and all circuits are closed walks.

Paths and cycles If a walk does not use the same vertex more than once (other than at the start and end), we call it a path. When a path begins and ends at the same vertex it is a closed path, which is usually called a cycle. All paths are walks (and trails) and all cycles are closed walks (and closed trails). Here are two different paths on the same graph: C

End D

B D

Start

A End

A F

E G

The open path on the left is D, G, C, F, E, A, and the closed path (called a cycle) on the right is B, C, F, D, E, A, B.

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Example 6 Consider this network:

a Find a path that starts at H, visits two other vertices, and then reaches B.

Create a strategy

Apply the idea

Start at vertex H and choose an edge where the If we start at H and move to X, we cannot make next vertex can make two more moves to reach it to B with only two more moves. vertex B. If we start at H and go to Y, then go to Q, then go to B, we achieve our goal. So the path is: Path = H, Y, Q, B

b Now ﬁnd a path that starts at H, visits four other vertices, and then reaches B.

Create a strategy

Apply the idea

Start at vertex H and choose an edge where the The following path starts at H, visits four next vertex can make four more moves without vertices, X, Y, Q, W, to reach B: repeating a vertex to reach vertex B. Path = H, X, Y, Q, W, B

Idea summary Walks Open start ≠ end

Closed start = end

Trail no repeat edges

Circuit

Path no repeat vertices

Cycle

Every sequence of edges is a walk. A walk can be open (start/end vertices different) or closed (start/end vertices same). If edges do not repeat, the walk is a trail. A closed trail is called a circuit If vertices do not repeat, the walk is a path. A closed path is a cycle. Walks, trails, circuits, paths and cycles do not need to use every edge or every vertex in the graph.

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Shortest path by inspection A common problem posed in networks is to ﬁnd the path of lowest weight between two vertices. For example, ﬁnding the best ( fastest, shortest, cheapest) route between two destinations. Let’s take a look at this network of towns.

A 4 2 C

9 3

3 E

• The red path AB will take 9 minutes. • The blue path ACEB will take 2 + 3 + 4 = 9 minutes. • The green path ADB will take 4 + 4 = 8 minutes.

A 4 2 C

9 3

3 4

This path does not look like the fastest but it takes 2 + 1 + 4 = 7 minutes, and is the quickest way to get from A to B.

A 4 2 C

9 3

This last path is close to the answer, but we can get from A to D even faster if we take a shortcut through C.

The weights represent travel time in minutes, and we want to get from A to B via the optimal ( fastest) path. Here are three paths we could take:

3 E 4

In theory, when trying to ﬁnd the shortest path by observation, we should check every path and its length. However, we can eliminate some options as possibilities by looking at the structure of the network carefully. Firstly, recall no path will pass through the same vertex or use the same edge more than once. Also, the shortest path will not take “the long way around”. For our example above the shortest path would never use the section C, E, D (which take 6 minutes) when the path could ﬂow directly from C to D in just 1 minute. We can perform the same sort of analysis on directed graphs as well, just be careful to follow the direction on the arrows when forming paths.

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A 6 B

4 11

6 5

Here is a directed graph. This time, the streets are all one-way.

6 5

Let’s try out a few paths that take us from vertex A to G: • The blue path AFG takes 10 + 11 = 21 minutes. • The green path ADG takes 4 + 12 = 16 minutes. • The purple path ABCEG takes 6 + 1 + 5 + 5 = 17 minutes.

11 1

We can now check more paths, thinking about taking detours on paths we already know about. Starting from the path ADG, the shortest one so far, we can shorten the trip from D to G by going through E.

11 6 5

E 5

C F

Going along the path ADEG only takes 4 + 6 + 5 = 15 minutes, and this is the fastest route.

In general, when solving shortest path problems, using trial and error is an appropriate method. However, we need to be careful as it can sometimes be difficult to ensure all paths are checked. Systematically recording the paths and eliminating paths that contain extended detours, from the shortest path found so far, can help avoid missing possible solutions. It is also effective for finding the shortest path on networks that are relatively small or not overly complex.

Example 7 F

For this graph:

11 A

13 5

B 7

G 12

a Use a systematic approach to ﬁnd the shortest path from vertex F to vertex E. Write the vertices in order.

10 C

D 7

Create a strategy Find the shortest path by selecting the edge with the smallest weight to travel to the next vertex. Be careful to avoid longer detours.

11.04 Walks, paths, and trails mathspace.co

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Apply the idea The answer is F, A, G, E.

Reflect and check Notice that the shortest path starts by going from F to A and not to B, even though the weight of edge FB is lower than that of edge FA. It is important to consider the whole path all together, and not just look at one vertex at a time. b What is the length of this shortest path?

Create a strategy

Apply the idea

Add the corresponding numbers from the edges. The answer is 11 + 5 + 9 = 25.

Idea summary To find the shortest path by observation will involve some trial and error. We can start by choosing any path that passes through one or more of the lowest weighted edges, then see if there are any faster ways to connect vertices along this path.

11.04 Practice questions Understanding Ex 1

666

Adventurer

1−3

Trailblazer 1−3

Using the diagram: a

Explorer

Are these valid walks? i

K, F, P, K

K, P, F, K, P

iii

K, P, F, R, M

R, P, F, M, R

K, P, F, K

F, R, M, P, F

K, P, F, K, P

Are these valid paths? i

K, P, F, R, M

iii

K, P, F, K

Match the deﬁnition with the correct term: i

Walk

A walk that does not use the same vertex twice, except possibly the start and end vertex.

A path that begins and ends at the same vertex.

A sequence of adjacent vertices.

A walk that starts and ends at the same vertex.

Path

iii

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Cycle

Closed walk

What are the similarities and the difference between a path and a cycle?

Fluency

4a, 5a, 6, 7, 8a, 9a, 10a, 12b–e, 14bc, 15, 16, 17b–d

4b, 5b, 7, 8b, 9b, 10, 11, 12c–f, 13, 14cd, 16, 17c–e

Find a valid walk that is also a path, it must cross each vertex once. a

Ex 2

4a, 5a, 6, 8a, 9a, 10a, 12a–d, 14ab, 15, 16, 17a–c

List the vertices that represent the described walk. a

A closed walk with 4 edges,

A closed walk with 3 edges,

starting from W. starting from B.

Find a trail from A to E with 4 edges that does not go through D.

What is the shortest possible trail from E to A through D and C?

11.04 Walks, paths, and trails mathspace.co

667

Ex 3

Find a valid circuit. a

Ex 4

Find the shortest possible closed trail using the information provided. a

The trail starts at A.

he trail starts at T and includes S and R T but does not include Q.

Ex 5

Find the longest possible closed trail using the information provided. a

The trail starts with vertices

F, D, C in that order.

The trail starts with vertices S, T, R in that order.

668

What is the shortest possible closed walk which includes these vertices? a

S and Q

R and T

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Find a valid path. a

How many unique paths are there from B to D that do not pass over the same vertex twice?

11.04 Walks, paths, and trails mathspace.co

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Ex 6

Find a path that satisﬁes the given information for each network: a

The path starts at B and ends at X.

The path starts at Q, has length 3,

and ends at P.

The path starts at X, goes through

The path starts at P and ends at Q

4 other vertices, and then ends at W.

For the given network, list a sequence of vertices, starting from A, that represents a cycle of 5 edges.

Using the diagram:

What is the correct term for this sequence of vertices? X, H, Y, Q, B, W, Q, Y, X

670

Find a cycle with 3 edges that starts at X.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Ex 7

Determine the shortest path for the given start and end point and ﬁnd its weight. a

Start at vertex F and end at vertex E.

Start at vertex E and end at vertex B.

9 5

10 11

3 5 10

Start at vertex E and end at vertex B.

Start at vertex X and end at vertex V.

3 8

12 9

12 13 17

10 11

Start at vertex T and end at vertex Z. 13 3 9

5 19

12 13

10 11

Reasoning

18, 19

Explain the difference between a walk and a path in a graph.

Describe a real life scenario where the shortest path algorithm would be useful.

11.04 Walks, paths, and trails mathspace.co

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Problem-solving 20−24 20

21−25

23−27

The map shows all possible routes in Mohamad’s town. Mohamad travels from his home, E, to his friend’s house, H, on a daily basis:

Determine the shortest path and its length between the two houses.

23 16 15 25

A grocery store is known for its fast deliveries across the town as shown in the map: If the store is at vertex A, and a delivery has to be made to vertex H, determine the shortest path the driver should take.

22 11

15 21 16 24

7 26 20

A firefighting department received an emergency call about a house fire across the town. If the ﬁreﬁghting department is located at vertex L, and the house is at vertex A. Using the graph of the town, determine the shortest route the fireman should take. 6

19 15

16 12

The police department received a call about a robbery in progress at a grocery store. The graph shows the map of the town: If the nearest police car is located at vertex D and the grocery store is at vertex I, determine the shortest route the police should take.

13 12

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Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7 18 6 5

26 25 10

29 11

13 7

12 24

The following table represents the mail route and times, in minutes, between houses. a

Draw a graph to represent the information in the table.

Find the fastest time for a mailman to get from house A to house E.

What is the fastest route?

Starting House A A A B B C C D D E

Ending House B F G F C E G E G G

The graph shows the time in hours taken to travel between various country towns: a

Find the least amount of time needed to travel between towns P and X.

What route must be taken?

Time 11 9 3 5 10 8 6 5 5 7

2.7 5.3

1.3

2.0

3.5

1.9

1.1

3.8

0.5 2.0

1.5

4.0

2.0

1.0

3.0

A group is organising a tour for the elderly in a park, ending with a picnic. The graph displays the map of the park: Calculate the path which gives the shortest distance from the start of the park at vertex A, to the picnic area at vertex J.

24 19

26 11

6 7

13 23

10 29

A rock band is planning a tour across several cities. The table represents the routes and distance (in kilometres) between the cities at which they will be performing: a

Draw a graph to represent the information in the table.

Find the shortest distance for the rock band to travel from city A to city E.

What is the shortest route?

Starting City A A B B C C E E

Ending City C F E F D F D F

Distance 9 6 11 5 3 7 5 8

11.04 Walks, paths, and trails mathspace.co

673

11 Chapter review 1

Are these networks valid or non-valid? a

For each network: i

How many vertices are there?

iii

What is the degree of each vertex?

B G

How many edges are there?

D F

A F

During the winter, ﬁve friends often go out running in pairs. These are the running pairs that have gone out so far.

674

Draw an undirected graph to represent the pairs of runners who have gone running together.

Which feature of the graph tells us how many runs a person has gone on?

Name all pairs of runners that should run together, so that all ﬁve friends have run with each other.

What type of graph will result if those pairs are added to the graph?

• Astrid and Bernice • Astrid and Dora • Astrid and Edwina • Bernice and Edwina • Christina and Dora • Dora and Edwina

Are these walks, paths, or neither? a

C, B, E, B, A

A, B, D, F

A, B, F, D

D, C, B, A, D

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

B A

E D

What is the shortest possible closed walk which includes these vertices? a

S and T

R Q

R and T

Find the path that starts at E, has length 3 and ends at C.

Find the shortest path that starts at A and ends at G, and ﬁnd its length.

D 3

Alicia takes a bus to school every day. The graph shows the map of the town’s bus routes and the time of the routes in minutes. If Alicia lives at vertex C and her school is at vertex I, determine the fastest bus route she should take.

I 6

7 B

12 29 C

2 B

13 23 10

16 J 12

For each graph: i

Draw an equivalent planar representation.

Determine the number of faces ( f ) for the representation.

S T

Chapter 11 review mathspace.co

675

676

For this graph: a

How many vertices does it have?

How many edges does it have?

Is the graph planar?

A connected planar graph has ﬁve vertices with degrees of 2, 2, 3, 3, and 4. a

How many edges does it have?

How many faces does it have?

How can we conﬁrm our answers to parts (a) and (b)?

An electrician is planning the work she wants to ﬁnish in one day. The table represents the routes and distance in kilometres between the homes at which she will be working: a

Draw a graph to represent the information in the table.

Find the shortest distance for the electrician to travel from city A to city D.

What is the shortest route?

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Starting City A A A B B C C C D E

Ending City C E B C D D F E F F

Distance 9 6 7 3 5 8 11 5 10 4

“I believe the process of mathematics is like weaving a tapestry, where each thread contributes to the whole pattern.” –Ada Lovelace

Answers

g 4p9 10 4

i 16u v

1.01 Index laws

n −9 p 64y12 12

q 16x 3

s x

2 3 a i Yes

ii Yes

iii Yes

iv No

b i False

c 2p14

d −5x2y14z11

f 9v5y11

g −6u11

h 4x2

c 6 b 9

e 11

f m4

c 2

d 0 Reasoning

Fluency 5 a a8

b a10 7

c b5 7

e 6y

f 12y

i u

j 2w

b 0 d

c 56m n

h am+n

l 72y17

k 15z

8 5

6 a 40g h

d y10 11

g −2a

7 6

7 a u11

c 3m7

d k13

e a5

f m3

g j7

k 9j2k3

8 a −32y15

b −96y11

5 5 5

c −30s t r

d 504y24

e −10

f 3y4

9 a j24

b w22

c y12

d 81a8

e 25h14

f x5y10

j x36

k −125x12 l

b k31

c 24r58

10 a w36v24 39 4

45 60

e 160r s f a b

g p

16 a The coeﬃcients of each term must multiply to 9, and the indices should add together when multiplying the same base. The solution to this problem does not multiply the coeﬃcients and instead multiplies the indices. The solution is 9s8.

h x60

c When dividing terms with the same base,

c 1

d 1

e 11

f 1

h 6

b p

15 Indira and Ernestino are both correct. Chad is incorrect. Answers may vary. The expression in the numerator is a product, so the division rule does not get applied to both parts of the numerator. We can check by substituting, for example, a = 2. By simplifying each expression, we see that Indira and Ernestino’s ﬁrst lines of work are equivalent to the original expression, while Chad’s is not.

d x67y15

b 1

12 a x y

14 Yes. The expression in the numerator allows the fraction to be separated into two parts, each with a denominator of a2. The division rule can be applied to both terms.

b When raising a product to a power, each part of the product must be raised to the power. Therefore, 2 must also be raised to the power of 5, leading to the solution 32x15.

11 a 24

28 8

t c86 b 10x2y11z15

ii True

4 a 10

r 1125y16

13 a n20r20

v No

678

l 22

m x60 o p17

1 Subtraction

j 81x9

k m

Understanding

h −13

c 0

d 81a8b20c4

f y7

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

subtract the indices, so not t1. The solution is 6.

d When raising a power to a power, we multiply the indices rather than add them which is done here. Then, we will have a different answer when we divide like bases. The solution is x2y22.

e The entire expression 4a must be raised to the power of zero, leading the expression to simplify to 1. Problem-solving 17 There will be 26 = 64 bacteria after 6 hours. 18 We know that 3x is not the same as x3, in general, but the three values x = 0, x = , and x= will make 3x = x3 be a true statement. 19 a = 3, b = 6 20 a True

b False

c False

21 16p = 24p 22 a 3a+6

b 343a−21

1.02 Negative indices

l −

k b an

7 a u−4

b 2r−6

c xy−2

d 4p−6q−7

8 a

9 a 20y6

b 28a2

h 25m3

10 a 6

c 23

d 11

g 4

h 4

11 a 32y8

b 60y5

e 6y12

f 4y11

12 a

o 3x8

13 a

6 a

e 8h7

Understanding 1 a True

b False

c False

d True

2 Less than 1. can be expressed as as a−1.

Fluency 4 a 2−4

2−3

2−2

2−1

10−3 10−2 10−1 100 101 102

103

104

b 10−4

10 100 1000 10 000

c −4

3−3

3−2

3−1

5 a

Reasoning

14 Both Oreste and Donovan are correct. Using

the index law

we can show that their

answers are equivalent.

Answers mathspace.co

679

15 Using the negative index law, we would get and dividing by zero has no

Problem-solving 19 There are many different representations, such as:

meaning as there is no number that, when multiplied by zero, gives 1.

•

16 The −1 index only applies to the variable directly in front of it, since there are no brackets. The

correct working is

17 The value of x−a would get very small and stay positive (x−a gets closer to 0). , since x > 1 and a is

We know that

21 a n = −3

getting very large, x gets very large. When we take the reciprocal, we are dividing 1 into a very large number of pieces, so each of the pieces would be very small and get closer and closer to 0 in size. For example, 10−10 = 0.0000000001.

Therefore,

• 2

−p

× 4p : −32, 0, 1,

b • For 2p and 4p: Since these expressions are directly proportional to p, as p increases, the values of 2p and 4p will also increase linearly. Thus, they will become very large as p becomes significantly large.

, which is less than the , which we

know is greater than the base . There are many examples of this same pattern, such as 3−1 < 3 and

23 4.1 minutes 24 a 3 × 10−6 Farads c

b 5 × 10−6 Farads

Volts or 0.03 Volts

Chapter 1 review

• For 2−p + 4p: As p increases, 2−p = approaches zero because the denominator becomes very large. Therefore, 2−p + 4p will be dominated by the 4p term and will increase linearly with p.

1 a 6x5

b −3a9

c 8y15 2 a u8

d 54m9 n8 b 25y8

c 16x12

3 a = 5, b = 4 4 a 53a−2

b 4v13 y

5 a

−p

• For 2 × 4p: Although 4p increases linearly, the term 2−p decreases exponentially towards zero. The product of a linearly increasing term and an exponentially decreasing term will approach zero. Therefore, 2−p × 4p will get closer and closer to zero as p becomes very large.

6 a 20y6

7 a

c 20736x4

8 a 9 a m2n6

680

c n = −3

base 2. However,

18 a • 2p : −4, 0, 8, 20 • 2−p + 4p : −4, 1,

b n = −4

• 4

22 The negative index law requires us to write the reciprocal of the base value and apply the positive index to the expressions.

• 4p : −8, 0, 16, 40

• 22

•

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b b

n5 m2

c m n

d 25m3

2.01 Factorise algebraic expressions Understanding 1 a nm

b 5x

c 4ab

d 2bn

e 4bx

f 3ab

g (x + 1)

h (4 − x)

2 a 2x (x2 + 2x + 1)

b 5y ( y2 + 3y + 2)

d 4k (x2 + 3x + 4)

c a (x − 3x + 6) 3 3pr Fluency 4 a 3a (x2 + 2x − 2)

b 2x (x2 − 8x + 1)

c 2ax (x + x + 2)

d 5x ( y2 + 2y + 3)

e 4b (b − 4b + 5)

f 6q (r − 6r − 2)

g 3xz ( yz + 7y + 3)

h 4ac (−bc + 2b + 1)

i −6ehg ( fh + 5f + 2)

j 10rtu (su + 7s − 4)

5 a (2y − 3) (x + 2) c 2 (n − 3) (2m2 + 3) 3

b (3m2 + 5) (n + 2) d (5y3 + 7) (z + 5)

e q (r + 4) (2p + 5)

f 2b (c − 2) (3a2 − 2)

g ( y + 4 + x) ( y + 7)

h (g + h) (2f + g + h)

i −(7 − x )(2y + 1)

j (8 − c2)(7 − 5d)

k (4m + 7)(m − 5)

l 2(z + 2)(10z − 3)

6 2x − 1 7 a 5j (4 − 3k − 11jk) b p2 q2 (3 + 4p2 q2 + 5p4 q4) c 2y (z − 5x + 6xyz) d pqr (1 + pq + p2 q2)

11 When factorising an algebraic expression, you can check your answer by expanding the factors and verifying that they multiply back to the original expression. Consider the algebraic expression 16x3 + 12x2 + 8x. Its factorised form is 16x3 + 12x2 + 8x = 4x(4x2 + 3x + 2). We can check if our factorisation is correct by multiplying the factors back together 4x(4x2 + 3x + 2) = 4x(4x2) + 4x(3x) + 4x(2) = 16x3 + 12x2 + 8x. The result matches the original expression, confirming that our factorisation is correct. Problem-solving 12 a 3 (x2 + 2x + 1) b The entire expression 3(x2 + 2x + 1), is the length of the rectangle. It is 3 times the width. This means that the term x2 + 2x + 1 represents the width of the rectangle. 13 a Let the width of the smallest field be x so that its area is x2. The largest field has an area of 9x2 and the middle field has an area of x (x + 5). b The area of the entire field is the sum of each smaller field: x2 + 9x2 + x (x + 5). The factored form of this expression is x (11x + 5) so the width of the field is x and the length is 11x + 5.

e h (−f − j + g)

14 It is a square when x = . For all other values of x, the rectangle is just a rectangle.

f 5m (5 − 3n − 2mn)

15 Example answer: 8x + 30 = 2(4x + 15)

Reasoning 8 The expression has not been fully factorised, as the terms of 2x3 + 3x2 − 4x contain a common factor of x. 9 Xander’s error lies in factoring the highest common factor. He hasn’t considered the last term −5xy in the original expression. Consequently, the correct factored form of the expression should be 5xy(7x + 2y − 1). 10 The highest common factor will have the lowest possible index of the terms because if it were any higher, not all the terms could be divided by that factor.

2.02 Expand binomial products Understanding 1 a 6x2

b 20y2

d −10p2

c −21z

2 a (x + 2) (x + 1) = x2 + 3x + 2 b (x − 3) (x + 2) = x2 − x – 6 c (x + 2) (x − 5) = x2 − 3x – 10 d (x − 3) (x − 4) = x2 − 7x + 12 3 a (x + 3)2 = x2 + 6x + 9 b (x − 3)2 = x2 − 6x + 9 c (x + 6)2 = x2 + 12x + 36 d (3x − 8)2 = 9x2 − 48x + 64

Answers mathspace.co

681

4 a (x + 5) (x − 5) = x2 – 25 2

b (x + 2) (x − 2) = x − 4 c (x − 8) (x + 8) = x2 − 64 d (x + 14) (x − 14) = x2 − 196 Fluency 5 a x2−9x + 8

b x2 + 5x + 4

c x2 − 9x + 18

d x2 − x − 12

f x2 − 6x + 5

g x − x − 30

h x2 − 7x − 18

6 a 6x2 + x − 12

b 12x2 + 7x − 10

c 4x − 7x − 2

d 12x2 − x − 20

e 2x2 + 2x − 12

f 6x2 + 9x − 60

e x − 12x + 35

g 10x − 15x − 10

h −18x2 − 3x + 6

7 a

8 a x2 + 14x + 49

b c2 + 12c + 36

c e2 + 10e + 25

d y2 − 10y + 25

e 49 − 14z + z

f 36 + 12a + a2

g a2 + 2ab + b2

h a2 − 2ab + b2

9 a 9x2 + 42x + 49

b 4y2 − 20y + 25

c 4y2 − 12y + 9

d x4 + 2x2 + 1

10 a 5m2 − 50m + 125 b 6n2 + 72n + 216 c 8q2 + 48q + 72 d 147x3 − 210x2 y + 75xy2 11 a x2 − 49

b 64 − m2

Reasoning 14 No. Explanations may vary. One possible answer: To expand perfect squares, we square the first term, square the last term and double the product of the terms. 15 36 × 24 can be written as (30 + 6)(30 − 6). We can apply the formula of difference of two squares which states that for any two numbers a and b, (a + b)(a − b) = a2 − b2. In our case, a = 30 and b = 6. So, (30 + 6)(30 − 6) simplifies to 302 − 62. This gives us 900 − 36 = 864. 16 (x + 3)(x + 3) = x2 + 6x + 9 and 13 × 13 = 169. As we can see, the structure of the results in both cases follows a similar pattern, reflecting the general formula for squaring a binomial. The pattern emerges when a binomial expression (in the first case, a mathematical expression, and in the second case, a number) is squared. The first term in the result is the square of the first term of the binomial, the last term is the square of the second term of the binomial, and the middle term is twice the product of the terms of the binomial. 17 The student’s expression is incorrect because they didn’t apply the correct formula for squaring a binomial. The correct formula for the square of a binomial is (a + b)2 = a2 + 2ab + b2, not a2 + b2. Therefore, when you square a binomial, you must remember to include the 2ab term which represents the twice the product of the two terms. 18 a We can expand the expression using the square of a binomial formula, which is (A − B)2 = A2 − 2AB + B2. Our expression is in this form, with A being and B being 3b.

c 49 − 36p2

d 81y2 − 1

e 81x2 − 4y2

f 0.04x2 − 9

Let’s start by squaring A, which is

12 a

Then, we calculate 2AB, which is :

13 a 75x3 − 108xy2 b 243x2 − 192y2 c 71 d 25t2 + 90t − 36r2 + 81 e 384x2 − 486y2

Lastly, B2 is (3b)2 B2 = (3b)2 = 9b2

f 20x3 − 405xy2

682

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Now, substituting A2, 2AB, and B2 back into the formula, we get:

v No. This creates one of the side lengths to be negative, which creates a negative volume. c Maximum occurs at length = 7.145 and Volume = 525.365, so any answer around length = 7 would be acceptable.

Simplifying, we get:

2.03 Factorise binomial products Understanding 1 a 2

b The resuting expression is not a trinomial because it contains terms with variables in the denominator and the expression has more than 3 terms.

2 a 81 = 92

b 100 = 102

d 25 = 52

c 16 = 4 3 x+4

5 a Yes

19 a P = 4x + 16 metres

b No

b A = x2 + 8x − 20 square metres.

Fluency

c The length of the ﬁeld would be (6 + 10) which equals 16 metres.

6 a (m − 6) (x + y)

The width of the field would be (6 − 2) which equals 4 metres. The perimeter of the field would be 24 + 16 which equals 40 metres. The area of the field would be 16 × 4 which equals 64 square metres. 2

20 a 6x + 36x + 54 units

b x3 + 9x2 + 27x + 27 units3

c Yes

b 62x2 − 25y2

c (4w + 3x) (5y − z)

d ( y − 4) (8y − 3)

e (g + h) (3f + g + h)

f (2x2 + 3z) (2y − 1)

g (2c − d) (c + 5d + 3)

h (3x2 − 5y) (x + 4y)

7 a (x + 2) (x + 5)

b (x − 3) (x + 8)

c (a + 8) (a2 + 1)

d (x − 2y) (8 + z)

e (3 + x) (8 + y)

f (x + z) (7y + w)

g (p + 2) (2m + 3)

h (4x + 3z) (1 + 8y)

i (2x + 3z) (1 + 6y)

j (df + rz) (3f − 11y) b (8 + e)2

d (u − q)2

c (7 − p)

e 6 (c − 2)

f 5 (q + t)2

23 n + 3n + 2n

g (4w − 5)2

h (8r + 3)2

24 a −y3 + 4y2 + 96y

j (b − r)2

b i N o, because we cannot have a negative length and y = −10 would create 2 negative lengths, even though the volume calculation would be viable. ii No, because you cannot have a zero length, and zero length yields zero volume. iii Yes, all the side lengths are viable and the volume is 200 cubic units. iv No, because the side length 12 − y becomes zero and zero length yields zero volume.

d No

b ( y − z) (x − w)

8 a (x + 3)2

21 (4x2 − 8x + 4) cm2 22 a 24x2 − 49

d 2d2

4 n+6

Problem-solving

c (c + 5)

b 5x

9 a (x + y) (x − y)

b (n − 5) (n + 5)

c (v − 1) (v + 1)

d (11 − v) (11 + v)

g (4 + 3y) (4 − 3y)

h (xy − 7) (xy + 7)

i (5m + 7) (5m − 7)

j (9x + 4y) (9x − 4y)

k 7 (x − 3) (x + 3) 2

l 5 (3t + 2) (3t − 2) 2

10 a 3y (x − 2y + 3x y ) b (4xy + 3z)2 c (11c − 7d) (11c + 7d)

Answers mathspace.co

683

20 a

d (x2 + y) (3y + z) 2

e 4b (2a b + 3c ) x

f 7 (2b + 1) 2

g 7hp (2h r + 3h + 7p )

h 2 (8w + 3t) (w − 4t )

i (4q2 + 3r)2 j 3d (4f − h) (4f + h)

b (x + 6) (x + 4)

Reasoning

2.04 Factorise monic quadratic trinomials

11 No, if we multiply (a + b) (a + b), the result is a2 + 2ab + b2. The sum of two squares is not equal to (a + b) (a + b).

Understanding

12 a The error is in the constant term of the factors. The correct factorisation should be x2 + 4x + 3 = (x + 1)(x + 3). b The error is in the signs of the factors. The correct factorisation should be x2 − 4x + 4 = (x − 2)(x − 2). 13 a i Perfect square trinomial ii (x + 1)2 b i Common factor then difference of squares ii n (n + 11) (n − 11)

1 a 9 and 10 c −6 and −4 2 a • (1 × 18)

• (2 × 9) •

(2 × 32)

• (−2 × −9) •

(−2 × −32)

• (3 × 6) •

(4 × 16)

• (−3 × −6) •

(−4 × −16)

•

(8 × 8)

•

(−8 × −8)

• (1 × −30)

e i Difference of squares

• (−3 × 10)

ii (x + 16 + y) (x + 16 − y)

• (3 × −10)

• (2 × −15)

f i F actor by grouping then difference of squares

• (−5 × 6) • (5 × −6)

ii (x + 4) (2x + 1) (2x − 1) 14 The trinomial cannot be factorised using perfect squares because the middle term, −72a, is not equal to twice the product of the square roots of the ﬁrst and last terms.

d 660

b 4 (t + 7) units

b (a − 3)2 (a + 3)2

19 (x − 1) (x + 1) (x2 + 1)

684

b m + 10

c y+2

d x−4

Fluency

17 x2 + 12x + 36 18 a (x2 + 4)2

3 a m+6 4 a = 12, b = 7

Problem-solving

16 a t + 7 units

(1 × 64) (−1 × −64)

• (−2 × 15)

c 261

b •

• (−1 × −18) •

d i Perfect square trinomial ii (3x − 2y)2

b 392

d −8 and 11

c • (−1 × 30)

c i Factoring by grouping ii (2x + 5)(x − 3)

15 a 256

b −9 and 7

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 a (x + 2)(x + 3)

b (x + 4)(x + 5)

c (t + 7)(t + 8)

d ( y + 6)( y + 12)

6 a (x + 8) (x + 3)

b (x + 4)2

c (x + 9) (x + 8)

d (x + 8)2

e (t + 5) (t + 2)

f (t + 7)(t + 1)

g ( y + 4)( y + 5)

h (x + 3)(x + 5)

i (x + 3)(x + 4)

j ( y + 7)( y + 2)

k (x + 5) (x + 8)

l (x + 6)2

13 a Exactly one of p or q must be negative.

7 a (x − 3) (x − 1)

b (x − 6) (x − 10)

c (x − 7) (x − 12)

d (x − 5) (x − 7)

b Both p and q must be positive.

e (x − 7) (x − 1)

f ( y − 2) ( y − 8)

c Both p and q must be negative.

g (t − 6) (t − 7)

h (x − 3) (x − 17)

i ( y − 7) ( y − 11)

j (t − 12) (t − 8)

k (x − 11) (x − 1)

l (x − 15) (x − 3)

8 a (x + 2) (x − 3)

b (x + 5) (x − 6)

c (x + 5) (x − 4)

d (x + 6) (x − 9)

e (x + 7) (x − 10)

f (x + 13) (x − 9)

g ( y − 16) ( y + 3)

h (t + 14) (t − 6)

i ( y − 15) ( y + 3)

j (x + 14) (x − 4)

k (x + 12) (x − 2)

l (x + 7) (x − 4)

9 a 2 (x + 2) (x + 5)

b 4 (x + 2) (x + 4)

c 4 (x − 9) (x + 8)

d 3 (x − 5) (x − 2)

e 5 (x − 2) (x + 3)

f −2 (x − 2) (x + 11)

g −4 (x − 5) (x + 2)

h −3 (x − 2)2

i −(x + 4) (x + 2)

j −(x − 3) (x − 4)

k x (x + 3) (x + 4)

l 2x (x + 5) (x + 3)

Reasoning 10 Octavia is correct, since the terms share the highest common factor of −3. The trinomial factorises to −3 (x2 + 5x + 6) = −3 (x + 3) (x + 2) first. 11 Omar used the fact that 144 is a perfect square and checked to see if it was a perfect square trinomial. Since it was, he efficiently factored the expression without needing to find a pair of numbers. Dima found two numbers that had a sum of 24 and a product of 144. She then used those to fill in the factors and then simplified.

14 a No, this would be the area of the whole square and would be equivalent to x2. b Yes. x2− Area II is the same as Area I. Area I has length of x and width of x − 4, so it would have an area of x (x − 4). c Yes. Area I has length of x and width of x − 4, so it would have an area of x (x − 4). d No, this would be Area II which would be equivalent to 4x. 15 a (x + 9)(x + 8) = x2 + 17x + 72 b (x + 2)(x + 1) = x2 + 3x + 2 c (x + 1)(x − 9) = x2 − 8x − 9 16 a a and b must be a factor pair of 36. To maximise the length of the entire ﬁeld you would increase the length by 36 and the width by 1. b a and b must be a factor pair of 36. We want to minimise the difference of the factor pair. The differences are: • 1 and 36 = 35 • 2 and 18 = 16 • 3 and 12 = 9 • 4 and 9 = 5 • 6 and 6 = 0 If both portions are extended by 6 then the areas in the bottom left and top right will be the same, which would minimise the difference in area. Problem-solving 17 (m + 5) (m + 9)

Both strategies work and lead to the same answer. Omar needed to recognise the pattern, while Dima needed to recognise that she could simplify since the factors were the same.

18 a (x + 6) (x + 3)

12 a The error is in the signs of the factors.

19 a (u − 8)(u − 3) b (x + 4)(x − 2)(x + 3)(x − 1)

The correct factorisation should be x2 + 7x − 18 = (x − 2)(x + 9). b The error is in the signs of the factors. The correct factorisation should be x2 − 11x + 30 = (x − 5)(x − 6).

b (m + 5) (m + 10)

20 a (x + 3) (x + 5) b 2 c 9x2 + 72x + 135

Answers mathspace.co

685

21 a (x − 2) (x + 2) (x2 + 3) b ( y − 2) ( y + 2) ( y − 1) ( y + 1) c (a + 1) (a + 8) d (d − 7)2 22 To determine the values of c and d, use factoring to get the area into the form of ( y + c) ( y + d). The expression y2 + 8y + 15 can be factorised into ( y + 3)( y + 5). This means that the values are c = 3 m and d = 5 m. 23 x = 7

Reasoning 12 a i 15 iii −16

ii The same iv Opposite

b i −45 iii −4

ii Opposite iv Opposite

13 Gabriella did not factorise the common factor correctly. Since there is a factor of 2 in both factors of their expression, they need to factor out a highest common factor of 4. The correct working would be: 16x2 + 48x + 36 = (4x + 6)2 = [2 (2x + 3)]2 = 4 (2x + 3)2

2.05 Factorise non-monic quadratic trinomials Understanding 1 Factorise −8 from the expression.

14 a Each polynomial has a leading coeﬃcient a ≠ 1 and none of the polynomials can be factored.

2 a 8x2 + 11x + 3 = (8x + 3) (x + 1)

b Find the product a × c, then look for factors of ac that can add or subtract to be equal to the value of b. If no such factors exist, the polynomial is not factorable.

b 2x2 + 3x − 20 = (2x − 5) (x + 4) c 2x2 − 19x + 45 = (2x − 9) (x − 5) d 10x2 + 29x + 21 = (2x + 3) (5x + 7) 3 (3x + 1) (x + 6)

Problem-solving

Fluency

15 a −5 (t + 1) (t − 2)

4 a (3y + 1)( y + 9) c (3x − 4) (x − 7) e (3t − 4)(4t + 1) g (4x + 5) (3x − 2)

b (2x + 5)(x − 8) d (8x + 3) (x − 3) f (9x + 4)2 h (6x − 7) (x + 8)

5 a 3 (x − 9) (x + 2)

b 4 (x + 5)2

c 2(s + 3)(4s − 9)

d 3 (3x + 1) (x − 5)

e 4 (2x − 1) (x − 2)

f 6 (4x + 1) (3x − 1)

g 5 (7x − 1) (3x − 1)

h 8 (3x − 2) (7x + 1)

b 0 c 0 d 2 seconds after jumping off the board, Tara hits the water. 16 a (4x2 − πx2) units2

b x2 (4 − π )

17 a 3x + 4, 2x + 3

6 a −(2x − 3) (3x + 2)

b −(2x − 7)(3x − 2)

c −3 (2x + 5) (2x − 3)

d −2(3x + 4)(x − 5)

e −(4x − 3) (x − 2)

f −(3x − 4)2

b To get the expression of the perimeter of the rectangular ﬁeld, we use the formula of the perimeter of the rectangle which is given by 2 × length + 2 × width. This means that the perimeter of the rectangular ﬁeld is P = 2(3x + 4) + 2(2x + 3) = 10x + 14.

g −(5x + 4) (x + 2)

h −(3x − 2)(x + 4)

c x = 10

7 a 10y − 112y + 22

b 2 (5y − 1) ( y − 11)

8 a 10x2 + 105x + 245

b 5 (2x + 7) (x + 7)

9 a 6 (x + 3)

18 a She might use a variable, such as x, to represent the width of her border if she did not know yet what border size she wanted.

b (x + 3) units

10 2x + 5, 3x + 4 11 −10 (t + 2) (t − 7)

686

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b The expression for area can be factorised as (2x + 5) (2x + 7) square units, so, the dimensions of the photograph are 2x + 5 units and 2x + 7 units.

c If the border is a width of x units, you would subtract 2x from each dimension since there is a border on both sides. That leaves the dimensions as 5 units and 7 units, so, the area of the photograph is 35 square units. 19 4x2 + 12x + 5 = (2x + 1)(2x + 5) is one possible answer. 20 k = 9 21 m = −6, n = −8 or m = −8, n = −6

2.06 Complete the square

8 a b c (x − 1) (x − 5)

d (x + 10) (x + 1)

e (x − 5) (x − 6)

f (x + 21) (x + 3)

g (x − 1) (x − 19)

h (x − 5) (x − 23)

9 a 3 (x − 3) (x + 1)

b 2 (x + 9) (x − 1)

c 4 (x + 1.5) (x + 2.5) = (2x + 3) (2x + 5)

Understanding 1 a 9

b 16

c 49

d 25

2 a Yes

b No

c No

d Yes

d −2 (x − 5) (x − 13) e −4x (x + 4) f

Fluency g

3 a 3x2 + 6x − 8 = 3 (x2 + 2x) − 8 b 2x2 − 10x + 1 = 2 (x2 − 5x) + 1

c 4x2 + 12x − 6 = 4 (x2 + 3x) – 6 2

d 6x + 24x − 7 = 6 (x + 4x) − 7 Reasoning

4 a 5x2 − 10x + 4 = 5 (x2 − 2x) + 4

= 5 (x2 − 2x + 1) + 4 − 5

= 5 (x − 1)2 – 1

1 x2 + 6x + 13 = x 2 + 6x + 32 − 32 + 13 dd and subtract the square of half of A

b 6x2 + 36x − 1 = 6 (x2 + 6x) − 1 = 6 (x2 + 6x + 9) − 1 − 54

10 When completing the square, we could get:

the coeﬃcient of x

2 = (x + 3)2 − 32 + 13

5 a (x + 9) − 81

b (x − 4)2 − 16

3 = (x + 3)2 + 4

c (x + 5)2 + 6

d (x + 7)2 − 2

e (x − 5)2 + 5

f (x − 9)2 − 4

= 6 (x + 3) − 55

Factor the perfect square trinomial

11 In step 1, Alessandro added the 32 = 9 instead of subtracting it.

6 a (x − 4) (x − 2) b (x + 7) (x + 1)

12 On line 3, she has expanded 3[(x + 3)2 − 9] − 7 incorrectly. She forgot to multiply the 9 by 3.

c (x + 12) (x − 2) d 2

Simplify the constant

This is a sum of squares, not a difference of squares. We cannot factor a sum of squares, so this quadratic is not factorisable.

7 a 2 (x + 2) − 15

b 5 (x + 2) – 29

c 4 (x − 2)2 − 13

d 10 (x + 1)2 − 21

e 7 (x − 1)2 − 6

f 9 (x + 3)2 – 83

13 y = 3(x − 4)2 + 2 means that y is equal to 2 plus an expression, 3(x − 4)2, which cannot be negative. Therefore the minimum value of the function occurs when 3(x − 4)2 is equal to zero, which happens when x = 4.

Answers mathspace.co

687

14 a The quadratic expression x2 − 18x + 81 can be factorised in two ways:

1. x2 − 18x + 81 = (x)2 + 2(x)(−9) + (−9)2

= (x − 9)2

2. x2 − 18x + 81 = x2 − 9x − 9x + 81

= (x2 − 9x) + (−9x + 81)

= x(x − 9) − 9(x − 9)

= (x − 9)(x − 9)

= (x − 9)2

b The quadratic expression 4c2 − 12c + 9 can be factorised in two ways:

1. 4c2 − 12c + 9 = (2c)2 + 2(2c)(−3) + (−3)2

= (2c − 3)2

2. 4c2 − 12c + 9 = 4c2 − 6c − 6c + 9

= (4c2 − 6c) + (−6c + 9)

Problem-solving

= 2c(2c − 3) − 3(2c − 3)

15 a x − 5 cm

= (2c − 3)(2c − 3)

= (2c − 3)

c The quadratic expression 3a2 + 14a + 8 can be factorised in two ways:

1. 3a2 + 14a + 8 = 3a2 + 12a + 2a + 8

= (3a2 + 12a) + (2a + 8)

= 3a(a + 4) + 2(a + 4)

= (a + 4)(3a + 2)

b x > 5 since the side lengths must be positive. 16 x − 4 17 a y = −(x + 4)2 + 36 b y = 36 when x = −4. 18 Answers may vary. x2 + 6x + 4 = (x + 3)2 − 5 and x2 + 8x + 7 = (x + 4)2 − 9 are two possibilities.

Chapter 2 review

1 a 5x

b 4xb

2 a k (k − 8) b 5 (y2 + 2y + 3) c 4c (−bc + 2b + 1) d p2 q2 (3 + 4p2 q2 + 5p4 q4) 3 The student’s error is that they haven’t considered the last term, −5xy in the original expression. Consequently, the correct factored form of the expression should be 5xy (7x + 2y − 1). 2

d The quadratic expression 2x + 5x − 3 can be factorised in two ways:

1. 2x2 + 5x − 3 = (2x2 − x) + (6x − 3)

= x(2x − 1) + 3(2x − 1)

= (2x − 1)(x + 3)

4 a x2−10x + 16

b x2 + 4x + 3

d x2 − 5x − 36

c x − x − 30 5 a Yes

b No, (y − 3)2 = y2 − 6y + 9 6 a m2 – 6 c 9x2 − 0.49

688

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b 64 − 25p2 d

7 a −24x2 + 14x + 24

8 9x2 − 72x + 144 cm2 9 a (k + 8) (k2 + 1)

b x = −7, y = 7

7 a m = 2, c = 1

c m = 0, c = −3

b 5 (r + s)2

8 a Yes

c 7 (5m2 − 7)

e (a + b) (5 + x)

f 5 (x + 4y) (x2 − y)

10 a (x − 5)2

6 a x = 6, y = −2

b No

c Yes

−1

−4

−3

−2

c (1 − x)

d (x + 9) (x + 8)

e (x + 13) (x − 9)

f (y − 16) (y + 3)

g (x + 12) (x − 2)

h (x + 5) (x − 4)

11 a (3t − 4) (4t + 1)

b (6x − 7) (x + 8)

c (9x − 5)2

d −2 (3x + 4) (x − 5)

4 3

12 5x + 6, 4x + 3

−4 −3 −2 −1 −1

13 a 10x2 + 75x + 125

d No

Fluency 9 a i

b (x + 8)2

b 5 (2x + 5) (x + 5)

−2 −3

14 k = 9

−4

15 a (x − 4)2 − 16

b (x − 2)2 + 2

b i

c (x + 5)2 − 4

e 10 (x + 1)2 − 21

f 2 (x + 2)2 − 15

−1

y 4

16 a

b (x + 11) (x + 3)

c (x + 37) (x + 5)

x −2

d (x − 5) (x − 6) 17 a y = 36 − (x + 4)2

c i

Understanding Gradient

y − intercept

y = 3x − 4

−4

y = −5x + 2

−5

y=4

x=3

undefined

none

2 B 3 x=6 4 y = −1

−1

b y = 36 when x = −4.

3.01 Linear graphs

−1

−9

−7

−5

1 −3−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

y x 1 2 3 4 5 6 7 8 9

5 y = 0, x = 0

Answers mathspace.co

689

d i

−1

y 4 3

4 3

−4 −3 −2 −1 −1

−1

−2

h i

−3 −4

−1

ii e i

−1

6 5

3 2

10 a

f i

−1

−3

−1

−2

x −2

y 2 1 x

−2

−1

4x −2 3

−1

11 a

−2 −3

−1

9 y 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1 −2 −3 −4

690

−2

1 −2

1 −1

g i

−1

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

x 1

c (0, 3)

d (−1.5, 0)

e y=9

f x = −3.5

12 a

−1

y 2

−2

y b

1 −1

4 2

−3

−4

−4 −3 −2 −1 −1

−5

−2

−3 −4

c (0, −2)

d (3, 0)

f 3 y 2 1 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13

9 8 7 6 5 4 3 2 1 −1

−1

−2

13 a

x 1

y x

−9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

x 1 2 3 4 5 6

14 a

y 6 5 4 3 2

1 x −1

3 x 1

−1 −2

1 −3

7 6 5 4 3 2 1 −6−5−4−3−2−1 −1 −2 −3 −4

−2

−1

x 1

−1

−2 −3 −4

1 2 3 4 5 6 7

3 2 1 x −4 −3 −2 −1

Answers mathspace.co

691

d i

ii (0, 1)

iii

4 2

4 3

−4 −3 −2 −1 −2

−4 −6

−4 −3 −2 −1 −1

−8

−2

x 1

−3

15 a i 2

−4

ii (0, 4)

iii

e i

iii

ii (0, 3) 4

1 −2

−1

x 1

−1

−3 −2 −1 −1 −2 −3

−3

f i 4

ii (0, 4)

iii

3 2

1 2 3 4 5 6 7

−4 −3 −2 −1 −1

−2 −3

ii (0, −2) 4

−4

16 a i x = 1.5

iii

1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

ii y = −3 4 3 1

−4 −3 −2 −1 −1 −3 −4

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−2

692

−4

c i 5 iii

ii (0, 5) 7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2

b i −1 iii

x 1

b i x = −1.5

ii y = 3

iii

5 4 3 2 1

iii

y 2 1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1 2 3 4 5

−2

−1

ii y = 3

ii y = −4

g i

iii

3 1

−4 −3 −2 −1 −1

−3 −4

−1

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−2

d i x = 2.5

5 4 3 2 1

ii y = 5 7 6 5 4 3 2 1

−2

iii

1 −1

c i x = 1.5 iii

ii y = 1

f i

1 2 3 4 5

ii y = 3

h i iii

3 2 1 x 1

−1 −2

−4 −3 −2 −1

x 1

−1

−2 −3 −4

ii y = −2

e i iii

17 a

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

x 1

2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7

y x 1

Answers mathspace.co

693

b 7

y 2

−1

4 3

−1 −3

−4

x 1 2 3 4

y = 4x − 3

−2

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1

x 1

−5

−2

−3

2 1

4 3 2

−3

−4

−4 −3 −2 −1 −1

−6 −5 −4 −3 −2 −1 −1 y = −2x − 4 −2

−5

−6

−2

−3

6 5 4 3 2 1

−4

5 4

−3 −2 −1

−1 −2 −3 −4

3 2 1

−4 −3 −2 −1 −1

y = 2x + 7

x 1

20 a i x = 2 7 6 5 4 3 2 1

−6 −5 −4 −3 −2 −1 −1 −2

694

y = −3x + 6

19 x = 7

−2

18 a

iii

ii y = 3 4

3 2 1 −4 −3 −2 −1 −1 x 1 2 3 4

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−2 −3 −4

x 1

b i x=7

ii y = −4

iii

7 y 6 5 4 3 2 1 −7−6−5−4−3−2 −1 −1

3 2 1

−4 −3 −2 −1 −1

−2

1 2 3 4 5 6 7

−3

−2 −3 −4 −5 −6 −7

−4

c i x = −6

ii y = 8

iii

y 8

−4 −3 −2 −1 −1

−2 −3

−4

−8 −6 −4 −2

d i x = −9

−2

Reasoning

−4

22 The gradient or slope of a line determines its steepness. A positive gradient slants upwards from left to right because for every step we move to the right (positive x-direction), we also move upwards (positive y-direction). On the other hand, a negative gradient slants downwards from left to right because for every step we move to the right, we move downwards.

ii y = −10

iii

y 2

−10 −8 −6 −4 −2

2 −2 −4

23 A line is uniquely determined by two points. The x-intercept and y-intercept of a linear equation provide two distinct points on the line (unless the line is vertical or horizontal). By connecting these two points, we can sketch the entire line.

−6 −8 −10

21 a

3 2 1 −4 −3 −2 −1 −1 −2 −3 −4

x 1

y = kx + d

Gradient k

y − intercept d

y = −px + r

−p

y=k

25 Changing m to 2m effectively doubles the gradient or slope of the line. This will make the line twice as steep, but it will still cross the y-axis at the same point, c. This is because the slope of a line determines its steepness and direction, while the y-intercept locates it on the y-axis.

Answers mathspace.co

695

Problem-solving 26

550 500 450 400 350 300 250 200 150 100 50

5 a

Sales

e 6 a e 7 a Inﬁnitely many solutions Price

10 15 20 25 30 35

c Inﬁnitely many solutions

Based on the graph, it is estimated that 350 toys would be sold if the price was $25.

d e Inﬁnitely many solutions

27 The cost equation is C = 60 + 0.60x, where x is the number of kilometres driven.

Substituting x = 250, it would cost $210 to rent a car for a day and drive 250 kilometres.

g h x = −8

28 29 y = x + 2

Reasoning

30 C = 5q + 150

8 a Clara provided the correct solution. Here’s why:

31 y = 1.25x + 0.5

3.02 Equations with more than 3 steps Understanding 1 5 − 2(3x − 4) = 3(2x + 1) − 7x Start with the equation 5 − 6x + 8 = 6x + 3 − 7x

Expand the brackets

13 − 6x = −x + 3

Combine like terms

13 = 5x + 3

Add 6x to both sides

10 = 5x

Subtract 3 from both sides

2=x

Divide both sides by 5

2 a Incorrect

b Correct

c Incorrect

d Correct

• Clara started with the given equation: 2z + 6 = 3z − 5 • She subtracted 3z from both sides: −z + 6 = −5 • She multiplied both sides by −1: z − 6 = 5 • She added 6 to both sides: z = 11 David made a mistake in his working: • David started with the given equation: 2z + 6 = 3z − 5 • He subtracted 3z from both sides: −z + 6 = −5 • He subtracted 6 from both sides: −z = −11 • He switched the right-hand side and the left-hand side and then changed the sign of z: −11 = z (this is the mistake)

3 a 60 ( f + 4) + 10 = −12 ( f + 1) − 7

b The mistake made by David was in the fourth step, where he should have multiplied both sides by −1.

b 48 (g − 2) + 9 = 10 (g + 3) − 4 c 32 (x + 5) − 6 = −28 (x − 1) + 5 Fluency b

e d = 10

g q = −2

i y=8

4 a

696

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

9 a His mistake was in Step 4. Instead of subtracting 5 from both sides, he added 5. The correct step should be subtracting 5 from both sides to get 2x = 5.

b 1 4(2x + 5) − 3 = 37

Start with the equation

4(2x + 5) = 40

Add 3 to both sides

2x + 5 = 10

Divide both sides by 4

2x = 5

Start with the answer

a+2=5

Add 2 to both sides

2a + 4 = 10

Multiply both sides by 2

2a + 4 + a = 10 + a

Add a to both sides

Subtract 5 from both sides Divide both sides by 2

10 a Alice provided the correct solution. She correctly expanded the equation, simpliﬁed it, and solved for x, resulting in x = 5. b Bob made a mistake in the third step of his working. After reaching −x + 17 = x + 7, he should have added x to both sides of the equation to get 2x = 10, just like Alice did. Instead, he subtracted x from both sides, resulting in −2x = 10, which led to the incorrect solution of x = −5. 11 a

2a − 7 + 4 + a = 10 + a – 7 Subtract 7 from both sides

3a − 3 = a + 3 b

Simplify Add 8p to both sides Add 16 to both sides Divide both sides by 16

(14p − 28) = −8p + 16

Start with the equation

4(14p − 28) = −56p + 112

Multiply both sides by 7 to clear the fraction

56p − 112 = −56p + 112

Expand the brackets

112p − 112 = 112

Add 56p to both sides

112p = 224

Add 112 to both sides

p=2

Divide both sides by 112

c Preference for either method is subjective. Method 1 may be considered easier due to fewer steps and simpler arithmetic, while Method 2 involves larger numbers and more calculations. It depends on individual comfort with fractions and arithmetic.

Simplify

Start with the answer Add 1 to both sides Double both sides Add 2b to both sides Simplify

Start with the equation Expand the brackets

a=3

12 a

Problem-solving 13 x = 18 14 a Let $r represent the cost of a rugby ball: Price of a cricket ball = r + 1.36 b 8(r + 1.36) + 9r = 119 c $6.36

d $7.72

15 a Let y cm represent the length of the longer side. Length of shorter side = y − 4 cm. b 2( y − 4) + 2y = 12

c 5 cm × 1 cm

16 x = 5 17 y = 8 18 y = 18 years old

3.03 Equations involving fractions Understanding 1 a 2

b 42

c 6

d 12

2 a Yes

b No

c No

d No

3 a True

b False

c True

d False

4 a

Answers mathspace.co

697

Fluency

12 a Yes b y = −1

c x = −1

b Yes, the equation has inﬁnitely many solutions.

h c=1

c x=2

6 a

g q = −1

7 a

d w = −9

16 k = 60

f p=9

g q = −18

17 r = 0

5 a

e n = −1 8 a e n = −4

Problem-solving 13 a b x = $7500 14 x = 20 15 u = −17

18 The equation for the total funds raised is

Reasoning 9 a Dylan made an error in Step 1. He missed the constant term and he should have made sure that had the same denominator. This should be the Step 1 before proceeding to cross-multiplication. By writing the binomial with the same denominator we have Next should have multiplied 2 by the binomial 32 − x, giving us the answer of 64 − 2x. b 1

. Solving for x gives x = $1153.85.

3.04 Rearrange equations Understanding

1 The purpose of rearranging equations is be able to evaluate for different variables when doing repeated calculations. 2 We should make c the subject

Write the right-hand side as a fraction

Cross multiply

Expand the brackets and simplify

Add 2x to both sides and subtract 48 from both sides

Divide both sides by 26 and simplify

3 C 4 a False

b True

c True

Fluency

5 a x=y−z

b x = −8y

f x = −7y

11 Many possible answers, for example: . When we multiply both sides of

6 a m = ghy

the equation by the common denominator, 12, we get 2 (3x + 6) = 3 (2x + 6). Expanding both sides gives 6x + 36 = 6x + 18. If we subtract 6x from both sides, we get 36 = 18.

However, this equation is impossible to satisfy as no value for x can make both sides equal.

698

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

d False

7 a

b y = 4 − 3x2

d y = 2x3 + 3

8 a

-5 -4 -3 -2 -1 -1

5 3 1

x 1

-2

-3 -4

-5

11 a y = x − 1

b y = −2x + 3

d m = 3, y = 6

c y = −2x + 3

d y = 4x − 5

e y = 3x − 5

f y = −2x + 3

g y = 2x + 4 33 99 i yyy==−− xx+++ 22 22 12 a m = − 5 , y = 5 2

h y = 2x + 3 4 j yy = x − 4 3

5 4 3 2 1

y 6 5

-5 -4 -3 -2 -1 -1

-2 -3

2 1

-4 -3 -2 -1 -1

-2

14 a

15 a

g z = x2

i c = sn2 + 6

-3

b m = 4, y = −16 y 2 -2 -1 -2

x 1

-4 -6 -8 -10 -12 -14

16 17

-16 -18

c m = −2, y = 4

18 19

Answers mathspace.co

699

Reasoning 20 Both equations can be rearranged to solve for x. When 3x + 8 = 2 is rearrange to solve for x the answer is numerical. Whereas when 3x + y = z is rearrange to solve for x the answer has unknowns, y and z, in it. However if you substitute y = 8 and z = 2, you get the same numerical answer as solving 3x + 8 = 2. 21 Alice is correct in her rearrangement, while Bob made a mistake. Alice rearranged the equation to make a2 the subject by subtracting b2 from both sides: 1

25 a

b r = 2.79 cm

26 a

b P = $2027

27 a

b r = 3 mm

3.05 Parallel and perpendicular lines Understanding 1 a Their gradients, the values of m, are the same.

c2 = a2 + b2

2 c2 − b2 = a2 + b2 − b2 3

b All graphs have the same gradient or rate of increase.

a2 = c2 − b2

2 −1

Bob, on the other hand, made a mistake in his rearrangement. He should have subtracted a2 from both sides to make b2 the subject, like this:

c2 = a2 + b2

b One pair: y = 3x + 15 and

2 c2 − a2 = a2 + b2 − a2

Another pair: y = 3x + 15 and

b =c −a

4 Perpendicular

However, he added a2 instead, yielding an incorrect equation:

5 D

b2 = c2 + a2

6 D

So, Alice is correct, and Bob made a mistake in his rearrangement.

Fluency

22 Pamela’s mistake was in Step 2 of her algebraic working. When she divided both sides of the equation by L, she incorrectly wrote the right-hand side of the equation as of

instead

. The correct equation should be

23 Ruﬁno has an error in his work. The equation for KE as given

Now he should have divided by v

Problem-solving

700

7 a Yes

b No

c Yes

d Yes

e No

f No

g No

h No

8 a Yes

b Yes

c No

d Yes

e No

f No

g No

h No

9 a −5

b y = −5x + 40

10 a y = −2x − 5

b y = 8x + 5

c y = −3x – 4

d y=2

e x = −7

Multiplying both sides by 2

24 a

and

11 a

c −8

d 1

12 a No

b No

c Yes

d Yes

13 a No

b No

c No

d Yes

14 a No

b No

c Yes

d Yes

e No

f Yes

g Yes

h No

b h = 3 cm

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

15 a

16 a

c No

17 a

c Yes

18 a

33 a i

iii 1

iv 1

b A parallelogram. This is because the opposite sides have the same gradient, making them parallel. It is not a rectangle, as the adjacent lines are not perpendicular.

c y = 2x + 6 e y = −8

3.06 Linear models

d x = −8

Understanding 1 A

Reasoning 19 T he ﬁrst two lines have the same gradient of 3, so they are parallel. The last two lines have the same gradient of

, so they are parallel.

Additionally, the product of the gradients of the ﬁrst and third lines is

2 B 3 a Horizontal

b Vertical

c Horizontal

d Vertical

4 a 250

b 12.5

d −5

c 8

, so they are

perpendicular. Since opposite sides are parallel and adjacent sides are perpendicular, the enclosure is a rectangle. 20 y = 2x + 4 21 y = 2x + 4

Fluency 5 a 115

b B

6 a The straight line graph starting at the origin is the cost function. The curved graph with y-intercept of 50 is the revenue function. b The intersection is the break-even point where cost is equal to revenue.

Problem-solving 22 a

23 a −1

b y = −x + 18

24 a 6

7 a Linear

b Non-linear

c Linear

d Non-linear

8 a

25 y = 0.5x + 5 26 27 a

b k = 10

28 a

b (5, 10)

120 110 100 90 80 70 60 50 40 30 20 10

M 1

b No

29 c = −2 30 a −1

b 1

c −1

d Yes

31 a 0

b −1

c No

d Yes

e A trapezium

9 a

Number of minutes 0 passed (x) Heart rate ( y)

93 99 105 108

32 a Gradient of: b Yes

c No

Answers mathspace.co

701

12 a 5° C

111 y 108 105 102 99 96 93 90 87 84 81 78 75 72

x 1 2 3 4 5 6 7 8 9 10 11

c 3 beats per minute 10 a

Depth of diver in metres ( y) 3.6

28 y 26 24 22 20 18 16 14 12 10 8 6 4 2

x 1

Number of minutes passed (x)

b 3° C

c y = 5x + 3

13 a

T 25

0 0.8 1.6 2.4 3.2

3.2

2.8

2.4 2

1.6

1.2 0.8

b 5 °C

0.4

d T = 5x + 5

x 1

c 0.8 m/min 11 a

14 a

d 19.2 m

48 C 45 42 39 36 33 30 27 24 21 18 15 12 9 6 3

d $90.00

c 5 °C

28 T °C 26 24 22 20 18 16 14 12 10 8 6 4 2

20 40 60 80 100 120 140 160 180 x

b 21 °C

5 10 15 20 25 30 35 40 45

b C = 1.20x

c 190 chirps per minute

c $1.20

d 120 chirps per minute 15 a 18

16 14 12 10 8 6 4 2

d 1

702

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

2 3 4 5 6 7 8 9

b 2 cm

c 26 cm

d 11

e h = 2d + 2

16 a

19 a The variables are income and number of hours worked. b 800

8500 8000 7500 7000 6500 6000 5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500

700 600 500 400 300 200 100

b 6000 m d A = −5t + 8500

c For example, y = 25x d $625 e 5 hours

b For example, y = 150x c $3300 d 36

1.6

Reasoning

1.4

21 a Litres of petrol used per kilometre

1.2 1

b Dollars earned per hour

0.8

c Heartbeats per minute

0.6

d Ascent or descent in metres per second

0.4 0.2

d 1 2 3 4 5 6 7 8 9 10 11

22 a 4 of B b 4 of A c Example answer:

b F = 0.10d + 0.30

Exchange rates between different countries’ currencies change on a daily basis. So this graph may give accurate conversions one day, but not the next.

c $1.70 d 3 km 18 a

10 15 20 25 30 35

20 a The variables are operation cost and number of hours operated.

c 1700 seconds

1.8

x 5

200 400 600 800 1000 1200 1400 1600 1800

17 a

C 6

23 a

5 4

120

−8

b The baseball is moving downwards instead of upwards at T = 4.

3 2

c Example answer:

1 t 2

8 10 12 14 16

b $2.80 c 4 minutes

The model is useful to track the velocity of the ball while it is in the air, but once it hits the ground or if it hits an object, the model will not be accurate. Also the velocity may change due to weather conditions.

d C = 0.30t + 0.40

Answers mathspace.co

703

24 a

Number of minutes passed, x

Problem-solving 0

20 50

Amount of fuel 250 225 200 175 150 left in tank, y

b y = 250 − 5x c The amount of fuel in the car is decreasing at a constant rate of 5 litres per minute.

27 a 16 km 28 a 104 °F

25 a is the acceleration - that is, the rate of increase in the velocity. b 18 is the velocity at which the ball is initially rolled.

30 a 50 °F

31 a −300 is the amount the ﬁsh population decreases by each year. b F = 6000 c F = −300t + 6000 d 3300 ﬁshes e 14 years 32 a 2 is the increase in earnings for each additional shirt ironed. b y = 4 is the earnings independent of how many shirts are ironed. c $32 d 12 shirts 33 a 20 mi/hr 34 a

is the amount of water that is ﬂowing out

b 30 is the capacity of the bucket. c d 3L 35 a 10 is the variable cost of manufacturing per chair. b y = 85 is the ﬁxed cost of manufacturing.

c For example, y = 10x + 85

d 13 °C

d $335

e Example answer: The model is useful to track the temperature for a couple of hours, but eventually we can assume the heater will be turned off, or a window or door will be opened, which will affect the temperature and the model will no longer by accurate. For instance after two hours the temperature according to the model will be 43 °C, so the heater will most likely be turned off.

704

b 80 mi/hr

of the hole every minute.

26 a is the amount by which the temperature is rising every minute. b y = 3 is the temperature of the room before the heater has been turned on.

d F = 1.8C + 32

e 95 °F

d 46.9 m/s The model is useful to track the velocity of the ball as it rolls down the hill, but once the hill ﬂattens the velocity will decrease as time increases. Therefore, the model will no longer be accurate.

b 1.8 °F

c Below

c e Example answer:

c −29 °C

b 10 °C

29 $77.08

d Example answer: The model is useful to track the fuel consumption of the car while it is moving at a constant speed. But as the car changes speed the fuel consumption will change, and the model will not be accurate.

b 1.6 km

3.07 Graph linear inequalities Understanding

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

1 A 2 a False

b False

c True

d True

e False

f True

g True

h True

3 C 4 a Solid

b Dashed c Solid

d Dashed

5 a Below

b Below

d Below

c Above

Fluency

b y=4

12 a

6 a No

b Yes

c No

d Yes

7 a i No

ii No

iii No

iv Yes

b i No

ii Yes

iii No

iv Yes

c i Yes

ii No

iii Yes

iv Yes

4 3

ii y ≤ 2

b i x = −3

ii x ≥ −3

c i y = −3x − 5

ii y < −3x − 5

−3 3

d i y = 4x + 7

ii y < 4x + 7

−4 4

e i y = −2x

ii y > −2x

f i

ii b y < −2x − 6

c y ≤ −3

d y > −5x + 6

5 4 3 2 1

d i No

iii No

9 8 7 6 5 4 3 2 1

iv No

−9 −8 −7 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4

1 2 3 4

ii No iii

9 8 7 6 5 4 3 2 1

x 1 2 3 4 5

−7 7 −6−5 −8 8− −6 −4 −3 −2 −1 −9 − 9− −1 −2 −3 −4

b i

4 3 2 1

b y=3

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

ii Yes

13 a i

1 2 3 4 5

11 a x = 4 5 4 3 2 1

−2 2

−5 −4 −3 −2 −1 −1 −2 −3 −4 −5

−5 − 5− −4 4− −11 −3 3− −2 2− −11 − − 2 −2 −3 3 −4 − 4 −5 −5

e Yes

9 a y ≥ 4x – 5

5 4 3 2 1

−4 − 4 − −11 −3 3 − −2 2 − −11

8 a i y=2

10 a

x 1 2 3 4 5

−4 −3 −2 −1 −1 −2 −3 −4 −5 −6

x 1 2 3 4

x 1

2 3 4

ii Yes

d 3x + 4y < 12 or 3x + 4y ≤ 12

Answers mathspace.co

705

iii

4 3 2 1

6 5 4 3 2 1 −4−3−2−1 −1 −2 −3 −4

−4−3−2−1 − 4−3−2− −1 −11 − −2 − 2 −3 − 3 −4 − 4

14 a

x 1 2 3 4 5 6 7 8 9

9 8 7 6 5 4 3 2 1 −4 −3 −2 −1−1

x 1

−2 −3 −4

ii No

706

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 4 3 2 1

−4 − 4 − −11 −3 3 − −2 2 − −11 − −2 2 −3 3 −4 4 −5 − 5 −6 − 6 −7 −

x 1 2 3 4 5 6 7 8 9

5 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 4 −5 5 −6 6 −7 7

c d i

−2 −3 −4

b 6 5 4 3 2 1

−4 −3 −2 −1−1

ii Yes iii

9 8 7 6 5 4 3 2 1

−4 − 4 − −11 3 − 2 − −3 −2 −11 −2 2 − 3 −3 − 4 −4 −5 − 5 −6

c i

iii

2 1 −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 8 −9 9 −10 − 10 −11 − 11

y x

7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

2 1 −4 −3 −2 −1−1

−2 −3 −4 −5 −6 −7 −8 8 −9 9 −10 − 10 −11 − 11

j y

−4 − 4 − −11 −3 3 − −2 2 − −11 − −2 2 −3 3 −4 −5 −6 −7

y x 2

5 4 3 2 1

7 6 5 4 3 2 1

9 y 8 7 6 5 4 3 2 1 −4 − 4 − −3 3 − −2 2 − −11− −11 −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

−4 −3 − −2 2 − −11 −11 − −2 − 2 −3 − 3

−4 −3 −2 − −11 −11 − −2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6 −7 − 7

−4 −3 −2 −1 −1 −2 −3 −4 −5 5 − −6 6 − 7 −7

−4 −3 −2 −1 −1 −2 −3 −4 −5

i e

7 6 5 4 3 2 1

11 y 10 9 8 7 6 5 4 3 2 1 −4 − 4 − −3 3 − −2 2 − −11− −11 −2 − 2 −3 − 3

Answers mathspace.co

707

15 a i 2

b i

ii y = 2x − 3

ii y = 4x − 4

iii y ≤ 2x − 3

iii y ≥ 4x − 4

16 a y = 4

b y=3

17 a y = 2

c y = 13

22 a By substituting the values of 12 tool sets and none of the plant bundles into the inequality, we get 12(12) + 21(0) = 144 + 0 = 144, which is less than $150 leaving $6. Thus, the club can’t buy more than 12 tool sets. Similarly, by substituting the values of 0 tool sets and 7 of the plant bundles into the inequality, we get 12(0) + 21(7) = 0 + 147 = 147, which is less than $150 leaving $3. Thus, the club can’t buy more than 7 plants.

18 a b

Therefore, the constraints of the budget make it impossible to exceed 12 tool sets or 7 plant bundles.

12 10 8

b Substituting 5 tool sets into the inequality, we have 12 (5) + 21p ≤ 150, Simplifying, we get 60 + 21p ≤ 150, and thus 21p ≤ 90. Solving for

6 4

, which is approximately p, we ﬁnd that 4.28. Since we can’t purchase a fraction of a bundle, the club can buy at most 4 plant bundles.

2 x 2

c i No

ii Yes

iii No

iv No

c Let’s say the club buys the same number of tool sets and plant bundles, i.e., t = p.

19 a b

Substituting this into the inequality, we get 12t + 21t ≤ 150, or 33t ≤ 150. Solving for t, we

. As we can’t purchase get fractions of tools or plants, the club can buy at most 4 tool sets and 4 plant bundles.

42 36 30

d If the budget is raised to $180, and they decide to buy an equal number of tool sets and plant bundles, the inequality becomes 12t + 21t ≤ 180, or 33t ≤ 180. Solving for t,

24 18 12 6

x 5

10 15 20 25 30 35

c i No

ii No

iii No

. Since we can’t buy we get fractions of tools or plants, the club can afford at most 5 tool sets and 5 plant bundles.

iv Yes

d 30 books Reasoning 20 a The inequality 4x + 5y ≥ 20 includes the line 4x + 5y = 20 as part of the solution set whereas 4x + 5y > 20 does not. In the graph, this will change the line from being dashed (exclusive) to being solid (inclusive).

Problem-solving 23 x + y ≥ 43 24

y 70 60 50 40

b The region 4x + 5y ≤ 20 is on the other side of the line from the region 4x + 5y ≥ 20. 21 a No, the inequality x + y > 10 cannot have integer solutions where both x and y are less than 3. If x and y were both 3, the sum would be 6, which is less than 10. b Yes, for example x = 2 and y = −9.

708

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

30 20 10 −40−30−20 −10 −10

x + y ≥ 50

x 10 20 30 40

25 19x + 9y ≤ 855 26

29 a 2x + 5y ≤ 30 b

19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 −8 − 8 − −6 6 − −4 4 − −2 2− −11 −2 − 2 −3 − 3

15 12 9 6 3 x x 2

c The x-intercept of 15 represents the greatest number of games of Table Tennis that Oprah could play if she decides not to play any games of Frog Days. d The y-intercept of 6 represents the greatest number of games of Frog Days that Oprah could play if she decides not to play any games of Table Tennis.

27 a b

y 18

y 45 40

30 a y ≥ 30 − 0.75x or 30x + 40y ≥ 1200

35 25

b She earns $30 per hour as a tutor and $40 per hour as a sales associate.

c She needs $1200 to cover expenses.

Chapter 3 review

10 5

1 a

5 10 15 20 25 30 35 40 45 50 55

c B 28 a 6x + 4y ≤ 48 b 14 y 13 12 11 10 9 8 7 6 5 4 3 2 1

−1

16 14 12 10 8 6 4 2

−4 −3 −2 −1 −2 −4

c There are 61 different ways, one for each integer coordinate inside the solution set of the inequality.

b (0, 1)

2 a x 1 2 3 4 5 6 7 8 9 10 11 12 13 14

3 2 1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Answers mathspace.co

709

7 6 5 4 3 2 1

−4 −3 −2 −1

130 C 120 110 100 90 80 70 60 50 40 30 20 10

x 5 10 15 20 25 30 35 40

c $3.20

−8 −6 −4 −2 −2

−4

d $76.80

e 14 kg

8 a

9 a

−6 −8

5 a

120 110 100 90 80 70 60 50 40 30 20 10

10 a

d b r = 2.52 cm

11 a M 1

b No 6 a 3°C

c y = 3x + 5

b 5°C

d 28 y 26 24 22 20 18 16 14 12 10 8 6 4 2

12 a No

b Yes

c No

d Yes

13 a No

b No

c Yes

d Yes

14 a 3 m

b 18 m

15 a No

b Yes

16 a y > 0.2x

3x + 4y ≤ −12 17 a (5, 0) c

x 1

7 a Number of litres (x)

20 30 40

Cost of rice (C)

64 96 128

b (0, 3) 6 5 4 3 2 1

−3 −2 −1 −1 −2 −3 −4 −5

x 1 2 3 4 5 6 7 8

d 3x + 5y < 15 or 3x + 5y ≤ 15

710

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

18 a

Fluency

6 a Concave down

b Concave up

c Concave down

d Concave up

1 1

b k = ±5

c x = ±11

e x = ±11

f y = ±5

g x = ±5

i k = ±2

k x = ±15

7 a

−4 − 4 − −3 3 − −2 2 − −11 − −11

−2 − 2 − 3 −3 −4 − 4

8 a z = 4, −10 b i No

ii Yes

iii No

iv No

c Yes 19 a

7 6 5 4 3 2 1

−4 −3 −2 −1 −1 −2 −3 −4 −5

5 4 3 2 1

−4 − 4 − −3 3 − −2 2 − −11 −11 − −2 −2 −3 −3 −4 − 4 −5 − 5 −6 −7

b y = −4x − 3

h p = ±10 b

c m = 0, −5

d m = 0, −5

h e = 4, 2

i s = −6, −7

10 We can ﬁnd the solutions for x by taking the positive and negative square roots of k, that is, . The equation can be rewritten as x2 = 9. As both 32 and (−3)2 are equal to 9, there are two solutions. 12 Answer may look like:

21 14x + 18y ≥ 210

4.01 Quadratic equations Understanding 1 5 2 A quadratic equation can have 0, 1, or 2 solutions. 3 a a = 2, b = −7, c = 5

b a = −3, b = 8, c = −7

c a = 1, b = 0, c = −16

d a = −1, b = −4, c = 3

5 x = −12

f x = −2, 16

11 Answer may look like:

c y ≥ −4x − 3

c a = 0, b = 4

d p = 11, −5

Reasoning x

20 a −4

4 a a = 3, b = 5

c e n = −7, 11

9 a j = 0, −7

b m = 16, −2

b a = 6, b = −3

To write the equation in the form (x − a) (x − b) = 0, we can use the given solutions to ﬁnd the factors. Since x = 4 is a solution, one factor is x − 4. Similarly, since x = −3 is a solution, the other factor is x + 3. Therefore, the equation can be written as (x − 4)(x + 3) = 0. 13 Jade’s answer is incorrect. When using the null factor law, the correct equations should be x − 4 = 0 and x + 6 = 0. Solving these equations, we get x = 4 and x = −6 as the correct solutions, not x = −4 and x = 6. 14 Rearranging gives x2 = −4. There are no real numbers that will give −4 when squared.

Answers mathspace.co

711

Problem-solving

Fluency

15 a Concave down

b Maximum

5 a i H orizontal shift 3 units to the right and vertical shift 4 units down

c t = 14, t = 10

ii (3, −4)

16 k = −16

iii

17 a = 13 18 a x = 0, 13

b Minimum

c x = 6.5 cm 19 p = 2, q = −3

−1 −2 −3 −4 −5

20 a b i a = 0, c = All real numbers and a > 0, c < 0 or a < 0, c > 0

x 1

iii a > 0, c > 0 or a < 0, c < 0

iii

x −5

1 a A parabola is a ∪-shaped curve that opens either upward or downward.

−6

b i I f the coeﬃcient a is positive, the curve is ∪-shaped, opening upward. If a < 0, then the curve is ∩-shaped, opening downwards. ii Changing the constant term c in a quadratic equation shifts the entire graph vertically. If c is positive, the graph moves upward, and if c is negative, the graph moves downward. The magnitude of the shift depends on the value of c. 2 The graph is concave upward.

−7 −8 −9

c i H orizontal shift 1 unit to the left, vertical shift 4 units up and dilated (compressed) by a factor of 3. ii (−1, 4) iii

11 10 9 8 7 6 5 4 3 2

axis of symmetry

10 8

y-intercept

6 4

x-intercepts

2 −2

x 1

−5 −4 −3 −2 −1

3 Turning point 4 The x-intercepts

x 1 2 3 4

d i H orizontal shift 6 unit to the right, vertical shift 1 unit up and dilated (compressed) by a factor of 0.5.

turning point

712

y 1

−5 −4 −3 −2 −1

ii (−3, −6)

Understanding

b i R eﬂection about the x-axis, horizontal shift 3 units to the left and vertical shift 6 units down

ii a ≠ 0, c = 0

4.02 Quadratic graphs

7 6 5 4 3 2 1

ii (6, 1)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

iii

4 3 2 1

d i Concave up

ii More steep

iii (−1, 2)

iv (0, 5)

18 16 14 12 10 8 6 4 2

x 1

−1 −2 −3 −4 −5

2 3 4 5 6 7 8 9

6 a i Concave up

ii More steep

iii (1, −3)

iv (0, −1)

−4 −3 −2

−1

7 a i y = (x + 3)2 − 9

y 3

ii (−3, −9)

iii

−1 −2 −3 −4 −5 −6 −7 −8 −9

−3 −2 −1

−1 −2 −3

b i Concave down

ii Less steep

iii (2, 2)

x 1

b i y = −(x − 2)2 + 1

ii (2, 1)

iii

y 1

1 −1

−6 −5 −4 −3 −2 −1

x 1

1 2 3 4 5 6 7

−1

−2

−3

−4

c i Concave down iii (−5, −4) v

ii More steep iv (0, −79) x y

−9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12

c i y = − (x + 4)2 – 2 ii (−4, −2) iii

y x −9 −8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18

Answers mathspace.co

713

b i y=0

d i

ii x = 0, 4

iii x = 2

iv (2, 4)

ii iii

4 3

1 −2 −1

x 1

−1

−2

−3

−4

c i y = 12

ii x = 2, 6

iii x = 4

iv (4, −4)

x −3 −2

−1

3 2

8 a y=x +4

b y = (x − 1)

c y = −x2 + 6

d y = 3(x + 2)2

e y = 0.5(x + 2)2 − 3

f y = −2(x − 3)2 + 4

10 8

9 a y = (x − 2) − 6 b

6 5 4 3 2 1 −1

−1 −2 −3 −4 −5 −6

−2

x 1

−4 1

d i y = −3

ii x = 3, −1

iii x = 1

iv (1, −4)

10 a i y = 0

ii x = 0, −4 iv (−2, −4) 4

2 1 −2

−1

−3

−4

−4 −3 −2 −1 −1 −2 −3

x 1

−2

x 1

e i y=5

ii x = 5, −1

iii x = 2 v

−2 −1

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

iv (2, 9) 9 8 7 6 5 4 3 2 1

−4

714

iii x = −2 v

x 1

f i y=3

ii x = 1, 3

iii x = 2 v

d y = −0.5(x − 1) (x + 7)

iv (2, −1) 6

28 24 20

5 4

1 −1

−1

x 1

−8 −7 −6 −5 −4 −3 −2 −1

1 2

−2

11 a y = (x − 1) (x + 7)

y 2

−8 −7 −6 −5 −4 −3 −2 −1 −2 −4 −6 −8 −10 −12 −14 −16

1 2

b (2.5, 12.25)

13 a

and y = x2 − 3

and y = −x2 + 3

and y = 2x2 − 6

and y = −3x2 + 9

14 a i y = (x − 4) (x − 8), (4, 0) and (8, 0) ii y = (x − 6)2 − 4 iii (6, −4) iv

b y = −2(x − 1) (x + 7)

12 a y = −(x − 6) (x + 1)

28 24

−1 −2 −3 −4 −5

20 16 12 8 4 −8 −7 −6 −5 −4 −3 −2 −1

5 4 3 2 1

x 1

b i y = −2(x − 1) (x + 5), (1, 0) and (−5, 0)

1 2

ii y = −2(x + 2)2 + 18

c y = 3 (x − 1) (x + 7)

iii (−2, 18)

y −8 −7 −6 −5 −4 −3 −2 −1−4

−8 −12 −16 −20 −24 −28 −32 −36 −40 −44 −48 −52 −56

18 16 14 12 10 8 6 4 2 −6 −5 −4 −3 −2 −1

x 1

2 3

Answers mathspace.co

715

15 a i Concave up

ii (x − 1) (x − 3), (1, 0), (3, 0) iii y = 3

iv x = 2

v (2, −1)

−4

−3

−2

−1 −1

−1

e i Concave up

−1

ii (x + 5) (x + 1), (−5, 0), (−1, 0)

−2

iii y = 5

iv x = −3

v (−3, −4)

b i Concave up iii y = −36

iv x = 0

v (0, −36)

y −6 −4 −2 −4 −8 −12 −16 −20 −24 −28 −32 −36

x 2

6 −6 −5 −4 −3 −2 −1

iii y = 15 vi

iv x = 0

v (0, 4)

y 4 3 2 1 −2

−1

16 14 12 10 8 6 4 2

x 1

g i Concave down iii y = 24 vi

d i Concave up ii (x + 3) (x + 1), (−3, 0), (−1, 0) v (−2, −1)

iv x = 1 24 21 18 15 12 9 6 3

−4 −3 −2 −1 −3

716

v (4, −1)

ii −(x + 4) (x − 6), (−4, 0), (6, 0)

−2

iv x = −2

iv x = 4

−2

−1

iii y = 3

−1 −2 −3 −4

ii (x − 3) (x − 5), (3, 0), (5, 0)

ii (2 + x) (2 − x), (−2, 0), (2, 0) vi

f i Concave up

c i Concave down iii y = 4

7 6 5 4 3 2 1

ii (x + 6) (x − 6), (−6, 0), (6, 0)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

v (1, 25)

x 1 2 3 4 5 6

h i Concave down

ii −(x + 4) (x − 2), (−4, 0), (2, 0) iii y = 8

iv x = −1

−1 −2 −3 −4 −5 −6 −7 −8 −9

ii (2x + 3) (x + 3), (−3, 0), iii y = 9

v (1, −9) 3

10 9 8 7 6 5 4 3 2 1

−4

16 a i x = 2 iv

iv x = 1 9 8 7 6 5 4 3 2 1

−2

−1

−3

−2

−1

ii (2, −3) 4

iii y = 1

v (1, 9)

1 −4 −3 −2 −1 −1

x 1

−2 −3 −4

x 1

k i Concave up ii (x + 3) (x − 1), (−3, 0), (1, 0) iii y = −3

ii −(x + 2) (x − 4), (−2, 0), (4, 0) vi

x 2

j i Concave down iii y = 8

l i Concave up x

y 1

−3

iv x = 1 −1

−4

i i Concave up

−2

x 1

−2

ii (x + 2) (x − 4), (−2, 0), (4, 0) vi

−4 −3 −2 −1 −1

−4 −3 −2 −1

iii y = −8

v (−1, 9) y

9 8 7 6 5 4 3 2 1

iv x = −1

v (−1, −4)

b i x = −3 iv

ii (−3, −2)

iii y = 7 9 8 7 6 5 4 3 2 1

−9−8 −7 −6−5−4 −3 −2 −1 −1 −2 −3 −4

x 1 2 3 4

Answers mathspace.co

717

c i

iii y = 1

4 3 2 1 −4 −3 −2 −1−1

g i x=4 iv

x 1

−1 −2 −4 −6 −8 −10

−2 −3 −4 −5 −6 −7

d i x = −4

iii y = −11

ii (−4, 5)

5 y 4 3 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12

e i x = −3

ii (−3, −11)

2 1 −7 −6 −5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14

f i x=5 iv

4 2 −1 −2 −4 −6 −8 −10 −12 −14 −16 −18 −20 −22

718

ii (5, 2)

8 6 4 2

h i x = −1

iii y = −9

ii (4, 7) y

x 1 2 3 4 5 6 7 8 9

ii (−1, −12)

iii y = −11

y 2 −4 −3 −2 −1 −2

x 1

−4

1 2

−6 −8 −10 −12

Reasoning iii y = −2

y x 1 2 3 4

17 For all forms, to find the y-intercept, we can substitute x = 0 and evaluate fully using the order of operations. It’s easiest to find the y-intercept using the standard form y = ax2 + bx + c since when x = 0, ax2 and bx become 0. This leaves just y = c and so the y-intercept is (0, c). 18 a False. The turning point of a quadratic graph can be located at any point in the Cartesian plane, not necessarily at the origin. It depends on the values of the coefficients in the quadratic equation.

iii y = −23

y x 1 2 3 4 5 6 7 8 9 10 11

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b False. If the coeﬃcient of the x2 term is negative, the graph will open downwards. The sign of the coeﬃcient determines the direction of the graph. c True. All quadratic graphs have a line of symmetry that passes through the turning point. The line of symmetry divides the graph into two symmetric halves. d True. The constant term in the quadratic equation represents the y-intercept of the graph. It is the value of y when x equals zero.

e True. The roots of a quadratic equation are the values of x where the graph intersects the x-axis. They can be found by locating the x-intercepts of the quadratic graph. 19 a a > 0 means the curve will be concave up and so has a minimum turning point, k < 0 means the turning point will be below the x-axis, so the curve will cut the x-axis at two points. h can take any value. b a < 0 means the curve will be concave down and so has a maximum turning point, k = 0 means the turning point will be on the x-axis, so the curve will touch the x-axis at one point. h can take any value. c a > 0 means the curve will be concave up and so has a minimum turning point, k > 0 means the turning point will be above the x-axis, so the curve will not cut the x-axis at all. h can take any value. 20 a (x − 2)2 + 5

b (x − 4)2 − 2

c The equation would become (x − (2 + a))2 + (5 + b). The turning point would move from (2, 5) to (2 + a, 5 + b) 21 a They have the same x-intercepts, both are concave up, and they share an axis of symmetry. b The y-intercept and turning points are different. 22 a y = ax2 − 2ahx + ah2 + k b y = ax2 + (−2ah)x + (ah2 + k) so b = −2ah c y = ax2 + (−2ah)x + (ah2 + k) so c = ah2 + k d Vertex form and standard form are equivalent quadratic equations in different forms. The equations give us different information more efficiently. We can find the vertex, (h, k), efficiently using vertex form. We can find the y-intercept, c, efficiently by plugging in 0 for x and evaluating the corresponding output using standard form. We can also find the

where a and b are the coeﬃcients of the x2 and x terms, respectively. 24 a If two quadratic curves have the same intercepts, it means they both pass through the same points on the x-axis and y-axis. If these are three different points, then the two quadratics are the same. If they have less than three intercepts, however, there can still be differences in their shapes, such as the steepness or direction of their parabolas. For example, the curves represented by y = x2 and y = 2x2 both have a single intercept at (0, 0), but the latter is a steeper parabola. b If three quadratic curves have the same turning point, it means they all have the same vertex (the highest or lowest point of the parabola). Additionally, if they share one common intercept, it means they all pass through the same point on either the x-axis or y-axis. If the shared intercept is different to the turning point, then the three quadratics are the same. If the shared intercept is also the turning point, however, then there can still be differences in their shapes. For example, the curves represented by y = (x − 1)2, y = 2(x − 1)2 and y = −1(x − 1)2 share a single x-intercept of (1, 0) which is also their turning point, but they have different steepness and directions of their parabolas. They also have different y-intercepts. Problem-solving 25 a

9 16 21 24 25 24 21 16 9

and use it to ﬁnd the y-value of the vertex.

23 The turning point of a quadratic graph is the point where the graph reaches its minimum or maximum value. The x-coordinate of the turning point can be found using the formula

9 10 0

In both forms, the leading coeﬃcient, a, is the vertical stretch/compression factor and tells us if the parabola will open up or down.

, in standard form

axis of symmetry,

10 5 t 1

c 24.75 m

2 3 4 5 6 7 8 9

d 25 m

Answers mathspace.co

719

26 a

4.03 Solve by factorisation

500 450 400 350 300 250 200 150 100 50

Understanding 1 A and D 2 a a = 3, b = 2 or a = 2, b = 3 b a = −2, b = −9 or a = −9, b = −2 c a = 4, b = 4 d a = −3, b = −3

s 5 10 15 20 25 30 35 40

b s = 19 27 a 6 m

e a = 1, b = −2 or a = −2, b = 1

c 524

f a = −3, b = −4 or a = −4, b = −3

c x = −4

b 3m

g a = 3, b = −3 or a = −3, b = 3

h a = 5, b = −3 or a = −3, b = 5

3 a 24

b 11

c 8 and 3

d x2 + 11x + 24 = (x + 8) (x + 3)

4 (m + 5) (m + 9)

5 a x = −2, 3

b x = −4, 1

c x = −3, 3

d x = −1, 2

x −5

−4

−3

−2

Fluency

−1

28 29 a Elise should expect to pay $900 b The price would decrease by $393.75 c

1000 900 800 700 600 500 400 300 200 100

cost ( y)

b m = 0, m = 8

c c = 0, c =

d y = 6, y = −11

e k = 3, k = 5

g x = 5, x = −5

7 a x = ±8

metre (x) 2

d The decrease in price per square metre makes the parabola wider. The x and y-intercept, domain and range, turning point, axis of symmetry, end behaviour, and intervals of the domain for where the curve is positive/negative and increasing/decreasing stay the same. 30 Since the turning point lies on the axis of symmetry, we know that 8 is the average of 12 and m. So then, we must have m = 4. If we substitute the coordinates of the turning point into the equation, we get: 15 = a(8 − 4)(8 − 12) If we solve this equation we get

720

6 a x = 0, x = 9

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b x = ±3

c x = 0, −12

d x = 0, 8

e x = −4, −1

f x = −5, −2

g x = 5, 3

h x = 6, −4

i x = 5, −4

j x = −5, 2

k x = 4, 10

l x=7

8 a x = 0, 16

b x = −3, 4

c x = −5, −1

d x=5

e x = −2, 9

f x = 7, 8

g x = −2, 4

h x = −12, 11

9 a m = 0, 14

b m = 5, −4

c x = 9, −9

d m = 0, −2

e m = 14, 13

f m = 0, 10

g n = −4, −2

l x = 12

10 a x = 6, −8

b x = 2, −10

c x = 10, −10

e x=8

i y = 7, 1

j y = −1, 3

11 a i x = −1, 6

ii (0, 6)

iii y = −(x − 6) (x + 1) b i x = 2, 6

ii (0, 12)

iii y = (x − 2) (x − 6) c i x=4

ii (0, −16)

iii y = −(x − 4)2 d i x = −5, 0

ii (0, 0)

iii y = x(x + 5) Reasoning

d If c is negative, the y-intercept would be below the x-axis. Since the curve is concave up, or U-shaped. It must cross the x-axis. 15 If it can be factorised, there will be one or two x-intercepts. If it factorises to the form y = a(x − p)2, then there is one x-intercept, since there is only one value (x = p), that can satisfy the equation when y = 0. If it factorises to the form y = a(x − p) (x − q), then there are two x-intercepts, since x = p or q satisfy the equation when y = 0. 16 When we have (ax + b) (cx + d) = 0, we know that zero can be produced by either (ax + b) = 0 or (cx + d) = 0. If (ax + b) (cx + d) is equal to a non-zero number, there are inﬁnite possibilities for two things multiplied together to give that number. We have to do further work to get a factorised equation equal to zero to get the correct values. Problem-solving

12 a x2 + 10x + 24 = (x + 4) (x + 6) b k = 25, 14, 11, 10, −10, −11, −14, −25 c x = (−24, −1), (−12, −2), (−8, −3), (−6, −4), (4, 6), (3, 8), (2, 12), (1, 24) 13 a When we factorise as (x − a) (x − b), the a and the b must be a factorisation of 28. So, as a pair we could have 28, 1, 14, 2, 7, 4. k is the sum of these factors, so the highest value is 28 + 1 = 29. b Using the value of k found in part (a) gives x2 + 29x + 28 = 0, with solutions x = −28, −1. The difference between the solutions is 27. 14 a To find the integer values of c that allow the expression to be factorised, we need to find all pairs of factors of c that have a sum of 4 (the coefficient of the x term). The possible pairs are (1, 3) and (2, 2), which correspond to the values of c = 1 × 3 = 3 and c = 2 × 2 = 4. 2

b If c = 4 then the equation is x + 4x + 4 = 0, leading to x = −2, −2. If c = 3 then the equation is x2 + 4x + 3 = 0, leading to x = −1, −3. c If c is negative, then we could use two integers, one positive and one negative, such that the sum of the two numbers is 4 to factor x2 + 4x + c. For example, 5 and −1, 6 and −2, 7 and −3 and so on. There are inﬁnite possibilities.

17 a x = 4, −7

b x = 8, −3

c x = 7, −4

d x = 9, −5

h x = −7, 3

18 x = ±4 19 m = 16, n = 6 20 a t = 3, 9

b 3 hours

c 6 hours

21 t = 5 s 22 a (x − 3) (x + 3) = y

6 4 2 −8 −6 −4 −2 −2

x 2

−4 −6 −8

Answers mathspace.co

721

b (3x + 5) (x − 2) = y

4.04 Solve by completing the square

4 y 3 2 1 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11

Understanding x 1

18 16 14 12 10 8 6 4 2

2 a 25

d Error b 16

3 9 4 Fluency

5 a (x − 1)2 + 45 2

x 2

d −(x + 2) (x − 7) = y

y 20

b (x + 1)2 + 1

c (x − 7) + 7

d (x − 3)2 − 4

h (x − 1)2 + 3

6 a x = −4, x = −8

c x = 1, x = 5

7 a x = −2, x = −16

c x = 8, x = 1

8 a x = 8, x = −2

5 x −8 −6 −4 −2

b x = 4, x = 2

9 a

−5

e 3(x + 10) (x + 1) = y

b Correct

c Correct

c (5x − 4)2 = y

1 a Error

−20 −30 −40

20 10 −12 −10 −8 −6 −4 −2 −10

x 2

Reasoning 10 Halve the coeﬃcient of x to get

and then

−50

square it to get

which would need to be

added to x2 + bx to make it a perfect square trinomial.

722

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

11 We can use square roots to solve any quadratic equation of the form a (x − h)2 = k. We can solve these equations by taking the square root of both sides. Completing the square is a method of solving that allows us to write a quadratic equation as a perfect square, and we can then use the square roots to solve. 12 Some quadratics are not factorable, and if the solutions are not integers, and especially if they are irrational, then drawing the graph of the equation will not give exact solution, but only an approximation. For example: has solutions of

Chapter 4 review 1 a Concave up

b Concave down

c Concave up

d Concave down

2 a x = ±9

ii (4, 0) iv

of Line 2, he should have added , but he has used subtraction sign before the expression. Going from Line 3 to Line 4 he has factorised x2 − 4x − 4 as (x − 4)2, but (x − 4)2 = x2 − 8x + 16. Given

Complete the square

Simplify fractions and exponents

Factorise the perfect square part

Evaluate and simplify

Collect like terms

Take the square root of both sides

Add 4 to both sides

Evaluate 4 + 1 and 4 − 1

Problem-solving

b t = 6, t = 18

4 a i D ilated by a factor of 2 and horizontal shift 4 units to the right

13 Tyler has made two mistakes. On the left hand

d x = 0, 8

3 a Maximum

Corrected solution:

b x = ±10

c n = −4, 8

11 10 9 8 7 6 5 4 3 2 1

iii (0, 32)

x 1 2 3 4 5 6 7 8 9 10 11

b i R eﬂection about the x-axis, horizontal shift 1 unit to the left and vertical shift 7 units up. ii (−1, 7)

iii (0, 6)

1 2 3 4 5 6 7 8 9

b (1, 0), (−3, 0)

c x=1

15 a 100 m

−9−8−7−6−5−4−3−2 −1 −1 −2 −3 −4

5 a (0, −3)

14 x = 12

11 10 9 8 7 6 5 4 3 2 1

d (−1, −4) 4

16 The turning point form of the equation is . This means the turning point is at (9, 27), and the parabola has end points at (0, 54) and (18, 54), so the width is 18 centimetres, and the height is 27 centimetres. 17 a Two solutions

b Two solutions

c Zero solutions

d One solution

2 1 −5 −4 −3 −2 −1 −1

x 1

−2 −3 −4

Answers mathspace.co

723

6 a y = (x + 2)2 − 5

b (−2, −5)

4 3 2 1

b i

10 8 6 4 2

−6 −5 −4 −3 −2 −1

−1 −2 −3 −4 −5

−10−8 −6 −4 −2 −2 −4 −6 −8 −10

y y=x+9

x 2 4 6 8 10

y = −x − 9

ii (−9, 0) 7 a x = 12, 0

b x = 1, −10

c x = 12

c i

8 a i x = −2, 4

y = −x + 3

ii y = −2(x + 2) (x − 4)

b i x = 0, 7

ii y = x (x − 7)

9 a x = 4, x = 10

b x=3

x 2 4 6 8 10

b x = 12 ± 56 m

10 a 144 m

ii (0, 3)

5.01 Solve graphically

d i

10 8 6 4 2

Understanding 1 The x- and y-coordinates at the point of intersection satisfy both equations.

3 a 1

b 0

e 0

f 1

c 1

d Inﬁnite

c (2, −4)

d (3, −6)

Fluency 4 a (5, −1)

b (−6, 4)

e (6, −3)

f (−4, −2)

5 a x = 6, y = 4

ii (4, 4) e i

b x = 0, y = −2

c x = 4, y = 0 10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10

y y = 4x − 4 y = 8 − 2x

x 2 4 6 8 10

x −10−8 −6 −4 −2 2 4 6 8 10 −2 −4x + 2y = −8 2x − 4y = −8 −4 −6 −8 −10

2 The lines would be parallel and never intersect.

y 10 8 y = 3x + 6 6 4 2 y = −x + 2 x −10−8 −6 −4 −2 2 4 6 8 10 −2 −4 −6 −8 −10

ii (−1, 3)

ii (2, 4)

724

−10−8 −6 −4 −2 −2 −4 −6 y=x+3 −8 −10

6 a i

10 8 6 4 2

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

f i

j i

10 y = −x − 3 8 6 4 2

x 2 4 6 8 10

−10−8 −6 −4 −2 −2 y = −1 −4 −6 −8 −10

y 6

y = 4x − 3

−4 −2

2 −2

k i

y = 4 − 3x

y y=

3 x−1 2

−4

−6

−6 −4 −2 2 4 6 −2 y = 1 x − 2 2 −4

ii (1, 1) h i

−6

y 8

−8

y = 3x + 2

ii (-1, -2.5)

y = 7 − 2x

2 −4 −2

l i

x 2

−2

−4

y = 4 − 0.5x 10 8 6 4 2 −10−8 −6 −4 −2 −2 −4 −6 −8 −10 y = 1.5x

−10−8 −6 −4 −2 −2

ii (1, 5)

ii (2, 3)

2 4 6 8 10

ii (4, 6)

i i

−10−8 −6 −4 −2 −2 −4 −6 −8 y = 2.5x − 4 −10

ii (−2, −1) g i

y 16 14 12 10 8 6 4 2

y = 12 − 1.5x

y = 1 x+3 2 2

2 4 6 8 10

−4

y = 5 x − 1 −6 2 2 −8 x 2 4 6 8 10

ii (1, 2) 7 a

500 400 300 y = 50x + 250 200 100 −5 −4 −3 −2 −1 −100 −200 −300 −400 −500

y = 100x + 50 x 1 2 3 4 5

b (4, 450)

Answers mathspace.co

725

c Renting either studio A or B for 4 hours costs the same amount, $450. 8 a

24 22 20 18 16 14 12 10 8 6 4 2

c No

d Yes

13 a No

b No

c Yes

d Yes

Reasoning

y = 2x + 10

14 No, it’s not possible. Simultaneous linear equations represent two lines. Two lines can either intersect at a single point (one solution), coincide resulting in inﬁnite solutions, or be parallel (no solution). They can’t intersect at exactly two points.

y = 4x

15 These equations actually represent the same line, just written in different forms. This means there are inﬁnite solutions. Every point on one line satisﬁes both equations.

b (5, 20) c Non-members and members pay the same amount for 5 classes, $20.

16 No, because there are an inﬁnite number of lines that go through that point. The gradient could be any real number.

9 a (775, 2325) b After making 775 items, the company’s revenue will equal their costs. This means they will not make a proﬁt and this is the break-even point. 1100 1000 900 800 700 600 500 400 300 200 100

17 a To intersect at (0, 3) the line must have a y-intercept of 3, therefore c = 3. b The value of m ≠ 5. If m = 5 and c = 3 then the second line is the same as the ﬁrst and there is an inﬁnite number of solutions.

Dollars

18 No, the lines have the same gradient, so they are parallel and will never intersect. C

19 Non-integer solutions represent fractional quantities or parts of a product, which may not be practical in real-life scenarios. In such cases, the company might need to round to the nearest whole numbers. This may not result in the ideal way of distributing the goods, but it ensures that the goods are distributed in whole units, which is more practical.

n 50

100 150 200 250

b (200, 800) c This is the break-even point. For 200 toys, both the cost and revenue are $800. 11 a

b No

1 2 3 4 5 6 7 8 9 10 11

10 a

12 a Yes

30000 27000 24000 21000 18000 15000 12000 9000 6000 3000

Problem-solving

20 a y = x + 3, 2x + 2y = 18

y = 10 000 + 772.60x

8 6 4 2

y = 1200x x 2 4 6 8 10 12 14 16 18 20 22 24 26 28

b (23.4, 28 076.7) c After about 23 weeks, Laura will be spending more money than she has in her bank account.

726

2x + 2y = 18 10

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

−8 −6 −4 −2 −2 −4 −6 −8

c 6 m and 3 m

y y=x+3

x 2 4 6 8 10

21 a Let x = Mohamad’s age and y = Rochelle’s age

5.02 Simultaneous inequalities

Equation 1: y + x = 21 Understanding

Equation 2: y = 4x + 11 b

22 20 18 16 14 12 10 8 6 4 2

1 a No

y = 4x + 11

c No

d No

2 (1, 3), (1, 0), (4, 0)

y + x = 21

3 C 4 a False y 2 4 6 8 10 12 14 16 18 20 22

c Rochelle is 19 years old and Mohamad is 2 years old. 22 a Let x = the number of parts Elvia completes and y = the number of parts Kang completes.

−2 −1 −1 −2 −3 −4

c True

5 a x < 1, y ≤ 3 b y ≥ 2x − 3, y ≤ 1 c x < 1 and y < 1 and y ≥ −x − 5 d y + 3x > 0, y + 3x ≤ 3 6 a

4 3

Equation 2: y = 2x − 6 10 y 9 8 7 6 5 4 3 2 1

b True

Fluency

Equation 1: x + y = 9 b

b Yes

e No

2 1 6x + 3y = 0

−3 −2 −1 −1

x+y=9

1 2 3 4 5 6 7

−2

y = 2x − 6

2x − y = 4

−3 −4 x

1 2 3 4 5 6 7 8 9 10

b Yes c

7 6 5 4 3 2 1

c Kang did 4 parts of the assignment and Elvia did 5 parts. 23 Answers will vary. For example: Let x be the number of days and y be the weight of the rice in kilograms. Maria has 15 kgs of rice and cooks 5 kgs of rice every 7 days as represented by the purple line in the graph. Arjun has 12 kgs of rice and cooks 1 kg of rice every 2 days as represented by the green line in the graph. The solution (14, 5) means that after 14 days, Maria and Arjun have the same amount of rice remaining. 24 The point of intersection on the graph has negative x and y-values. In terms of the height of a plant over time, only positive values make sense. Since the two lines do not intersect in the first quadrant, they will never reach the same height on the same day.

−4 −3 −2 −1 −1

1 2 3 4 5 6 7

−2 −3 −4

7 a i 5 iii

ii 3 6 5 4 3 2 1 −4−3−2 − 4 −3 −2 − −1 11 −1 −

x 1 2 3 4 5 6 7 8

−2 − 2 −3 − 3 −4 − 4 −5 − 5 −6 − 6

Answers mathspace.co

727

ii −1.857

b i 0.714 iii

7 y 6 5 4 3 2 1

6 5 4 3 2 1

−7−6−5−4−3−2−1 5− −4−3−2 −− 11 −1 −2 − 2 −3 − 3 −4 − 4 −5 −6 −7

−6−5−4−3−2 −1 −1

1 2 3 4 5 6

−2 −3 −4 −5 −6

d 25 units2 9 a

c i No point of intersection

ii No point of intersection iii

2 1 −4 −3 −2 −1 −1

iii

6 5 4 3 2 1

10 a x ≥ 17, x + y ≥ 37 b

x 1

36 32 28 24 20 16 12 8 4

x 4

c i No

8 12 16 20 24 28 32 36

ii Yes

iii No

11 a x ≥ 10, y ≥ 7, x + y ≤ 23

x 2

−4 y=x−5

b Yes

y 22 20 18 16 14 12 10 8 6 4 2

x 2 4 6 8 10 12 14 16 18 2022

c 9

728

iv Yes

d 13

−8

−3

−5 −4 −3 −2 −1−1 −2 −3 −4 −5 −6 −7

−6

−4

ii 0

y=0 −8 −6 −4 −2 −2

b 6 units2

d i 2

y = −x − 5

x 1

−2 x

−4 −3 −2 −1 −1 −2 −3 −4

7 6 5 4 3 2 1

8 a

1 2 3 4 5 6 7

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

12 a x ≥ 0, y ≥ 0, 100x + 150y ≤ 3900 and y ≤ 32 − x b

17 a x ≥ 0, x ≤ 0.35 d

30 25

1 0.9 0.8

0.7

0.6 0.5

0.4

0.3

x 5

0.2

10 15 20 25 30 35

0.1

x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

c B, D 13 a 0 ≤ 5x + 8y ≤ 40

b 0 ≤ 10x + 4y ≤ 40

14 a 3x + 2y ≤ 18

b 4.5x + 1.5y ≤ 18

c x ≥ 0, y ≥ 0 d

12 y 11 10 9 8 7 6 5 4 3 2 1

e i Yes

ii No

iii Yes

iv No

f Yes, because this would satisfy all of the inequalities and the point (0, 1) lies within the region. 18 a 2x + 4y ≤ 12

b 3x + 1.5y ≤ 12

c No, x and y cannot have negative values as they are measuring number of pallets made in a day, of which the smallest value would be 0. This gives the inequalities x ≥ 0, y ≥ 0. d x 1

15 a 4x + 2y ≤ 18

b 6x + 1.5y ≤ 18

c x ≥ 0, y ≥ 0 d

b y ≥ 0.2

c x+y≤1

7 6 5 4 3 2

12 y

11 10 9 8 7 6 5 4 3 2 1

x 1

Reasoning 16 a No solutions. The regions are on opposite sides of the line y = 4. b Infinitely many solutions. The regions have the infinite set of points on the line x = −2 in common. c No solutions. The regions are on opposite sides of the boundary line. d Infinitely many solutions. The regions have the infinite set of points on the boundary line in common.

19 Yes, it is possible for a coordinate point to solve one part of the linear inequality pair but not the other. In the diagram, the dashed lines represent the two inequalities y = x + 2 and y = 3x + 2. The darker shaded region is the intersection of the regions that satisfy the inequalities y < x + 2 and y < 3x + 2. There are points like (−2, −2) that solves y < x + 2 but not y < 3x + 2. 5 4 3 2 1 −5 −4 −3 3− −2 2− −11 −11 −2 − 2 −3 − 3 −4 − 4 −5 − 5

x 1 2 3 4 5

Answers mathspace.co

729

20 a x ≥ 0, y ≥ 0

b x + y ≤ 200

25 a x ≥ 5, y ≥ x + 8, x + y ≤ 26, and x + y ≥ 20

c The number of ﬂuorescent light bulbs produced must be at least half as many as the number of incandescent light bulbs.

21 18 15

21 a y ≥ 35, y ≤ 50

12 9

55 50 45 40 35 30 25 20 15 10 5

6 3

c (5, 15), (5, 21), (6, 14), (9, 17) d 9 sessions 26 a x ≥ 5, y ≥ 0, y ≤ 2x, x + y ≤ 18, and x + y ≥ 13 b

x 2

c Yes

d False

14 12

22 a x ≥ 0, y ≥ 0, y ≤ 12 − x

6 4

x 2 4 6 8 10 12 14 16 18

c (5, 8), (5, 10), (6, 12), (18, 0), (13, 0)

4 2

d 12 lessons

x 2

c i Yes

ii No

5.03 Solve using substitution

iii No

iv Yes

Understanding

d 6 cm × 6 cm

1 a Answer varies:

23 a x ≥ 0, y ≥ 0, x + y ≤ 29, 15x + 18y ≥ 270 b

x 3 6 9 12 15 18 21 24 27

Problem-solving b

y 30 25

Tomi will get y = 2(6 − y) − 3, which simplifies to y = 12 − 2y − 3, then finally 3y = 9. This approach is efficient as it results in a simple equation where we can solve for y directly. b Answer varies:

20 15 10 5

x 5

20 25 30

Carla will get x + 2x − 3 = 6, which simplifies to 3x − 3 = 6, then finally 3x = 9. This approach is also efficient because it results in a simple equation where we can solve for x directly. c Answer varies:

c 10 hours 24 a x ≥ 0, y ≥ 0

d Yes b 55x + 65y ≤ 800

c 2x + 7y ≤ 200

730

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Regie will get 6 − x = 2x − 3, which simpliﬁes to 3x = 9. This approach is eﬃcient because it results in a simple equation where we can solve for x directly.

d Answer varies:

Reasoning

Junard will get

, which simpliﬁes

to y + 3 + 2y = 12, then finally 3y = 9. This approach is slightly less efficient because the equation involves a fractional coefficient, which can be a bit harder to work with. Once the fraction is handled, the resulting equation is also simple. 2 Yes, it is a solution. 3 a True

b True

c False

d True

e False 4 No, it is not a solution. Fluency 5 a x = −8, y = 8

b x = −5, y = 3

c x = −8, y = −3

d x = −6, y = −3

e x = −3, y = 5

f x = 6, y = 4

g x = 1, y = 6

h x = 8, y = 4

i x = −5, y = −8

j x = 8, y = 1

l x = 12, y = 8

12 If the process of substitution leads to a false statement, it indicates that the simultaneous equations have no solutions. In other words, the lines represented by the equations do not intersect at any point. 13 If you get a true statement, it indicates that the simultaneous equations have inﬁnitely many solutions. This could mean that the equations represent the same line. 14 Example explanation: In this case, it may be easier to make y the subject in the ﬁrst equation and then substitute it into the second equation. This would leave us with an equation in terms of x, so we could then solve for x. We would then substitute the value of x and solve for y using either equation. Finally, we could substitute the values of x and y into both original equations to check our answer. This would avoid fractions and unnecessary rearranging. 15 No. If we substitute one equation into the other, we do not end up with a true statement: 3(x + 5) = 3x + 5 3x + 15 = 3x + 5 15 ≠ 5

6 a b Inﬁnitely many solutions 7 a x = 7, y = 4

b x = −6, y = 5

c x = 6, y = −5

d x = 2, y = 3

e f = 0.4, g = 1.2

f a = 1.5, b = 2.5

8 The son’s current age is 4 and the mother’s current age is 40. 9 a a = $1.68 and k = $2.24 b The cost of one apple is $1.68 and the cost of one kiwi fruit is $2.24. 10 a x = 250 000 and y = 810 000 b In 2008, there were 250 000 jobs created and in 2013, there were 810 000 jobs created. 11 a x = 8, y = 14 b In the group, there are 8 adults and 14 children.

Since this is always false, the two equations would not have any solutions. 16 a x = 4, y = 2 b Yes c Yes, this is the only solution to the simultaneous equations. The solution (x = 4, y = 2) is the point of intersection for all three lines represented by these equations. As these are linear equations, and each pair of them intersects at this point, there can’t be another solution for all three equations simultaneously. This means the simultaneous equations has exactly one solution. d

3 2 1 −2 −1 −1

x 1 2 3 4 5 6 7 8 9

−2 −3 −4

Answers mathspace.co

731

Problem-solving

8 a x = 4, y = 5

17 a The fixed charge is $5 and the charge per km is $10. b $255

19 35 sausages and 52 soft drinks.

d (−5, 3)

e (3, −2)

g (−1, 1), (2, −8)

h (1, 0),

f x = 2, y = 2

g x = 5, y = 4

h x = 6, y = 4

10 a x = $2, y = $4 b One pen costs $2 and one ruler costs $4. 11 a R = 5, C = 4

b Marjan purchased 5 rhododendrons and 4 chrysanthemums.

5.04 Solve using elimination

Reasoning

Understanding 1 C 2 Multiply the ﬁrst equation by 7 and the second by 5. 3 a True

b True

c False

Fluency b x = −3

4 a 7

5 a Multiply the ﬁrst equation by 5 and the second equation by 7. b x = −1, y = −3 6 a i 3x + 11y = −48 and −6x + 21y = −76 ii

and y = −4

b i 2x + 3y = 5 5x + 4y = 2 ii x = −2 and y = 3 c i 4x + 3y = 20 and 8x − 3y = 4 ii x = 2, y = 4 d i 6x + 5y = 240 and x + 3y = 66 ii x = 30, y = 12

12 You would add the two equations. Because the y terms have opposite coefficients, they would negate each other out, leaving you with an equation in x to solve. Once you’ve found the x-value, you can substitute it into either equation to find y. 13 If you add the two equations, both variables would be eliminated, resulting in a true statement 0 = 0. This means there are infinitely many solutions. Each original equation represents a line, and these lines coincide (i.e., are the same line), hence there are infinitely many solutions. 14 Answer may look like: If the coefficients of one variable are the same or opposite in both equations, it’s often easier to use the elimination method because you can add or subtract the equations to eliminate that variable right away. With substitution, you would first have to solve one equation for one variable, which could be more work. 15 a When Elias substituted y to solve for x, he only found one value for x because he forgot that −1 could also be a solution to x2 = 1.

b x = 2, y = 8

7 a

732

d x = 5, y = 15

e x = −3, y = 1

b The solution is not valid because it is impossible to have a negative value of a passenger cab.

b (8, 24), (−3, 24)

c (2, 14)

c x = 3, y = 1

9 a There are 17 four-passenger cabs and −2 six-passenger cabs.

18 Each racket is $50 and each ball is $5

20 a (0, 0), (−5, 25)

b x = 3, y = −1

c x = 2, y = −9

d x = 8, y = 3

e x = 40, y = −2

f x = −8, y = −2

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b (−1, −3)

b i Elimination

Problem-solving

ii One variable already has the same coeﬃcient.

16 a x + y = 30 6x − 3y = 90

c i Elimination

b x = 20, y = 10

ii The coeﬃcients of the variables are multiples of each other.

17 x = 10, y = 7 18 x = 5, y = 5 19 a 2016

d i Substitution b 68%

20 a (−1, 3) and (2, 6) b (−3, −4) and (2, 1) c (−1, 1) and (1, 1) d (−8, −6) and (8, −6) e (−4, 3) and (4, 3) f (−6, 3), (0, 3) and (−3, −6)

5.05 Applications of simultaneous equations

ii One equation is already solved for a variable. 3 a False b True c True Fluency 4 a x + y = 6 y = 2x b $2 000 000 c $4 000 000

Understanding 1 a Answers may vary. For example, one could use technology to ﬁnd the accurate intersection point. Alternatively, one could use the substitution method as the ﬁrst equation is already in the form y = mx + c. b Answers may vary. For example, one can simply read the solution from the graph. Alternatively, one could use the substitution method as both equations are already in the form y = mx + c. c Answers may vary. For example, one could use the substitution method as Equation 1 is already in the form y = mx + c. Alternatively, one could use the elimination method by multiplying Equation 1 by 2 and then adding the two equations to eliminate the y variable. d Answers may vary. For example, one could use the elimination method by multiplying Equation 2 by 3 and then adding the two equations to eliminate the y variable. Alternatively, one could use the substitution method by solving Equation 2 for x or y and then substituting it into Equation 1.

5 a x + y = 184 x − y = 22 b x = 103 c y = 81 6 a

Call time (in minutes)

Total cost for GO SMALL plan

Total cost for GO MEDIUM plan

44.00

47.00

48.00

50.00

52.00

53.00

56.00

b 40 minutes 7 a x + y = 70 5x + 10y = 385 b x = 63, y = 7 8 a T = 50n

b T = 25n + 1200

c 48 months 9 There are 10 part-time employees and 26 full-time employees. 10 a x + y = 34 x = y + 6

2 a i Substitution

b 20

ii An equation is already solved for a variable.

c 14

Answers mathspace.co

733

18 a 190 student tickets and 160 general admission tickets

11 a x + y = 4000 0.07x + 0.08y = 303 b y = $2300 12 a x = y + 12 x + y = 28 b x = 20 units, y = 8 units 13 a x = 6, y = 4

b 396 units

c 108 units 14 0.9 km

b Answers may vary.

b x = $8300, y = $7700 c Qin invested $8300 in stock A, and $7700 in stock B. 4500 4000 3500

y Machine

3000 People

2500 1500 1000 500

b x = 60

x 20 30 40 50 60

c x > 60

By choosing the substitution method, we can rewrite the ﬁrst equation as y = 38.49 − 2x and substitute this into the second equation, 5x + 3y = 99.22. This is an eﬃcient method to solve for x and once x is known, it can be substituted into either original equation to solve for y. Problem-solving

2000

By choosing the elimination method, we can multiply the equation x + y = 350 by 3 and combine it with the equation 3x + 5.5y = 1450 by means of subtraction. This is an eﬃcient method to solve for y and once y is known, it can be substituted into either original equation to solve for x. 19 a Oxtail stew costs $16.25 and sweet potato pie costs $5.99.

15 a x + y = 16 000 0.14x − 0.06y = 700

16 a

b Answers may vary.

c x = $1700

d 0 < x < 60

Reasoning 17 a C = 10 000 + 15x, R = 12x b No, Barry cannot reach a break-even point with this plan because the cost to make each meal is higher than the revenue he is earning from selling the meals.

20 This equation cannot be solved algebraically, so we can use technology to ﬁnd the approximate solution: x ≈ 0.458, y ≈ 1.373. 21 (1.29, 0.29) and (2.71, 1.71) 22 a 23 24 (−5.11, 7.30) and (22.93, 16.64)

Chapter 5 review 1 a i

c Answers may vary. Under the current plan, Barry’s restaurant would not be successful because it cannot reach a break-even point or make a profit. Every meal sold would actually increase his losses due to the negative contribution margin. To turn this situation around, Barry might consider increasing the price he charges customers for the meals or finding ways to decrease the cost of producing each meal. This would help him cover his fixed costs and eventually reach a break-even point where the restaurant becomes more sustainable.

734

3 2 1 −6 −5 −4 −3 −2 −1 −1 −2 −3 −4

ii (−4, 3)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

x 1

b i

6 5 4 3 2 1

y ≤ −2x + 7

y≤5 9 a x

−6 −5−4−3−2 −1 −1 −2 −3 −4 −5 −6

1 2 3 4 5 6 y=1

−7−6−5−4−3−2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9

ii (2, −1) 2 a x = −5, y = 2

b x = 0, y = −3

c x = −4, y = −1 3 a (1280, 2500) b After selling 1280 dresses, the boutique’s revenue from sales is equal to their expenses. This is their break-even point, implying any dresses sold beyond this quantity will lead to a proﬁt.

6 a y = x + 2, x + 2y = 22 y x + 2y = 22

x 1 2 3 4 5 6 7 y=x−4

y = −x − 4

x 1 2 3 4 5 6 7

d 25 units2

10 a 5x + 3y ≤ 24

y=x+2

b 4x + 6y ≤ 24

c x ≥ 0, y ≥ 0

2 x 2

c Base is 6 m and the other side lengths are 8 m each. 7 a −3 c

5 4 3 2 1 −7−6−5−4−3−2 4− −3 −2 − −11 −11 − −2 − 2 −3 −3 −4 4 −5 −6 −7 −8 −9

5 No, the lines have the same gradient, so they are parallel and will never intersect.

b Yes

4 D

5 4 3 2 1

−4 −3 −2 −1 −1 −2 −2 −3 −3 −4 −4 −5 −6 −7 −8 −9

7 6 5 4 3 2

b −7 3 2 1

x 1

Answers mathspace.co

735

11 a x ≥ 0, y ≥ 0, y ≤ 8 − x b

Fluency

4 a No

b No

c Yes

d Yes

e Yes

f No

g No

h Yes

5 a c=5

5 4 3 2

d c = 17

g Hypotenuse:

h Side:

x 1

c i Yes

ii No

iii No

i Side: iv Yes

6 10 m

d 4 cm × 4 cm

7 a 14.1 cm

b 11.2 mm

c 29 m

d 13.6 cm

8 a 10.8 cm

b 19.2 mm

12 a x = −10, y = 0 b Inﬁnitely many solutions c No solution

9 132 m

d f = 4, g = 9 13 a (−1, 7) and (−1, 1)

b (1, 6) and (−4, −14)

Reasoning

14 a x = −2, y = 4

10 The triangle can be a right-angled triangle since 32 + 52 = 72 (9 + 25 = 34 ≠ 49). Hence, by the Pythagoras’ theorem, it’s not a right-angled triangle.

c x = 6, y = −15

d x = −14, y = 8.5

15 a (4, 7) and (−3, 0)

b (−4, 8) and (4, 8)

16 a x + y = 10 000

∴ BD = 6 cm

y = 3x

102 = 82 + AC2

b $2500

∴ AC = 6 cm

c $7500

∴ BD = DC = 6 cm, so D is the midpoint of BC.

x + y = 20 000

17 a

12 a 152 = 225

0.10x + 0.04y = 1340

92 + 122 = 225

b x = $9000, y = $11 000

∴ 152 = 92 + 122 and Pythagoras’ theorem holds.

c Alfred invested $9000 in bond C, and $11 000 in bond D.

b i Yes

18 1.8 km

6.01 Pythagoras’ theorem

1 a z2 = x2 + y2

b x2 = z2 − y2

2 a x = 8.49

b x = 3.61

c x = 6.24

d x = 8.06 b 49

c No, not a right-angled triangle, 45 ≠ 49.

736

ii No

iii Yes

iv No

b x=

Problem-solving 13 a 25 cm

Understanding

3 a 45

102 = 82 + BD2

14 h = 5.61 m 15 a x = 4.9 m

b x = 1.8 cm

16 a $210

b $420

17 205 cm 18 Beth, by 69 m

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

6.02 Trigonometric ratios

13 1.96 m 14 h = 4.4 m

Understanding 1 a AB 2 a AC

15 θ = 9.88°

b BC b BC

c BC

e ∠ACB

d AC

16 a x = 155.43 m

b 425.86 m

17 a h = 21

b f = 22

c h = 154, x = 72

d x = 34

3 a i

iii

b i

iii

Reasoning

c i

iii

d i

iii

18 An isosceles triangle can be split into two right-angled triangles. Therefore, the adjacent side can be seen as half the length of the base.

e i

iii

f i

iii

4 a i

b i

c i

e x = 12

Using the formula: Simplify the numerator and make A the subject of the identity. Take the inverse cos of both sides. 19 Using the formula:

Fluency 5 a i

iii

b i

iii

Multiply both sides of the equation by b, and make b the subject of the equation by dividing both sides by 0.3.

Problem-solving 20 21

7 a f = 4.08

b f = 6.12

c g = 9.46

d a = 11.59

8 b=8

22 a 135°

b I = 135°

23 a 131.2 m

b 101.3°

24 31.8°

9 d = 3084 10 29.0° 11 a i x = 10

ii 53.1°

b i x=4

ii 53.1°

12 d = 1.48

Answers mathspace.co

737

6.03 Angles of elevation and depression

17 d = 1022.58 18 h = 17 m

Understanding

19 d = 10.86

1 It is an angle of elevation as it is the angle formed from the horizontal looking upwards.

20 θ = 45.97

2 a σ or ∠CBD

21 a 29.26°

b θ or ∠ACB

c They are the same size. 3 a α or ∠BCA

b 18.76°

b 1601 m

23 a PC = 22.65 m

b PA = 18.29 m

6.04 Bearings

c α

Understanding

Fluency b y = 52.77°

5 a y = 36.87°

b x = 53.13°

d 100 cm

22 a 60 m

b Draw a horizontal line from A to the left of A and the angle of depression is from the horizontal line to the line AC.

4 a x = 37.23°

c 4.61°

1 a NW

b NE

c SE

d SW

2 a 244° T

b 334° T

c 296° T

d 206° T

4 a 225° T

b 315° T

c 292.5° T d 157.5° T

e 045° T

f 135° T

3 C

6 θ = 52.52° 7 x = 182

Fluency

8 d = 229 9 a AB = 11.86 m

b AC = 16.24 m

c BC = 4.4 m 10 a AC = 9409 m

b BC = 5808 m

c 3601 m 11 a d = 7.81 m c 66.08°

b h = 13.53 m d 69.72°

13 Fiona is not correct as an angle of elevation is measured from the horizontal. 14 a The shadow gets longer as the angle of elevation gets smaller.

ii S 41° W

c i 320° T

ii N 40° W

d i 125° T

ii S 55° E

6 a 090° T

b 127° T

c 222° T

7 a 270° T

b 317° T

c 052° T b N 41°W

9 a 053° T

b N 53°E

10 a 132° T

b S 48°E

11 a 221° T

b S 41°W

12 NE 13 231° T 14 305° T

b The length of the shadow is zero.

15 a = 62°

c tan 90° is undeﬁned. The sun directly overhead means the angle of elevation is 90° and it’s not possible to draw a triangle, therefore the shadow has no length.

16 w = 90.1 km 17 a b = 53.6° 18 a x = 15.1

b S 53.6°W

c 233.6°T

b y = 13.1

19 d = 2130

Problem-solving 15 a x = 455.84 m

b 425.84 m

16 a h = 127.38

b 129.1 m

738

ii N 30° E

b i 221° T

8 a 319° T Reasoning 12 Viggo is incorrect as the angle of depression must be measured from the horizontal. The correct angle of depression is 23.9°.

5 a i 030° T

20 a h = 9.37

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b θ = 25°

Reasoning

30 a

21 a The true bearing is (360 − a)° T b The true bearing is (180 + b)° T A

c The true bearing is (180 − c)° T d The true bearing is (90 + a)° T 17.6 m

22 a Sample answer: At point A use the bearing of 45° to draw a right-angled triangle. The horizontal distance, a, is found using cos 45° =

At point B use the bearing of 60° to draw a right-angled triangle. The horizontal distance, b, is found using cos 30° =

Therefore the overall distance is a + b = 12 cos 45° + 16 cos 30°.

c 27.76 km

d 233.6°T

23 a 1131.7 m

b 294.5°T

31 68.96 km 32 a 1282 m

b 2033 m

c AB = 2403 m

Chapter 6 review b third side = 10

c hypotenuse = 2 No, to be a right-angled triangle we can check if (6, 8, 13) is a Pythagorean triple. 62 + 82 = 36 + 64

Problem-solving

= 100

24 Due west b b = 5.0

25 a 65° 26 a ∠BSP = 90°

133 = 169

c d = 11.8

d 2191.7 m

27 a i ∠BAC = 65°

ii ∠DCE = 44°

b 49.2 km c θ = 215°T b x = 7.31

28 a N a°W c a = 47°

Sum of squares of shorter sides Square of longer side, the hypotenuse

≠ 100

Compare to sum of squares of shorter sides

62 + 82 ≠ 132

Not a Pythagorean triple

b BS = 750.3 m

c SP = 525.4 m

b d = 22.3

1 a b=8

b 12 sin 45° + 16 sin 30°

29 a

38°

3 a 1083.07 m

b 1437.92 m

4 a f = 2.77

b f = 8.91

5 a

b θ = 30°

6 a 50°

b x = 19.30 m

c h = 20.50 m

7 a h = 20

b x = 26

8 θ = 54.69° 248 km

Start

9 a AC = 12 032 m

d 225 km

10 θ = 28.01°

Finish

b 227.8°T

b BC = 10 462 m

c 1570 m

c 334.9 km

11 a h = 92.97 m

b 94.6 m

12 a i 035° T

ii N 35° E

b i 218° T

ii S 38° W

13 a 319° T

b N 41° W

14 SW

Answers mathspace.co

739

15 301° T

19 103 m2

16 w = 88.9 km

20 11 998 cm2

17 Due west

21 109.78 m2

7.01 Surface area of prisms and cylinders

22 158.4 m2

Understanding

Reasoning b x2 cm2

1 a 6 faces

c SA = 6x2

2 7 faces 3 a 6 squares

23 Example answer: The formula depends on how many circular faces the cylinder has as each circular face has an area of π r2. A closed cylinder has two circular faces so the formula is SA = 2π r2 + 2π rh.

b 2 circles and a rectangle c 4 triangles and a square

A cylinder open at one end has one circular face and the formula is SA = π r2 + 2π rh.

d 6 rectangles and two hexagons

A cylinder open at both ends has no circular faces and the formula is SA = 2π rh.

Fluency 4 150 cm2 5 8 cm 6 a 288 m2

b 451.26 m2

d 392 cm2

c 96 cm

e 380 cm

f 204 cm2

7 a i 5 cm

ii 228 cm2

b i 8 cm

ii 608 cm2

8 a 216.49 cm2

b 380 cm2

c 174.49 m2

d 534 cm2

9 a

24 They are both correct as the formulas are the same. Scott’s formula is the factorised version of Lachlan’s formula. 25 a Option 2 has a smaller surface area (219.5 cm2) than Option 1 (about 245 cm2) so Quicker Oats should choose Option 2 to minimise the amount of materials used. b Maximising proﬁts means selling the smallest volume of chocolate chip oats per package. Option 2 has a smaller volume (185.25 cm3) than Option 1 (about 282.74 cm3) so to maximise proﬁts, Quicker Oats should choose Option 2 to minimise the amount of product used.

c 11.32 m2 10 a i No

ii No

iii Yes

iv No

26 a The surface area is made up of:

b 262 square units 11 a 23 184.95 cm2 2

b 452.39 cm2

c 207.35 m

d 439.82 m2

12 a 75.40 m2

b 75.40 m2

13 a 12.5664 m

b 50.27 m2

14 h = 69 15 a i x = 11.00 cm

ii 376.95 cm2

b i z = 4.90 m

ii 858.76 m2

16 a 91 m

b 50 m 2

17 a 312π cm

b 12.98 cm2

• Two rectangles for the curved walls, both with heights of 3 cm and lengths equal to the circumference of circles of radii 2 cm and 10 cm respectively. • A small circular base with a radius of 2 cm that ﬁts into the large circular base, making a complete circle with a radius of 10 cm. • A large circular base with a radius of 10 cm. b Surface Area = 854.51 cm2 27 Surface Area = 4π r2

18 596 cm2

740

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

7 Height = 5 cm

Problem-solving

8 a 785.4 cm3

28 2978.7 cm

29 a Any scale factor s which satisﬁes the equation s2(350) = a, where a is the new surface area of the box and 325 ≤ a ≤ 335. .

, and

For example,

b Any dimensions which result in a total surface area between 325 cm2 and 335 cm2. For example, 11.65 cm × 9.71 cm × 2.43 cm.

b 1335.2 cm3

c 678.6 cm

d 226.2 m3

e 45.9 cm3

f 70.7 cm3 3

g 307.9 cm

h 955.0 cm3

9 a 1696.5 cm3

b 2309.1 cm3

c 59.7 cm3 10 883.57 cm3 11 16 286.02 L

30 20.00 m2 31 a 11.13 m2

b 4 L costs $98

32 a h = 24

c 886 cm2

b h = 21 2

d 836 cm

e Bag B

12 864.19 cm3 Reasoning 13 a 1.782 m3

33 495.1 cm2

b 2 m3

34 186.32 cm2

c Oliver’s father’s rounded measurements give a volume that is 0.218 m3 larger.

35 Number of cans = 9 36 212.4 cm2

7.02 Volume of prisms and cylinders

15 a 37.31 m3 b 37.29 m3

Understanding 1 a True

14 Doubling the length and width of the right prism while keeping the height the same results in a four-fold increase in the volume.

b False

c True

d True

e False 2 The volume of both a rectangular prism and a cylinder involves ﬁnding the base area and multiplying it by the height. However the methods of ﬁnding the base area, the interpretation of the height, and the visual representation of the volume differ between the two shapes due to their distinct geometric properties. 3 Options B and D

ii 9 cm

ii 99 cm2

iii 495 cm3 5 a 520 cm3

b 114 cm3

c 64 cm3

d 231 cm3

e 840 cm3

f 90 cm3

6 Base Area = 80 cm2

e 37 310 000 mL

16 1 134 000 L 2

iii 90 cm3 b i Trapezium

d In these real-world situations, using rounded or approximated values can lead to small discrepancies in calculations. In the case of the cylindrical container, it could mean not having enough water for the experiment if a large number of cylinders needed to be ﬁlled. It is essential to understand the potential impact of approximations and make decisions accordingly.

Problem-solving

Fluency 4 a i Triangle

c The exact value is larger by 0.02 m3.

17 a Vidar should get the ﬂat base type. b Yolanda should get the ﬂat base type. 18 The second can with a diameter of 18 cm and a height of 16 cm. 19 a Morcom Aquatic Center b The pool at Morcom is 1.8 times the volume of the pool at Sanford Seminole. c

Answers mathspace.co

741

20 a 1335.18 cm3

b 4005.54 cm3

Fluency

c 28 L

5 a 672 cm3

b 171 cm3 3

21 37.5 L

d 3136 cm3

c 1920 cm

22 a 35 343 L 23 a 140 m2

b 35.343 kL

6 a 1608.50 cm3

b 883.57 cm3

d 609.47 cm3

c 263.89 cm

b 8820 L

7 0.57 m3

24 90 m3

8 a 1140 m3

25 67.5 L

b 424 m3 3

26 4.35 m3

d 272 cm3

e 982.35 cm3

f 601.65 m3

27 a 176 m

b 3.52 m

g 7164 units

h 1800 cm3

c 3520 L

d 880 L

9 a 1512 cm2

b 429.37 m2

28 a 2540 m2 29 a 7771 m2

c 1203.75 m

b 406.4 m3

7.03 Surface area and volume of composite solids Understanding 1 A composite solid is a 3D ﬁgure that is composed of simpler geometric ﬁgures. Examples could include a house shape (a rectangular prism with a triangular prism for the roof) or a cone with a hemisphere (like an ice cream cone) on top. 2 a To calculate the surface area of a composite solid, we need to calculate the area of all of the faces that are on the outside of the composite solid. b To calculate the volume of a composite solid, we need to calculate the volume of each of the solids that make up the composite solid and combine them appropriately. Sometimes this requires addition, sometimes subtraction. 3 a A cube and a square pyramid b Two rectangular prisms c A trapezoidal prism and a triangular prism d A triangular prism and a cone e A cylindrical hole in a rectangular prism f Two hemispheres 4 a A rectangular prism and a triangular prism. b Three cylinders, two of which are identical. c Two trapezoidal prisms. d A rectangular prism and a cylinder.

d 365.45 cm2 2

e 6755.25 mm

b 1631.91 m3

30 14 kL

742

c 228 cm2 2

g 120 cm

f 867.08 cm2 h 621.24 cm2

10 157π cm2 Reasoning 11 The simple solids and their required dimensions to ﬁnd the volume are: • A rectangular prism for the base. We need to know the length, width, and height. • A cylinder for the pole. We need to know the radius or diameter of the pole, and its height. • A rectangular prism for the backboard. We need to know the length, width, and height. One of these dimensions can be thought of as the thickness. • A cylinder that has had a cylinder cut out of it for the hoop. We need to know the inner and outer radii or diameters, and the vertical thickness of the hoop. 12 a i Volume = 300 cm3 ii Volume = 300 cm3 b Calculate the total surface area of the rectangular prism: 2 × 7 × 6 + 2 × 6 × 8 + 2 × 7 × 8 = 292 cm2 Subtract the triangle at each end of the triangular hole:

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Add in the three rectangular faces that are part of the triangular prism going through the shape. Pythagoras’ theorem is needed to work out that the width of two of these rectangles is

= 361 cm (nearest whole number) 13 a To ﬁnd the volume of a prism, use the formula: Volume of Prism = A × h, where A is the base area and h is the height. To ﬁnd the base area, subtract the area of the two rectangles from the square. Area of Rectangle = l × w Volume = (14 × 14 − 2 × 6.5 × 5) × 14 m3 Volume = 1834.00 m3 b Calculate the surface area of the original cube, subtract the four rectangles which are the ends of the holes and add in the rectangular surfaces which are parts of the prism going through the shape. 6 × 142 − 4 × 5 × 6.5 + 4 × 14 × 6.5 + 4 × 14 × 5 = 1690 m2

b 299.625 m3

c h = 0.5875 m c 1507.96 cm2

1 a True

b False

c True

d False

2 6.35, 6.36, 6.37, 6.38, 6.39, 6.40, 6.41, 6.42, 6.43, 6.44 3 5.44 4 0.15 5 12.715 Fluency 6 a • Length of the rectangular prism: 7.95 cm • Width of the rectangular prism: 5.95 cm • Diameter of the cylinder: 4.95 cm b • Length of the rectangular prism: 8.05 cm • Height of the rectangular prism: 8.05 cm • Width of the rectangular prism: 6.05 cm • Diameter of the cylinder: 5.05 cm c Total surface area = 393 cm2 d Total surface area = 405 cm2 7 a Total weight = 68.051 kg b Total truncated weight = 68.03 kg

Problem-solving

15 a 1658.76 cm2

Understanding

• Height of the rectangular prism: 7.95 cm

Area of Square = side × side

14 a 1198.5 m3

7.04 Estimation and accuracy

b 131.95 cm2 d 3298.67 cm2

c Total rounded weight = 68.05 kg d The truncated value is slightly below the actual total, while the rounded value is very close to the actual weight. 8 a 7.2%

b 14.6%

c 4.2%

d 5.3%

16 13 815.88 cm 17 a 68 m2

b 70 m2

c 100 m2

d 376 m2

e $16 544 18 a 16.3 cm2

b 614

19 a 36.47 m3

b 36 469 L

c 34 016 L

d 1717 L

e 21 20 a 127.1 m2

b 140 m2

c $2520

Reasoning 9 The company would overestimate the weight of the materials, which could lead to overpayment if the materials are bought or sold based on weight. They could reduce this error by recalibrating or replacing the weighing scale. 10 a Different rounding and level of accuracies being used. b Caleb rounded his to the nearest metre, Ida used to the nearest tenth of a metre and Lavinia to the nearest hundredth of a metre. c Yes, they have all used different levels of accuracy.

Answers mathspace.co

743

11 Truncating the interest rate could lead to slightly lower total interest payments. Over a large number of customers, this could result in signiﬁcant savings for the bank. 12 a Alex will use: Mass: 0.37 kg (the closest value to Liam’s measurement that Alex’s scale can measure) Velocity: Using Liam’s measurements, the time taken is 5 = 2.2321ss, which Alex will measure as 2 2.24 5 2 seconds, giving a velocity of = 2.5 m/s m/s. 2 b i 1.16 Joules ii 0.92 Joules c 26.09% d i 0.93 kg m/s

ii 0.82 kg m/s

e 13.41% f T he percentage error in Alex’s kinetic energy calculation is higher than the percentage error in his momentum calculation. This is because kinetic energy is proportional to the square of the velocity, so errors in the velocity measurement are ampliﬁed in the kinetic energy calculation. In contrast, momentum is directly proportional to velocity, so errors in the velocity measurement have a less drastic impact on the momentum calculation. Problem-solving

b 2261.95 cm3

8 a 3.42 m3

b 0.684 m3

9 a 180 cm3

b 824.67 cm3

10 a i 864 cm3

ii 648 cm2 3

b i 1526.81 cm

ii 894.58 cm2

11 a 58 904.86 cm3

b 2120.58 cm2

d 3063.05 cm2

c 2434.73 cm 12 a 488 m2

b $1802

13 a Trina’s

b Kareem’s

c 0.025 cm3 14 By rounding down, the total calculated viewing time will be slightly less than the actual time spent by users. Over a vast number of users, this could result in a signiﬁcant discrepancy between the tracked and the actual viewing times. 15 The products may be overcooked if the timer is consistently slow, which can affect the texture and taste. 16 a 525 Myr

b 475 Myr

c 5%

d 2%

8.01 Review Venn diagrams and two-way tables Understanding

13 $9 125 000

1 Probability of a storm occurring tomorrow.

14 a 40.6 thousand years ago b 33.0 thousand years ago

2 • P (B)

c 10.3%

• P (A′)

15 a 48.15%

• 1 − P (A)

b 1.48%

3 P (A ∩ B)

Chapter 7 review 1 a 544 m2

b 216 cm2

c 612 cm2

d 608 cm2

4 49

2 h = 71 3 Design B because it has a smaller surface area (494.12 cm2 ) than Design A (531.93 cm2 ). 2

4 a x = 12.10 cm

b 537.21 cm

5 a 490 cm3

b 200 cm3 3

c 2040 cm

d 140 cm3

6 Base Area = 110 cm2

744

7 a 552.92 cm3

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Support

Do not support

Males

Females

5 a

Reasoning

17 We want either A or B to occur, but not both. This could happen in two ways: • Event A occurs and event B does not occur • Event B occurs and event A does not occur c {Zebra, Tiger}

We can rewrite an event not occurring as its complement.

Fluency

• Events A and B′ occur

6 a B

• Events B and A′ occur

b i No

ii No

iii No

iv Yes

• A ∩ B′ occurs

7 a B b i No

ii No

iii No

8 a 0

iv No

P ((A ∩ B′) ∪ (B ∩ A′))

Play Rugby League

Don’t play Rugby League

Play Rugby Union

185

Don’t play Rugby Union

179

146

High emission

• B ∩ A′ occurs Now, A ∪ B represents event A or B occurring.

b 1

A ∩ B represents event A and B occurring.

Urban

If we were doing the calculation, we could also do P (A ∪ B) − P (A ∩ B). 18 a 58

c The “number of favourable outcomes” is the number of tourists who speak only one of the languages. This is made up of tourists who only speak Mandarin or who only speak Spanish.

The “total number of outcomes” is the total number of tourists.

The probability that a person speaks only 11

Microchipped

Not microchipped

Vaccinated

Not vaccinated

12 a 120 13 a

L′

Total

C′

Total

16 a

c b

19 Since there are no outcomes where both A and B can occur at the same time, the events must be mutually exclusive. A Venn diagram that represents this situation is shown:

b Yes, there is no intersection between sets A and B.

c The probability of selecting a student with no sisters. 15 a

a No, there is no intersection between sets A and B.

b 14 a P (S ∪ B)

one of the languages is

c Yes, the outcome is either in set A or set B, and there are no other possible outcomes. d No, the sets are not necessarily the same size. e Yes, if the outcome is not in set A then it is in set B. f No, set B is not an empty set.

Answers mathspace.co

745

Problem-solving 20 a 326

Fluency

b 37.7%

c 0.725

6 a

21 The number of people in set T ∩ C is 36.

7 a

22 a A Venn diagram or two-way table may be used. A two-way table is shown:

c Independent

Pescatarian

Not pescatarian

Total

Children

Adults

Total

b Independent

8 a Independent b i 6 possibilities out of 36: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). The answer is

23 a A Venn diagram or two-way table may be used. A Venn diagram is shown:

Assembly line

45%

b i

Quality control

20%

ii 6 possibilities out of 36: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6). The answer is iii

10 Yes, events A and B are independent.

11 a b

Understanding

c b Dependent

c Independent

12 a i

b i

c i

d i

2 Dependent 3 a Independent

b Dependent

4 a b Drawing an orange ball, then a pink ball c • Drawing a pink ball, then an orange ball • Drawing a pink ball, then a pink ball

13 a Dependent

d Dependent 5 a Two days

b No

c Rain on the ﬁrst day, sun on the second day (RS) and sun on the ﬁrst day, rain on the second day (SR).

b i

iii

v 14 a i ii

746

v 0

9 0.5

30%

8.02 Independent and dependent events

1 a Independent

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

22 a

b i

G G

ii B

c i

B B

d i b i

GG GB

BG BB

23 a {SR, SL, SC, GV, GM, GC}

e i

Reasoning

15 a

24 a

c No

d The draws are independent if the probability of each successive ball drawn is the same as the probability of the ﬁrst ball drawn.

c d

The draws are dependent if the probability of each successive ball drawn is not the same as the probability of the ﬁrst ball drawn.

In this case, it is dependent.

16 a b

25 3 of the cards are labelled 1, 2 or 3.

3 of the cards are labelled 9, 10 or 11. Since the card is replaced, there are always 11 cards to choose from.

17 a b 18 a b c 8 possibilities out of 90: (2, 10), (3, 9), (4, 8), (5, 7), (7, 5), (8, 4), (9, 3), (10, 2). The answer is 19 a

20 a

21 a

26 Independent would mean that P (A ∩ B) = P (A) × P (B) and since P (A) ≠ 0 and P (B) ≠ 0, it follows that P (A ∩ B) ≠ 0. For mutually exclusive events P (A ∩ B) = 0. 27 If the events A and B are independent, then P (A ∩ B) = P (A) × P (B). Here

Answers mathspace.co

747

34 a

Problem-solving 28 a

Recipient

Not

Recipient Recipient Not

Not Recipient

29 a

Recipient

Not Recipient

Not

d The probability is getting higher. e 25

30 a

c d

e Increase 31 a

Edge

C D

8.03 Conditional probability

Understanding

Probability b

1 a False Edge

C D

4 a Patricia is chosen second, after Aaron has been chosen ﬁrst.

ii Higher probability by

b Greater than

iii Lower probability by 32 a 1st race 0.8 W 0.2

Fluency

2nd race W

0.9 0.1 0.6

0.4 0.4 L

0.6

0.6 0.4 0.1

0.9

b 0.504

3rd race W WWW L WWL W WLW L WLL W LWW L LWL W LLW L LLL

5 0.8 6 0.18 7 a

b 0.4

8 a 0.4

c 0.678

33 a

Tea

9 a

Toast

10 a i 260

b ii 237

iii 111

Milo

Tea

11 a {0, 1, 2, 3, 4, 5}

Cereal Milo

b 0.42

748

d True

3 A

c i Higher probability by

0.3

c True

Probability

0.7

b True

c 0.88

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

iv 208

12 a

24 P (A∣B) is greater because:

13 0.6875 14 a

and

But according to the size of the circles on the Venn diagram:

15 a b Given that the ﬁrst die is a two, the possibilities are six: (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6). Out of them, there are only two possibilities with sum of 7 or more: (2, 5) and (2, 6). The answer is

Problem-solving 20 a

16 a

17 a

Comedy News

Comedy

Horror Animation Comedy

Comedy

18 a (0.4 × 0.8) + (0.6 × 0.7) = 0.74

Horror Animation

Current Affairs

Comedy

Horror

Animation Comedy

19 a 0.75 × 0.05 = 0.0375

Horror

b 0.25 × 0.95 = 0.2375

Animation

c 0.0375 + 0.2375 = 0.275 d 20 a 177

b b

26 a

0.2

21 a 0.8 × 0.5 = 0.4

0.7

0.8

b (1 − 0.8) × (1 − 0.5) = 0.1 c 1 − 0.1 = 0.9 d (1 − 0.8) × 0.5 = 0.1

0.85

0.3

e P (A∣B) = P (A) = 0.8 (A and B are independent events) Reasoning

0.15

C T

22 27 a

Since P (B ∩ A) = P (A ∩ B), therefore P (A∣B) × P (B) = P (B∣A) × P (A) 23 a

e Since B is inside of A, if the event B has occurred, the event A must have occurred. So, P (A∣B) = 1.

Answers mathspace.co

749

28 a

9 a

Labor (next election)

Liberal (next election)

Greens (next election)

Total

Labor (last election)

Liberal (last election)

Greens (last election)

Total

140

200

29 a

Believe in Tooth Fairy

Do not believe in Tooth Fairy

Total

100

b 38.75%

10 a 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 b 1, 2, 3, 4, 6, 8, 12, 16 c 4 numbers from set A: 10, 14, 18, 20 and 2 numbers from set B: 1, 3. In total, 6 numbers are unique.

e H and S

Do not Believe believe in Total in Santa Santa

8.04 Mixed probability concepts

11 a

Understanding 1 a Diagram iv

b Diagram i

c Diagram ii

d Diagram iii

2 a P (A ∩ B) = P (A∣B) × P (B) b P (A) + P (B) = 1

Read the book

Didn’t read the book

Total

104

166

b i

c P (A ∪ B) = P (A) + P (B) − P (A ∩ B) d P (A ∩ B) = P (A) × P (B)

Watched Didn’t watch Total the movie the movie

12 a i x = 40

iii

ii y = 400

iii z = 200

b 360

3 B

Fluency

13 a

4 w = 0.3, x = 0.3, y = 0.2, z = 0.2

14 a

5 6 a {23, 24, 32, 34, 42, 43} b i

iii

7 a

c 15 a

b 10

Y B

8 a 3

R Y B R

750

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

BY RB RR RY YB YR YY

b i

iii

v 0

vii

16 a

c Two events A and B are considered independent if P (A∣B) = P (A).

Using the fact that

, we

can rewrite this condition for independence as P (A ∩ B) = P (A) × P (B).

Serve in 0.55 Serve in 0.81

We calculated P (S ∩ H) and P (H) × P (S), and with these results we can conclude that these events are independent.

0.45 Fault

23 a

0.19

b The probability of picking an identical pair depends on how many cards are left. Since the number of cards decreases as pairs are picked, the probability of picking a pair is affected by how many cards have been removed.

Fault

b 0.45

c 0.0855

17 a Star Jet

Star Jet

Air Tiger

Star Jet Air Tiger

In this case, it is dependent. 24 a No second attempt is needed if the exam is passed on the ﬁrst attempt.

Air Tiger

b c Candidates will have a variety of abilities and preparedness for the examination.

b i

Problem-solving

ii 18 a

19 a

20 0.856

25 a

26 a 0.18

b 0.45

c 0.38

b 43

d 0.64

e 0.9

Reasoning

27 a 27

21 a P (A ∩ B) = P (A) × P (B) b P (A ∩ B) = P (A∣B) × P (B)

28 a (0.55 × 0.02) + (0.45 × 0.08) = 0.047

c The formula P (A ∩ B) = P (A∣B) × P (B) can be used for both independent and dependent events.

The answer is 4.7%.

For dependent events, we have to use the formula from part (b). For independent events, P (A∣B) = P (A) since it doesn’t matter whether B occurred or not. So, we can use the formula from part (b) and still get the same answer as the formula from part (a). 22 a i b

b 0.23 29 22.5% 30 10.58%

Chapter 8 review 1

iii 2 a

Y′

82 12

X′

41 53 b Independent

Answers mathspace.co

751

3 a

4 a

9.01 Representations of data Understanding 1 4 and 7

5 0.2

2 a Black

b Blonde b 72

6 a

3 a 5

4 a 5 employees b 58 years

7 a 0.7

b 0.65

8 a

c 6 students

c 20 years

5 35

6 a By counting the number of dots above a particular outcome.

9 j = 0.4, k = 0.4, l = 0.1, m = 0.1

b By ﬁnding the heights of the bars using the frequency axis.

10 a

c By counting the number of leaves for a particular stem.

240

120

Fluency

210

7 a Ungrouped

8 a 4 b 330 = 11

11 a

310

Heart rate

Class centre

Frequency

45–54

49.5

Went hiking

Did not hike

Total

Went swimming

Did not swim

Total

100

12 a

Egg sandwich

6 10 4 10 Muffin

b 0.3

15 a

79.5

Score

Frequency

16 ≤ Age < 18

14 10 12

Iced coffee

24 ≤ Age < 26

Orange juice

26 ≤ Age < 28

b 82 people Iced coffee

c 44 people

d 26 people 10 a 20

b 3

c 12

11 a 28

b 99

c 71

d 28

12 a 42 people

13 a Bird

b Cat

c 8 students

14 a Wednesday c Tuesday

752

d 40 people

22 ≤ Age < 24

75–84

5 10 5 10

14 a

18 ≤ Age < 20

Orange juice

69.5

20 ≤ Age < 22

5 10 5 10

13 a

59.5

65–74

c 51 people

iii

c 0.8

55–64

b 96 people 9 a

b i

b Grouped

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

d 36 students b 60

15 a 4.7

b $100 000

c 16

iii No, there are 10 people who travel an hour or more to work, even though it is true that most people live between 20 and 40 minutes away.

d 75%

e 16

iv No, there is no indication about what mode of transportation the people surveyed use to travel to work.

16 a Histogram, as this is continuous data with a large spread of values. b Stem-and-leaf plot, as we want to know individual data values across a medium range and small data set. c Dot plot, as there would be concentrated values in a very small range.

b $122.55

21 a 1.7

c 1

d 3.8%

Country

Oil Production (barrels per day)

log10 Oil Production

Saudi Arabia

10 625 000

7.0

Norway

1 763 000

6.2

Ecuador

555 000

5.7

Brunei

113 000

5.1

France

61 000

4.8

Albania

21 000

4.3

Finland

13 700

4.1

Portugal

7100

3.9

Japan

4000

3.6

Costa Rica

300

2.5

b Between 60 mm and 80 mm c Greater than 60 mm in length because ‘more likely’ refers to how frequent a group of lengths is compared to the others. In the frequency table, we can see that there are 92 leaves greater than 60 mm in length that were collected compared to 35 leaves less than 60 mm in length that were collected. 18 a Yes, the tallest person is Kate from the height graph and she is also the oldest person from the age graph. b No, the heaviest person is John and he is not the oldest. The oldest is Kate.

b 0.01 kg

c No, the heaviest person is John and he is not the tallest. The tallest is Kate. d No, the youngest person is Tom and the lightest person is Kate. e Yes, based on the data, Kate who is 10 years old is lighter than John who is 8 years old. John weighs 45 kg while Kate weighs 20 kg. 20

4 Frequency

17 a 35

Frequency

20 a 180.5 hours e 19

Reasoning

19 a

Problem-solving

3 2 1

3 4 5 6 7 8 Log Oil Production

9.02 Quartiles

Understanding

1 a 28

2 a Q1 = 1, Q3 = 9

b 11.5

c Q1 = 15, Q3 = 26.5 0 20 40 60 80 100 Commute time (minutes)

b i N o, the number of people who live less than 40 minutes away from work is greater than those who live more than 40 minutes away. ii Yes, since a higher proportion of people live within 40 minutes of their workplace.

c 49

d 7

b Q1 = 24, Q3 = 32 d Q1 = 16, Q3 = 24.5

3 10.5 4 a iv: Minimum: Lowest value b v: Lower quartile, Q1: At most 25% of the data is below this value c ii: Median: 50% of the data lies on either side of this value

Answers mathspace.co

753

d iii: Upper quartile, Q3: At most 25% of the data is above this value e i: Maximum: Highest value

16 a 75

b 8

c 3

17 a 79

b 64

c 84

d 20%

c 24

d 35

e 20%

Fluency

18 a 74

b 79.5

5 a 74

b 61

c 114

d 53

e 16.5

f 19.75

6 a 72

b 51.5

c 85.5

d 34

7 a i 6, 10, 15, 17, 24, 39, 40 iii 17

iv 10

ii 7

v 39

vi 29

b i −6, −4, −1, 7, 7, 9, 9 iv −4

iii 7

ii 7

v 9

vi 13

c i 4, 8, 10, 14, 15, 19, 20 iii 14

iv 8

ii 7

v 19

vi 11

d i 20, 22, 28, 32, 40, 42, 43, 54 iii 36

iv 25

v 42.5

iv 74.5

v 84.5

iv 101

v 114

vi 13

v 207

ii 6 vi 9

iii 29.6

iv 19.5

v 39.1

vi 19.6 ii 17 kg

iii 21 kg

9 a 51

b 7

c 1

10 a 47

b 7

c 4

b Set 1 appears to be symmetric as the differences between corresponding quartiles are nearly equal, indicating a symmetrical distribution.

c 2

11 a 5

b 16

c 11

d 17

12 a 19.5

b 26.5

c 7

13 a 72 hours

b 85 hours

c 13

14 a

Score (x)

Freq ( f )

Totals

Set 2 is skewed right because the difference between the third quartile and the maximum is much larger than the difference between the minimum and the ﬁrst quartile, indicating a few higher values causing a right skew. Set 3 is skewed left as the difference between the minimum and the ﬁrst quartile is much larger than the difference between the third quartile and the maximum, indicating a few lower values causing a left skew. 21 a For Group 1, the range is 39, the interquartile range is 23 (116.5 − 93.5), the median is 99 and the mean is approximately 104.24.

b 7

c 6

d 8

15 a 1

b 3

c 2

754

c 1

20 a Set 2 has the largest interquartile range. This indicates that the middle 50% of data in Set 2 is spread over a larger range of values compared to the other two data sets.

8 a 18.93 kg b i 19 kg

b 9 marks

d In this data set, the scores less than 7, while constituting the minimum values, don’t signiﬁcantly impact the mean and median due to their low frequency. However, they do increase the range and could potentially affect the ﬁrst quartile by shifting it lower.

ii 8

h i 15.8, 19.5, 22.4, 29.6, 35.4, 39.1, 46.2 ii 7

19 a 8 marks

ii 5

g i 197, 198, 202, 205, 207, 228 iii 203.5 iv 198

Reasoning

vi 10

f i 100, 102, 110, 113, 115 iii 110

ii 8 vi 17.5

e i 69, 71, 78, 79, 82, 84, 85, 88 iii 80.5

g The Emerald Eagles’ scores are approximately symmetrically distributed around the median, while the Sapphire Sharks’ scores show a slight right skew, indicating a majority of scores on the lower side and some higher scores pulling the mean upwards.

e 2

For Group 2, the range is 30, the interquartile range is 18 (105 − 87), the median is 100 and the mean is approximately 97.48.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b The mean of Group 1 is higher than that of Group 2, which suggests that Group 1, on average, scored higher on the EQ test. c The spread of Group 1 is larger than that of Group 2, indicating that there is a greater variability in EQ scores within Group 1 than within Group 2. 22 a 2.5

Understanding 1 a

b 8

c There are only 2 students who have more than 8 pets (the upper quartile), which is equal to 8% of the total number of students (2 out of 25). It is not exactly 25% because the quartiles are determined by dividing the data into 4 equal parts, but this does not guarantee that the percentage of values above or below a quartile will always be 25%. 23 Yes, the percentage of data values between the upper and lower quartiles is always exactly 50%. This is because the lower quartile (Q1) represents the 25th percentile of the data and the upper quartile (Q3) represents the 75th percentile of the data. The difference between the 75th and 25th percentiles is 50% of the data. For example, consider the dataset: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The lower quartile is 3 (25th percentile) and the upper quartile is 8 (75th percentile). The data values between Q1 and Q3 are 4, 5, 6 and 7, which represent 50% of the total dataset.

minutes. This means that about half of

the ﬂights had a delay less than

minutes

and half had a delay of more than c 45 minutes

d 35

e 83

f 83%

g 22

c 137 minutes

2 a i 19

ii 11

b i 34

ii 15

3 a True

b False

e True

f True

4 a i 50%

ii 25%

c True

d False

iii 50%

iv 75%

c 15

d 10

v 75%

minutes.

Fluency 5 a 3

b 18

e 7 6 a

h 22% 26 a 142 minutes

c First and third

24 65

b 10 minutes

20 b

b Second

Problem-solving

25 a

9.03 Boxplots

b 27 b

Minimum

Lower quartile

Median

Upper quartile

Maximum

Minimum

Lower quartile

Median

Upper quartile

Maximum

Answers mathspace.co

755

Minimum

b i

Minimum

Lower quartile

Median

31.5

Upper quartile

36.5

Maximum

Minimum

Lower quartile

Median

51.5

Upper quartile

Maximum

7 a

Data

20 22 24 26 28 30 32 34 36 38 40

iii Symmetrical c i

20 30 40 50 60 70 80 90 100

Data

Minimum

Lower quartile

Median

Upper quartile

Maximum

20 30 40 50 60 70 80 90 100

8 a 2

b 40

d 38

e 6

9 a i

c 2, 3, 40

Minimum

Lower quartile

Median

Upper quartile

Maximum

20 22 24 26 28 30

iii Left skewed

756

iii Somewhat symmetrical, with slight right skew

Data

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

d i

Minimum

Lower quartile

Median

Upper quartile

Maximum

g i

Data

Minimum Lower quartile

Median

Upper quartile

Maximum

Data

iii Somewhat symmetrical, with slight right skew e i

Minimum

Lower quartile

Median

50.5

Upper quartile

Maximum

iii Somewhat symmetrical, with slight right skew h i

Data

Minimum

Lower quartile

Median

Upper quartile

Maximum

Data

40 42 44 46 48 50 52 54 56 58 60

iii Somewhat symmetrical, with slight left skew f i

Minimum

Lower quartile

Median

Upper quartile

Maximum

iii Somewhat symmetrical, with slight right skew i i

Data

Minimum

Lower quartile

Median

Upper quartile

Maximum

iii Symmetrical

Answers mathspace.co

757

13 a 31 years

Data

c 25% 14 a 49

b 17 years d 7

e 1st

f 5

b 50

c 12

d Carl

e Carl Reasoning

iii Right skew j i

Minimum

Lower quartile

Median

15.5

Upper quartile

Maximum

15 The typical pulse rates increases from about 68 to about 108 and are spread out over a larger range after exercise. 16 Answers may vary. One possible explanation: A boxplot visually displays the median, providing a measure of the central value of the data set. The interquartile range, represented by the box, indicates the variability within the central half of the data, offering insight into the spread of data. Lastly, the ‘whiskers’ of the boxplot highlight the overall range of data, contributing to our understanding of the data’s distribution.

Data

17 a

Scores

iii Symmetrical 10 a The cyclist’s heart rates varied by 36 beats per minutes (bpm). b The middle half of the cyclist’s heart rates varied by 19 beats per minute.

100

Scores

11 a i D ata set A - skewed left, data set B - symmetrical ii Data set B

iii Data set B

b i D ata set A - skewed right, data set B - skewed left ii Data set B

iii Data set A

12 a

Class 10P

Class 10Q

Median

Lower quartile

Upper quartile

Range

Interquartile range

Outliers

None

b Class 10P

758

c Interquartile range

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

c Science. The highest score is the maximum value of the data. Science has a highest score of 91, while music has a highest score of 88. Science has a minimum score of 52, while music has a minimum score of 43. Science has a median on 74 and Music has a median score of 66.

18 a The median score of the Gamma Geckos is 5 points higher than the median of the Delta Dragons.

22 a

b The range of the Delta Dragons is 12 points wider than the Gamma Geckos. c The interquartile range of the Delta Dragons is 19 points wider than the interquartile range of the Gamma Geckos. d Answers can support either team, but should use summary statistics to justify the response. For example:

19 a Negatively skewed

95 100

b 20 c 25%

• We can expect the Gamma Geckos to win since the median number of points scored is 5 points higher than the median number of points scored by the Delta Dragons. • We can expect the Delta Dragons to win since in about 25% of games they scored more points than the highest number of points scored by the Gamma Geckos.

d 52 and 85 e There are now 17 scores, so there are 13 = 76.5%. At least

values below Q3 and

75% of the data lies below Q3. Problem-solving 23

b Positively skewed c Yes. The group that received treatement had a lower median and most people had symptoms for less than 9 days, whereas the control group had half the people having symptoms for more than 9 days. 20 a Yes, because correcting the error will not affect the positions of the Q1, Q2, and Q3 points. As the data point 23 is less than the maximum but greater than the Q3 division then the boxplot will remain unchanged.

24 a Any ﬁve-number summary with an interquartile range of 2 and a range of 8. For example: • Minimum: 1

• Lower quartile: 2

b This changes the range from 21 to 14, a decrease by 7.

• Median: 3

• Upper quartile: 4

c Because the positions of Q1 and Q3 were unchanged, there is no impact on the IQR.

This has the largest possible range of 8 and the smallest possible interquartile range of 2.

21 Answers should specify the expected minimums, maximums, or differences to justify the context. a For example, the price of items at the discount store. Most items will be $1, but some discount stores have items that also cost up to $5 giving them a range of $4. b For example, the ages of every person in a kindergarten classroom. Most of the people are the kindergarteners whose ages are close together giving an interquartile range of 1–2 years old, but the teacher is also in the room so the range could be anywhere from 15–70 depending on how old the teacher is.

• Maximum: 9

b Any ﬁve-number summary where the range is three times the interquartile range. For example: • Minimum: 1

• Lower quartile: 2

• Median: 3

• Upper quartile: 4

• Maximum: 7 This has a range of 6 and an interquartile range of 2. 25 • Boxplot 1/Histogram A • Boxplot 2/Histogram B • Boxplot 3/Histogram D • Boxplot 4/Histogram C

Answers mathspace.co

759

26 a

4 a True

Monthly salaries (in thousand dollars)

b False

c True

d True

e False Fluency 15

5 a

Score (x)

Scores of students

100

Debbie’s blog visitors

100

150

200

6 a

Number of pages read

9.04 Cumulative frequency Understanding

15 26

11 → 15

16 → 20

21 → 25

26 → 30

760

c 10 Score (x)

Frequency ( f )

Cumulative frequency (cf )

20 → 24

25 → 29

30 → 34

35 → 39

40 → 44

45 → 49

50 → 54

b 75

Score

Frequency

1≤x<2

2≤x<3

1→5

c 34

7 a 30

b 6

c 27

8 a 60

b 3

c 13

9 a 86

3≤x<4

4≤x<5

5≤x<6

6≤x<7

Score (x)

Frequency ( f )

Cumulative frequency (cf )

1≤x<5

5≤x<9

9 ≤ x < 13

13 ≤ x < 17

17 ≤ x < 20

Score (x)

Frequency ( f )

Cumulative frequency (cf )

12 ≤ x < 13

13 ≤ x < 14

14 ≤ x < 15

15 ≤ x < 16

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

100 90 80 70 60 50 40 30 20 10 0

b 14

Cumulative frequency

Cumulative frequency (cf )

6 → 10

b 82 0

Frequency ( f )

20 30 40 50 60 70 80 Class

d 15%

14 a

90 Cumulative frequency 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 10

Number of sightings

Cumulative frequency (cf )

b 19 locations

c 12 locations

Reasoning 15 a The difference between the heights of two adjacent columns. Time (sec)

b 4 ≤ Score < 5

16 a

50 45

Score (x)

Frequency ( f )

Cumulative frequency

40 35

30 25

Frequency values are ﬁlled based on the given data. Cumulative frequency is calculated by adding the frequency of the current row to the cumulative frequency of the previous row.

5 0

50 100 150 200 250 300 350 400 Mass (g)

Score

Frequency

75 ≤ x < 80

80 ≤ x < 85

85 ≤ x < 90

90 ≤ x < 95

95 ≤ x < 100

13 a 2 c

c 2, the median is the middle value when the data is ordered. Since there are 9 families, the median is the value corresponding to the 5th family in the cumulative frequency, which is 2.

b 12 Score (x)

Cumulative Frequency ( f ) frequency (cf )

2 3 4 5 6 7 8 9 10 11 12 d 7

b 9, the total number of families is the sum of the frequencies in the table.

1 2 5 5 5 9 7 5 8 1 2 e 11

1 3 8 13 18 27 34 39 47 48 50 f 18

d 2.44, the mean is calculated by dividing the sum of the product of scores and their frequencies by the total number of families:

Cumulative Frequency

Frequency

Number of locations ( f )

Flu Season

10 8 6 4 2 0

2 3 Score

Answers mathspace.co

761

The histogram shows the distribution of the number of times families got the flu. It helps to visualise the data and identify patterns such as the median and range of the data. f 4 , the number of families who got the flu more than twice can be found by looking at the cumulative frequency for scores greater than 2. There are 9 families in total, and 5 of them got the flu 2 times or less, so 9 − 5 = 4 families got the flu more than twice.

b 16 students

c 6 students

d 2 students

e 153

20 a i 2100 ii 1900

iii 2450

b i 675

Understanding b The boxplot

c The boxplot

d The histogram

2 a Negatively skewed

3 a • Frequencies for each group • Mode b • Quartiles

d Cumulative frequency histogram, this is the height of the bar with that value. We don’t need to add up the frequencies for each bar below like we do for a histogram.

• Outliers

c • Frequencies for each group • Mode d • Total

16:10 ≤ x < 16:20

16:20 ≤ x < 16:30

16:30 ≤ x < 16:40

16:40 ≤ x < 16:50

16:50 ≤ x < 17:00

19 a

Cumulative Frequency

b 25

c 16 22 20 18 16 14 12 10 8 6 4 2 0

• Shape • Rate of change

Fluency English Mathematics Median

Lower quartile

Upper quartile

Range

Interquartile range

Cumulative Frequency frequency (f) (cf ) 4

• Median • Outliers

• Range and interquartile range

4 a

16:00 ≤ x < 16:10

• Shape

• Outliers

c Histogram, the shape tells us about how the data values are spread out. We would have to look at the rate of change for the cumulative frequency histogram.

Time recorded

b Bimodal

c Symmetrical

Problem-solving

762

1 a The histogram

• Shape

18 a

iii 760

9.05 Compare data sets

17 a Cumulative frequency histogram, this is the height of the last bar. We don’t need to add up the frequencies for each bar like we do for a histogram. b Histogram, this is the height of the bar. We don’t need to subtract the cumulative frequencies like we do for a cumulative frequency histogram.

ii 640

d 48%

b Mathematics 5 a Upmarket restaurant

b $22

c Upmarket restaurant

d $36

e 26 items 6 a Weightlifter A b The range and interquartile range c The median d Weightlifter B 7 a Lower

145 150 155 160 165 170 Height (cm)

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b 0

c 2

d Positively skewed

e 4

f Block B

g No

8 • Histogram A/Boxplot 2

Problem-solving

• Histogram B/Boxplot 4

15 a

• Histogram C/Boxplot 1

Cooper

• Histogram D/Boxplot 3 9 • boxplot 1/Cumulative Frequency D • boxplot 2/Cumulative Frequency C • boxplot 3/Cumulative Frequency B

Marion

Minimum

44.8

47.1

49.1

54.7

Median

52.5

60.45

55.75

65.25

Maximum

66.6

68.5

• boxplot 4/Cumulative Frequency A 10 a Matt

b Matt

c Matt

b Yes, Cooper has an outlier of 66.6. c

Cooper

Reasoning 11 a 5.2

b 8.9

c 3

e No. The results are mostly 4 or 5 out of 10 and the mean is only 5.2 f Y es. The results are mostly 9 and 10 out of 10, and the mean is 8.9 g Answers may vary. One possible answer is that one class may have covered more material than the other class. 12 a Bookstore A, it has a lower range and interquartile range. b $10 c i True

ii True

13 a The boxplot has a long tail to the right which indicates positive skew, while the histogram does not appear to be skewed. The median for the histogram is roughly in the middle, while the median of the boxplot is located further to the left. b The boxplot indicates that the data is widely spread, while the histogram indicates that most of the data occurs near the median. 14 a

65 000 70 000 75 000 80 000 85 000 90 000 95 000 100 000 105 000

b 20 000

Marion

d 3

c 50%

d From the boxplot it appears the highest 25% all earn $100 000. However, this is using class centres, from the data provided we can only conclude that the top 25% earn in the range $95 000 to $105 000.

50 55 60 Time (seconds)

d Cooper e No, a high outlier does not change what the minimum score is, hence would not alter the selection for pole position. 16 a

Tim

Odi

Minimum

29.8

26.0

33.2

34.8

Median

35.0

38.4

37.4

40.0

Maximum

39.6

42.8

Mean (1 d.p.)

35.1

37.2

Range

9.8

16.8

Interquartile range

4.2

5.2

b Tim throws more consistently, his range, interquartile range and standard deviation are all signiﬁcantly smaller. c All of Tim’s scores are consistent, so no outliers for his data. Odi has an outlier of 26, which is far from the rest of his data set. d Odi appears to be the better hammer thrower. His mean is higher and other than the outlier the other values of the 5 number summary are all larger for Odi. In particular, his top 25% of throws are larger than any throw from Tim. Odi’s median is also signiﬁcantly higher (3.4 m) than Tim’s. e As the next lowest throw is more than 6 m larger than his outlier it appears this throw may have been a misadventure and it would be reasonable to remove it when looking at his mean. Including the outlier would distort how high he usually throws and what he would likely achieve in a competition.

Answers mathspace.co

763

9.06 Samples and populations

10 Fair 11 a Biased

Understanding 1 a False

b False

e False

f False

c True

d True

2 a No

b No

c Yes

d Yes

3 a No

b Yes

c Yes

d Yes

e Yes

f No

b The population is all sales at the bookstore, and the sample is the 20 receipts she looks at. c The population is all students at Antony’s school. The sample is the 85 students he surveyed.

d Not biased

12 a All products in the factory

b 30%

c 61%

d No

13 a All apples in the orchard

b 62%

c Yes 14 a Yes

4 a The population is all the residents of the town, and the sample is the 500 residents surveyed.

5 a Yes

b Biased

c Not biased

b No

c No

d No

e Yes 15 Approximately 610 bees 16 1 089 600 17 a Source A

b Source A

c No

Reasoning

b Yes

c Yes

d No

18 Jenny’s claim is more valid since she has surveyed more people.

b No

c No

d Yes

7 a ii

b iii

c i

d iv

8 a Yes

b Yes

c No

d Yes

19 This is a very convenient sample so would be very inexpensive and not take much time. However, a disadvantage is that there’s a chance that someone at the dog park doesn’t have a pet dog such as a professional dog walker or a friend of a dog owner. There is also the possibility that someone with accessiblity needs would not be able to access the park so would be excluded.

e No 6 a No e Yes Fluency

9 a Example answers: • All the seniors • The ﬁrst 30 students to arrive at school that day

20 a The sample is not representative of the target population. b The sample is from self-selecting participants, i.e. only those that made an effort to respond.

• All the students that were in the library at lunch time b Example answers:

c The sample is not representative of the target population.

• All students of a primary school in Sydney • All students in Year 7 in a particular suburb • 100 children chosen from the Sydney under 15s baseball tournament

21 a Example answer: Assign a number to everyone on the electoral role and use a random number generator to decide who to call.

c Example answers:

b The sample will not include anyone who was out for the evening for work or social reasons, and the results are based on less than half of the people they tried to contact, so the sample is not representative of the populations.

• The members of 3 teams chosen from the local hockey tournament • All students from a local school who compete in a school sports competition • All the active members of a local football club d Example answers:

22 a The proportion from the second day, 12% is twice the proportion from the third day, 6%.

• 30 people walking their dog at the local park • All the students from a local school who own a dog

764

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b The differences in sample proportions are due to the varying sample sizes (750 dingoes in the first sample and 100 in the second) and natural random variation. A larger sample size generally yields more representative results, and the uneven distribution of tagged dingoes across the area can lead to different proportions in each sample due to chance and spatial distribution factors. c The sample statistic from the second day is more accurate because sample size is larger, and thus more likely to be representative of the population of dingoes. 23 a The sample proportions are different due to sampling variability. It is impossible for every sample to have the exact proportions as the population. b Both samples are valid, so they could use either one. Often we will take the average of many sample proportions to get a more accurate estimate of the population parameter. Problem-solving 24 a Example answers: • Check additional reputable secondary sources to see which information is corroborated by other authors or creators. • Look at the sources or references cited by each source. A well-cited claim with primary sources to back it up is generally more reliable. • Check the date of publication. Newer sources, like the 2015 documentary, might have had access to more recent research or previously unavailable information. However, it’s essential to note that newer doesn’t always mean more accurate. • Evaluate the intent and potential bias of each source. For instance, a textbook might aim for neutrality and comprehensive coverage, whereas a documentary might focus on speciﬁc stories or narratives.

• Sources can sometimes contain errors, biases, or omissions, and crossreferencing multiple sources can help identify and rectify these issues. • Diverse sources can provide a richer context, helping one understand the broader implications and nuances of a historical event. • Relying on a single source may lead to a limited or skewed understanding, while multiple sources can fill gaps and provide a more in-depth exploration of a topic. 25 a Yes b The sample of fish were collected in several different places and one at a time, without large nets. 26 a Samples should focus on ways to reach only athletes without including non-athletes. For example, the sample could be all starting players from both teams or all of the players on the bench for one of the teams. b Samples that include non-athletes. For example, the first 50 spectators would not necessarily be athletes, and a sample that includes all of the people in the stadium, including coaches, officials, and fans, would have a mixture of athletes and non-athletes. 27 a If 10% of people experience side effects, we might expect about 500 000 residents in Sydney to experience side effects with the medication. b One possible answer: The way we share this prediction is important because causing unnecessary panic could lead to people avoiding the medication, but we also need to be honest about potential risks. c One possible answer: While 1000 people is a decent sample size, it may not be truly representative of Sydney’s diverse population of 5 million, so we should be cautious about making wide-reaching assumptions.

b Example answers: • Different sources may offer various perspectives or focus on different aspects of an event, giving a more rounded understanding.

Answers mathspace.co

765

Chapter 9 review 1 a Grouped

b Ungrouped

2 a 29

b 83

3 a

d Yes. The group that received the experimental fertiliser matured faster since it had a lower median, and most plants reached maturity in less than 12 weeks. In the control group, half of the plants took more than 13 weeks.

c 45

Minutes

Frequency

0 ≤ Minutes ≤ 10

10 < Minutes ≤ 20

10 a 290

20 < Minutes ≤ 30

30 < Minutes ≤ 40

11 a

Cumulative frequency (cf )

40 < Minutes ≤ 50

50 < Minutes ≤ 60

60 < Minutes ≤ 70

70 < Minutes ≤ 80

4 a 101

b 16

c 8

d 3

5 a This indicates that the middle 50% of ratings in Set B are spread over a larger range of values compared to the other two data sets. b Based on the ﬁve-number summaries, Set A appears to be symmetric, Set B is skewed right, and Set C is skewed left. 6 70 seconds

b 8 workshops

c 10 workshops

d 22 attendees 12 a 49

b 45

13 a Building X c Building Y

c 55 b 2 plants d Yes, Building Y

14 A

Lower Upper Minimum Median Maximum quartile quartile 65

15 a

Antoine

Benji

Minimum

6.5

6.0

6.9

6.8

Median

7.3

7.4

7.5

7.9

Maximum

7.8

8.1

Mean (1 d.p.)

7.2

7.3

Range

1.3

2.1

Interquartile range

0.6

1.1

60 65 70 75 80 85 90

c Somewhat symmetrical, with a slight left skew.

b Antoine

8 a The number of books read by students varied by 42 books. b 25%

c 13

d 4th quartile

e 10

c Given that Benji’s lowest jump is signiﬁcantly shorter than his next lowest jump, it might be considered an off-performance. It would be reasonable to exclude this outlier to get a more representative average of his usual performance in competitions.

9 a The median of the control group is 5 weeks higher than the median of the treatment group.

d No, Antoine’s range and interquartile range are still smaller than Benji’s, even after removing the outlier.

b The treatment group’s range is 3 weeks wider than the control group’s range. c The treatment group’s interquartile range is 1 week wider than the control group’s range.

766

c 31.0%

Number of workshops (f )

Number of attendees

b $610.00

7 a

b 60

16 a All cherries in the farm.

b 58.5%

c Yes

d 30 420

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

17 a Biased

b Biased

7 a

c Not biased 18 a The sample proportions are different because the researchers used samples of very different sizes. b The sample statistic from the second week is more accurate because sample size is larger, and thus more likely to be representative of the population of beetles.

10.01 Two-way tables

Vegetarian

Vegan

Neither

Children

Adults

Total

8 a 39

b 26

c 47

9 a 11% of junior students and 10% of senior students were very satisﬁed with the canteen food. b The most common rating among junior students is 4, while for senior students it is 3.

Understanding 1 a 14

c The data suggests that satisfaction with canteen food tends to decrease as students progress from junior to senior years, indicating potential issues with food quality or variety for older students.

b 16

c Outback Animal Care Centre 2 a i Marginal frequency

ii 451

b i Marginal frequency

ii 171

c i Joint frequency

ii 143

d i Joint frequency

ii 388

10 a 14 trains

3 a False

b False

c True

d False

4 a True

b False

c True

d True

e True

b 51 trains

c 127 trains

11 a 27.63%

b 44.47%

Fluency

d 34.61%

Reasoning

5 a

Running Basketball shoes slides

Work boots

Total

Teenagers

Adults

Total

169

Sprinkles Hot fudge Chocolate

Vanilla Total c

Total

Fish

Total

Beef

Chicken

Children

Adults

Total

6 a This value represents the number of cats with feline diabetes treated at Venetian Pet Clinic. b This value represents the number of cats with feline diabetes who received a special diet at Mohawk Animal Clinic.

12 a

Brushtail possum

Honey possum

Fruit

Insects

120

Total

Nuts

110

Total

130

180

310

b There were 130 brushtail possums and 180 honey possums, so there were more honey possums observed than brushtail possums. c There were 80 possums observed eating fruit, 120 possums observed eating insects, and 110 possums observed eating nuts. So, there were more possums eating insects than fruit or nuts. d There were 35 brushtail possums observed eating nuts and 70 honey possums observed eating insects, so there were signiﬁcantly more honey possums observed eating insects than brushtail possums observed eating nuts.

Answers mathspace.co

767

13 a True. Both sets of marginal frequencies will add to the total. b False. A single joint frequency can be equal to the marginal frequency, but only if all other joint frequencies in that row are equal to 0. This would only be possible if all values in a row were 0, in which case that row likely wouldn’t be included.

1. Participation in school sports during the hot week (using the same Likert scale: Always, Often, Sometimes, Rarely, Never) 2. Purchase of an ice block from the canteen during the hot week (Yes or No) Your two-way table would look something like:

14 a The 60 people include teenagers not just adults so it can’t be concluded that all 42 are adults who said either no or sometimes.

Problem-solving 15 a

No Casual Part-time Full-time job Men

Women 25

Often Sometimes Rarely Never Total Students would be surveyed on both their sports participation and their ice block purchases. By comparing the results, you could see if there’s a noticeable difference in sports participation between those who bought ice blocks and those who didn’t. If the coach’s hypothesis is correct, you might expect higher “Rarely” or “Never” responses in the “Bought ice block” column.

10.02 Associations between categorical variables Understanding

b 40 people

c 38.9%

e 21 people

1 a 43%

c No

3 The horizontal axis

b There are 70 animals in the study that carry the Agouti gene and have dark fur. 17 Answers will vary. The table provided is one such solution: Football

Soccer

Total

Period 1

Period 2

Period 3

Total

4 No Fluency 5

b Answers may vary. One possible answer: To investigate the hypothesis, you could design a follow-up survey that looks at two primary variables:

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

Adults

Children

Total

Math

42.86%

57.14%

100%

Science

45%

55%

100%

6 a

18 a 650 students

768

b No

2 A column percentage frequency table

16 a x = 70

Total

Always

b The 24 adults represent the total of adults. Using this information with the 18 adults saying “yes” tells us that 6 adults responded either “no” or “sometimes”. c The statement only indicates the relative distribution among the “yes”, “no”, and “sometimes” responses, not their absolute values. We would need more information to precisely determine the number of each response.

Did not buy ice block

Bought ice block

Brown Blonde

Black

Total

Straight 5.9%

Red

47%

5.9%

41.2%

100%

Curly

40%

30%

20%

100%

10%

Red

Brown

Blonde

Black

Straight

50%

66.7%

25%

77.8%

Curly

50%

33.3%

75%

22.2%

Total

100%

7 a

Comedy Action Drama Horror Total Teens

30%

20%

45%

100%

Children

30%

40%

10%

20%

100%

Comedy

Action

Drama

Horror

Teens

50%

33.3%

81.8%

20%

Children

50%

66.7%

18.2%

80%

Total

100%

Reasoning 15 It means that there is no clear association between the variables. 16 a Location b Column percentage table c

8 a Height b

Play tennis

Do not play tennis

Total

Height ≥ 170 cm

31%

69%

100%

Height < 170 cm

30%

70%

100%

Cardio

Weight

Total

Currently pregnant

30%

70%

100%

Currently postpartum

73%

27%

100%

c Yes, there appears to be an association as the numbers are quite different. Those who are pregnant seem to prefer weights, while people who are postpartum seem to prefer cardio. d No, association is not causation. There appears to be an association but we cannot say whether one variable causes the other. 10 a

38%

Low emissions

38%

62%

Total

100%

d Yes, there does appear to be an association as the numbers are signiﬁcantly different. 17 a 40

b 21

c Dominant hand e

Maths Music Science English Total Righthanded children

30%

20%

25%

100%

Lefthanded children

33%

20%

30%

17%

100%

f N o, association does not exist because values down the column are the same. 18 a Number of passengers b Column frequency table c No. of passengers:

5−9

Roll over

17.1%

30.0% 28.8% 28.6%

82.9% 70.0%

100.0% 100.0% 100.0% 100.0%

Lefthanded

Righthanded

Total

No roll over

Not female

14.5%

85.5%

100%

Total

Female

9.7%

90.3%

100%

Regional

62%

d Row frequency table

9 a Pregnancy status b

Urban High emissions

Left-handed

Right-handed

53.3%

42.0%

19 a Favourite subject at school

Female

46.7%

58.0%

b Column frequency table

Total

100%

b Yes

12 a Fitness level

b Yes

13 a Personality type

b No

14 a Dietary restriction

b B

c No

71.2%

71.4%

d No, because values across the rows are mostly the same.

Not female

11 a Parenting status

> 15

10−15

Maths

Science

Maths major 39.52% 38.36%

Music

Art

11.67% 23.33%

Science major 30.54% 27.40% 23.89% 13.89%

c No

Music major 22.75% 22.60%

37.78% 20.56%

Art major

7.19%

11.64%

26.67% 42.22%

Total

100%

d Yes, because values across the rows are different.

Answers mathspace.co

769

e No, association is not causation. There appears to be an association but we cannot say that one causes the other.

Percentage

3 or less 100

Problem-solving 20 a November

b June

c Book loans

d No, there does not appear to be an association as the corresponding segments are similar in size.

Employed

Unemployed

32%

17%

Does not own a smartphone

68%

83%

Total

100%

10.03 Bivariate numerical data Understanding

e No, association is not causation. There appears to be an association but we cannot say one variable causes another.

1 a Univariate data c Univariate data e Bivariate data g Bivariate data 2 a

Doesn’t own smartphone

Percentage

100 80 60

2 4 6 8 10 12 14 16 18

20 0

Employed Unemployed Employment

22 a Age b Column percentage frequency table 10 years 11 to 20 21 to 30 31 years and under years years or older Less than 3 hours

32.14%

22.86%

9.52%

14.71%

4 to 8 hours

7.14%

28.57%

9.52%

26.47%

9 to 15 hours

25.00%

22.86% 38.10%

29.41%

Over 15 hours

35.71%

25.71% 42.86% 29.41%

Total

100%

y 18 16 14 12 10 8 6 4 2

g Yes it does as the segments are different size in each column.

b Bivariate data d Univariate data f Bivariate data h Univariate data

18 16 14 12 10 8 6 4 2

Smartphone owners employment Owns smartphone

11 to 20 21 to 30 31 & older Age group

e Yes it does as the segments in each column are quite different. It suggests that older people tend to watch more hours of TV.

d Yes, association exists because values across the rows are different.

40 20 10 & under

b Column percentage frequency table Owns a smartphone

80 60

21 a Employment c

TV watching per age group 9 to 15 15 or over 4 to 8

x 2 4 6 8 10 12 14 16 18

y 18 16 14 12 10 8 6 4 2

x 2 4 6 8 10 12 14 16 18

770

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

9 a Height of the ball

y 18 16 14 12 10 8 6 4 2

b Amount of time passed 10 As temperature increases, soups sold decreases. 11 Height has no effect on salary. 12 a The independent variable is the number of immigrants and the dependent variable is the unemployment rate.

x 2 4 6 8 10 12 14 16 18

3 a Yes

b No

4 a Neither c Decreasing

c No

d Yes

b Unemployment rate in Australia Unemployment rate (%)

b Increasing

5.5

d Decreasing

Fluency

4.5

5 a i Number of dogs owned

ii Number of shoes destroyed

3.5

b i Size of a pizza ii Cost of pizza c i Time spent studying ii Exam mark 6 a Ticket sales affect the revenue made from a show. b Dependent variable - revenue from the show, Independent variable - number of ticket sales

Number of immigrants (thousands) 250

Climate change Global temperature anomaly (°C) 0.9

0.8

8 a Distance from the equator affects temperature.

0.7

b No c Both time spent travelling and distance covered affect each other, with speed playing a crucial role in their relationship.

f N umber of pages affects time spent to ﬁnish a novel.

325

13 a The independent variable is CO2 emissions and the dependent variable is the global temperature anomaly.

b Dependent variable - height of the kite, Independent variable - speed of the wind

e Time spent working affects wages earned.

300

c As the number of immigrants increases, the unemployment rate appears to decrease.

7 a The height of the kite and the speed of the wind.

d No

275

0.6 CO2 emissions (million MT)

35 000 36 000 37 000 38 000 39 000

c It appears that as CO2 emissions increase, global temperature anomalies also increase.

g No h Number of pets affects time spent caring for pets (hours). i T ime spent training (hours) affects athletic performance.

Answers mathspace.co

771

18 a Online shopper survey averages

14 a 18 y 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

Average amount spent (dollars)

x 1 2 3 4 5 6 7 8 9 10 11 12 13

b 11

b Based on the scatterplot, the statement does not seem to be accurate. In fact, the data suggests the opposite: as the average time spent on a website increases, the average amount spent also increases.

15 a Dependent variable - height, Independent variable - time b

Height 0.8 0.6 0.4 0.2 −0.2 −0.4 −0.6 −0.8

19 a Time 2 4 6 8 10 12 14 16 18 20 22

16 a 180 y

100 90 80 70 60 50 40 30 20 10

Exam mark

Books read per year 10

175 170 165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90

b 50 c Yes, the teacher is correct. Notice how the points follow a line from the bottom left corner to the top right corner. This means that if a student reads more books they receive a higher mark in their yearly maths exam. Problem-solving x 2

b 105 cm Reasoning 17 a 10 trees b No. There is no connection between the number of people in a student’s family and the number of trees in their garden because there is no trend in the data in the scatterplot.

772

Average time spent (minutes)

20 Looking at the scatterplot, the number that fulﬁls the required characteristics is 4. 21 a Student B

b Student C

c Student A

22 a Answers may vary. One possible answer is the relationship between the amount of study time and test scores. These variables may have a relationship because the more time a student spends studying, the better they might perform on a test.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b Answers may vary. One possible answer is that the amount of study time is the independent variable because it can be controlled by the student, while the test scores are the dependent variable because they can change based on the amount of study time.

c Answers may vary. One possible answer is the relationship between the colour of a car and its speed. These variables likely do not have a relationship because the colour of a car does not affect how fast it can go.

9 a

Accidents 40 35 30

10.04 Associations between numerical variables

25 20 Age

Understanding

1 a No

b Yes

c Yes

d Yes

b Strong, negative, linear

2 a True

b False

c True

d False

c No

e True

10 a

3 a The strength and direction of a scatterplot describe the correlation. b When two variables are related, we say there is an association. c To show causation, we must do a carefully designed study. d When correlation is not causation, some possible explanations are coincidence and contributing (third) variables.

40 35 30 25 20

Age 25

b Strong, negative, linear

Fluency

c No

4 a Linear relationship b Non-linear relationship c Non-linear relationship d Non-linear relationship e Linear relationship f Linear relationship

11 a

Price 16 15 14

5 a Positive and strong b Positive and weak c Positive and moderate-weak d Negative and strong 6 a As air temperature increases, sea temperature increases. b Strong, positive, linear 7 Negative 8 a

Hours 45

13 Supply (kg) 600 700 800 900 1000

b Strong, negative, linear c When very few bananas are available to be sold. 12 a

Time 65

Sport 90

Temp.

1 2 3 4 5 6 7 8 9 10 Maths 40 50 60 70 80 90

b Weak, positive, linear c Student 10’s

b 10

c Weak, negative, linear

d 8 °C, 69 seconds

Answers mathspace.co

773

13 a

19 a It appears that y can change signiﬁcantly without much corresponding change in x.

t 70 60

b i Yes

30 20 n 1

b Increases c Strong, positive, linear

21 a D

14 Strongly negative

22 a i No

15 Strongly positive

b B ii Yes

iii No

b Region 2

16 a No

b Yes

c No

d No

e No

f Yes

g Yes

h Yes

Reasoning 17 This data suggests that as height increases, there is an increase in the ability to play soccer. However, it does not necessarily mean that if you are tall, you are automatically better at soccer. Correlation does not imply causation, and other factors may also inﬂuence soccer skills.

23 a True. For example, if we know that influenza causes patients to develop a fever then there is a correlation between having influenza and developing a fever. However, it is possible for the correlation between influenza and developing a fever to be a strong or weak correlation. b False. For example, if there is a correlation between the number of people drowning in pools and number of people going to the movie theater, that does not mean that an increase in pool drownings causes more people to go to the movie theatre or vice versa.

18 a The explanatory variable is the number of questions and the response variable is the test result.

c False. A positive association means that as the value of the independent variable increases, the value of the dependent variable also increases, but does not mean that changes in one cause changes in the other. For example, suppose data collected from a study found that people who exercised more were also found to be at higher risk for skin cancer. These quantities demonstrate a positive association, but that does not mean exercising more causes a higher risk of skin cancer, there are likely other underlying factors inﬂuencing the relationship.

b 100 Test result

Number of questions 10 15 20 25 30 35 40 45 50 55

c Strong, positive, linear correlation; the larger the number of completed revision questions, the higher the mark on the test. d The relationship is not perfectly linear because of the different abilities of the students.

774

iv No

Problem-solving

d No

95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15

iii Yes

20 Answers may vary. For example, if there is a relationship between the average temperature and migration, the scatterplot might show a pattern where the points tend to increase or decrease together, indicating a positive or negative relationship between the two variables.

ii Yes

24 Answers may vary. One possible answer: Time spent studying and test scores could lead to a weak positive relationship.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

It could be that while additional study hours slightly benefit most students, other factors play a significant role in determining their final score. For instance, some students might have natural aptitude, meaning even with fewer study hours, they can achieve similar or even better scores than their peers who study longer. Others might benefit from private tutoring, have access to better resources, or engage in practical applications of their knowledge that aren’t directly measured by the study hours. Here is a table of values and scatterplot that would lead to a weak positive association:

Hours Studied

Test Score

Student A

Student B

Student C

Student D

Student E

Student F

Student G

Student H

Student I

Student J

Fluency 5 19 y 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

x 1 2 3 4 5 6 7 8 9 10 11 1213141516171819

6 a i

y 20 15 10 5 x 5

ii y = −x + 21 b i

Test score

y 20

15 10

5 55

x 5 Hours studied 1

ii y = 0.56x + 5

c i

10.05 Line of good fit

y 20

Understanding

1 Decreasing

2 Line 2 3 a No 4 a Negative

b No

c Yes b C

d Yes

5 x 5

ii y = 0.9x + 1.5

Answers mathspace.co

775

d i

y 20 15 10 5 x 12

9 a i 4.62 ii Positive

iii Increases by 4.62 units iv 7.58

b i −7.68

ii Negative

iii Decreases by 7.68 units

iv 5.49 c i −5.69

x 12

ii Negative

ii y = 1.5x − 17.5 f i

iii Decreases by 5.69 units iv 7.84

d i 3.16

ii Positive

iii Increases by 3.16 units

iv −2.72

10 a The vertical intercept indicates that when a family has no income, $27 is its average spending on dining out.

2 x 2

ii y = 0.9x + 0.5

b No 11 a

c 0.3

d $60

Price (thousands of dollars) 24

7 Example answer: 38 36 34 32 30 28 26 24 22 20

Price 18 20 22 24 26 28 30 32 34 36

Copies sold

ii y = −1.3x + 29.5 e i

190 180 170 160 150 140 130 120 110 100 90

16 12 8 4

Age (years) 2

b Answers will vary if calculated by hand, but should be around eighteen.

x 10 20 30 40 50 60 70 80 90

c A car that is 0 years old costs $18 400. d Answers should be around 11 years. 12 a −4

776

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b 3865

Reasoning

Problem-solving

13 a 27.7 kg

16 a

b The y-intercept indicates that when a child’s height is 0 cm, their average weight is 27.7 kg.

Ice blocks 140 120

c No, a child cannot have a height of 0 cm.

100

14 a A scatterplot can help visualise the relationship between the quantities, but is not necessary for a complete response.

100

60 40

Grades

Temperature

14 16 18 20 22 24 26 28 30 32 34 36 38 40

70 60

b i Approx. 91

ii Approx. 149

c Increases

40 30

17 a A

b x = 3682

18 a y = 0.5x + 4

b 7.45

Minutes 50 100 150 200 250 300 350 400 450

There is a strong, negative, linear association between minutes spent on LookAtMe and maths grades. A student who spends 0 minutes on LookAtMe is predicted to have a 99.6% grade in maths based on the line of good fit. b According to the linear model, for every additional minute spent on LookAtMe, a student’s maths grade is expected to drop 0.25% points. Or, for every additional 100 minutes spent on LookAtMe, a student’s grade is predicted to drop 25%. 15 a There are only 2 points above the line and 18 points below it, so this line does not go through the middle of the data. It is too heavily impacted by the outlier, so is not a line of good fit. b An outlier should not be completely ignored, but the line should go through the majority of the data. Lorena’s line has 9 points above it and 10 points below it, and goes through the middle of the data, so it is a line of good fit.

19 a A line of good fit: y = 3x + 15

H% 30 25 20 15 10 5 Y 1

b The new line of good fit would have a smaller slope and a higher y-intercept.

10.06 Interpolation and extrapolation Understanding 1 a False

b False

e False

f True

c False

d True

2 a Interpolation

b Extrapolation

c Extrapolation

d Interpolation

3 a y = 8.45

b y = −17

c y = −7.4

Fluency 4 a Interpolation

b Extrapolation

c Interpolation

Answers mathspace.co

777

5 y = −52.61, Extrapolation 6 a 3 seconds 7 a 20 years 8 a

225 200 175 150 125 100 75 50 25

b 2.5 hours b 1300 ﬁsh

c 11 years

People

17 Interpolation and extrapolation can be limited by the accuracy and validity of the data, as well as the assumption that the relationship between variables remains constant. In the context of climate change, factors such as human intervention, natural disasters, and technological advancements can inﬂuence the relationship between variables, making predictions less reliable. Problem-solving 18 a y = −0.005x + 49

Temperature °C 12 14 16 18 20 22 24 26 28

b 150 people

c 23 °C

d 140 people

e 19 °C

9 a 2448 Pa

19 a A = −25T + 750

b Interpolation ii No

10 a 90 c i No

iii No

20 a

11 a 8500 m

iv No

b 6500 m

c 300 seconds

d 1800 seconds b c = 100

12 a

iv No

iii No

Litter collected (kg) 5

b 22 °C ii Yes

Weeks 1

c $400 000

13 a Negative gradient

b −2.5

c y = −2.5x + 105

d y = 95

2 3 4 5 6 7 8 9

b Example answer:

Litter collected (kg) 5

Reasoning

14 a 52.5 m

b Yes, because the prediction is determined using interpolation.

2 1

15 Yes, if we examine the scatterplot and compare with the other points. It follows the same relationship.

Weeks 1

2 3 4 5 6 7 8 9

c Example answer: y = −0.2x + 4

16 a Interpolation. The speed of 70 km/h falls within the range of known data points, which is from 40 km/h to 100 km/h.

778

b A = 100 hectares

c No, because it is using extrapolation.

c Yes d i Yes

b y = 29 °C

c Yes, because it is using interpolation.

d Approximately 2.4 kg e No, because it is using extrapolation.

b Extrapolation. The prediction for the revenue for the next year is beyond the range of known data points (2015 to 2019).

21 a For each 1000 km increase in distance driven, the least-square line predicts that the maintenance costs increase by $990.

c Extrapolation. The number of hours of study at 15 hours does not fall within the range of known data points, which spans from 2 to 14 hours.

b A car that is not driven at all is predicted to cost $14 620 to maintain.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

c No, because for a car that hasn’t been driven, it should have little to no cost to maintain.

d $50 260

Fluency

e Yes, because it is an extrapolation close to the data range where the cost is likely to increase along with distance driven, and the correlation coeﬃcient is high. f Y es, a car that is driven more will have more wear and tear and would be likely to have greater maintenance costs. 22 a 95 m

b 50 m

c Option B

23 Any scatterplot with weak correlation or very few points. For example: 9 8 7 6 5 4 3 2 1

b Over-generalisation c Comparing apples and oranges d Cherry-picking 6 a False

b True

c True

d False

7 a Yes

b No

c No

d Yes

8 a Yes

b No

c Strategic choice of measure of centre (mode) d Yes 9 a Cherry-picking

x 1

2 3 4 5 6 7 8 9

10.07 Misrepresentations of data Understanding 1 a 100%

b 193%

c No

2 a Not representative, convenience bias and likely missing certain age demographics or those without reliable internet. b Not representative c Representative d Representative 3 a False

b False

Date

Cases

October 15

October 18

200

October 21

174

October 24

339

October 27

251

October 30

398

c True

b C

10 The claim is directed to any person that uses the product. However, the sample size is small and not representative of all people that might use the product. There is no information on whether the product was properly tested. There is no information on the sampling method used or the method for collecting the data.

24 Neither interpolation, not extrapolation would be overly reliable. This data set is more nonlinear, so a non-linear model would be more appropriate for making predictions.

5 a Causation and correlation

d True

11 a Biased question b When giving customers the menu, the servers can simply inform the customers that the mushroom risotto is a new menu item. 12 The tick marks on the vertical axis are not evenly spaced. 13 They do not correspond to the tourist numbers they are supposed to represent. In particular, on the corners the sizes are exaggerated. 14 The title is persuasive and the vertical axis scale doesn’t start at 0. It also does not give an idea of the trend overall. 15 By using a logarithmic scale, the graph appears to show a slight increase after taking the drug, and a steady decline of the drug in the blood over time. This may lead people to think that the drug concentrations in the blood are relatively stable over time. When converted to a normal scale, the inital increase would be much more signiﬁcant. 16 a Russia

b U.S.

c No

Answers mathspace.co

779

Reasoning

Problem-solving

17 Kang has used only the statistics that are favourable to him.

22 Graph 1 is misleading because the vertical axis does not start at zero therefore the difference in the number of days is exaggerated.

19 a The axis is broken too close to the top of the Mar 95 column, so it is not clear where the scale break is.

23 a The y-axis scale is not shown and therefore the Argentinian column appears taller than it should be. There is actually a large difference in the rates of the USA and Argentina. b

b It is not recommended for graphs such as column graphs where the viewer takes in the comparative areas of the different categories, using a truncated axis will exaggerate any differences in the data.

b The graph with the vertical axis that goes from 0 to 20.

15 700 14 100

15000 10000

7000

5000

330

258

Brazil Argentina

USA

Italy

Germany Norway

b Example answer: Although the scale on the y-axis does not begin from 0, it is reasonable because the difference in the data is small. If the scale began from 0, it would be diﬃcult to choose a scale that would allow the reader to identify the exact values for each year.

Chapter 10 review

b One possible answer: The survey, conducted by an anti-ageing skin company, suggests half of the regular smartphone users are showing early ageing signs. It’s unsurprising that the study recommends reduced screen time and speciﬁc skincare products. c One possible answer: The article’s headline is sensationalised, potentially causing unnecessary worry as it lacks context on the study’s broader use.

780

22 300

20000

24 a It has made the increases in the 4-year intervals 1992-1996 and 1996-2000 appear faster than they really are (relative to the rate in the 2-year interval 1990-1992).

c Since 1920, the temperature has increased by around 0.8 °C. The claim states that a further increase by 0.7 °C in the next few decades could lead to an increase in catastrophic events. This shows that increases by less than 1 °C are actually signiﬁcant increases in temperature. 21 a One possible answer: The survey claims to use over 2000 samples, but doesn’t clarify the selection process. This raises doubts about its accurate representation of all frequent smartphone users.

COVID-19 Testing by Country

25000

c A line graph with clear indication of a broken axis below the lowest point in March. 20 a The graph with the vertical axis that goes from 13 to 15.

Number of COVID-19 tests per million people

18 By compressing or expanding the horizontal axis it can appear as if the trend is gradual or quite abrupt.

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

1 a

Truck

Sedan

SUV

Total

Under 30

Over 30

Total

b 2 a 354 students c 27.94%

c b d 8941 students

3 a Living area b Yes, there appears to be an association. 4 a Time of day preference

b Yes

5 a Education level

10 a

b Column percentage frequency table c

High School or less

Sales (in thousands) 25

Some Bachelor’s Graduate university degree degree

Less than 48.08% 2 hours

36.36%

6.52%

3.51%

3 to 38.46% 6 hours

36.36%

21.74%

21.05%

7 to 13.46% 12 hours

21.82%

54.35%

54.39%

Over 12 hours

0.00%

5.45%

17.39%

21.05%

Total

100%

d Yes

5 Month 1 2 3 4 5 6 7 8 9 10 11 12

b Moderate, positive, linear c No d No, correlation does not imply causation. 11 a i

6 a January

b February

c Instrument sales

d No, there does not appear to be an association as the corresponding segments are similar in size.

15 10

7 a Yes, time spent studying a subject is the independent variable and the number of chapters is the dependent variable. b No c Yes, the duration of sunlight exposure is the independent variable, and plant growth is the dependent variable. 8 a

5 x 5

ii y = x + 4 b i

iii Positive

16 14

Days of exercise

12 10

8 7

x 2 4 6 8 10 12 14 16 18

2 1 5

Sick days 25

b Yes, the statement is correct. The scatterplot shows that the data forms a decreasing, linear relationship, meaning that as a student exercises more, they get sick less. 9 a Positive and strong

b Negative and weak

iii Negative

12 a The y-intercept suggests that when the radish is 0 days old, it has no height above the ground. b Yes, it is plausible that a seed will not sprout above the ground in the same day it is planted. c 19 days 13 a y = 33.75

b Interpolation

14 a M = $941.40 b No, because it is using extrapolation. c 25 days d Yes, because there is a strong correlation and interpolation was used for the prediction.

Answers mathspace.co

781

15 a Mistaking causation and correlation

c Directed

d Directed

b Over-generalisation

e Undirected

f Undirected

c Strategic choice of measure of centre

g Directed

h Directed

16 a Cherry-picking b The sector for Prefers AlphaModel is much smaller than the other two sectors, but it is only 3% smaller than the other two percentages. c A column chart would be more appropriate because it would show that all three phones are preferred by approximately the same percentage of people.

10 a i 9

ii 33

b i 7

ii 63

11 a 0

b 2

c 0

12 a i A = 1, B = 1, C = 2, D = 1, E = 1 ii Not connected b i A = 2, B = 2, C = 1, D = 3 ii Connected c i A = 3, B = 2, C = 4, D = 2, E = 3

11.01 Introduction to networks

ii Connected

Understanding

ii Not connected

1 a True

b False

e True

f False

c False

d i A = 0, B = 1, C = 2, X = 1 d True

c C

Reasoning 13 Jonah has counted the intersection of edges as cutting the edge into two pieces which is not correct. The number of edges is 4.

b A, C, D

2 a 3 edges

d D

3 a Non-valid

b Valid

c Non-valid

d Valid

e Valid

f Valid

g Valid

h Non-valid

14 Yes, it is possible if you connect the two odd vertices, A and E, all vertices are even. Problem-solving 15 a Undirected

4 a Edges = 3, Vertices = 4

d Directed

b Edges = 5, Vertices = 5 c Edges = 4, Vertices = 5 d Edges = 3, Vertices = 4 e Edges = 4, Vertices = 5 g Edges = 5, Vertices = 6 h Edges = 5, Vertices = 6

5 a 6 vertices and 9 edges b Directed c Vertex P

d S and U

e Undirected

f 5

16 a 3

b 53

c 100

17 a 1

b 20

c No

11.02 Network representations Understanding 1 a Not simple

7 a 5 vertices and 3 edges

c Not simple

b One edge 8 a Yes

b No

9 a Undirected

782

c Yes

c Directed

e Undirected f Undirected

This happens because each edge has two ends which are counted when ﬁnding the degree of a vertex.

Fluency

b 9 edges

b Directed

18 The sum of the degrees is 2 + 2 + 3 + 1 = 8. The number of edges is 4. Double 4 is 8 therefore the sum of the degrees is equal to double the number of edges.

f Edges = 5, Vertices = 6

6 a 6 vertices

d 2

d No

b Undirected

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

b Simple d Not simple

2 Complete as the graph is simple and each vertex is joined to every other vertex.

3 a i The ﬁber optic cables

b Complete as the graph is simple and each vertex is joined to every other vertex.

ii The cities iii V = 7

iv E = 11

v Yes

vi No

b i The wires ii The electrical components iii V = 8

iv E = 9

v No

vi No

c i The roads

ii The houses

iii V = 6

iv E = 10

v Yes

vi No

d i The roads

iv E = 10

v No

vi No

11 a 4 minutes b 8 minutes c 38 minutes

ii The towns

iii V = 7

c Since a loop refers to an edge that connects a vertex to itself, having one means the currency will be converted to itself. So, there are no loops because a currency doesn’t need to convert into itself. There would be no conversion fee for staying in the same currency.

12 a 124 likes b 117 likes c This graph is directed and weighed.

Young Temora

Boorowa Harden

b No

c Yes

d No

Reasoning

Cootamundra Junee

13 a Yes

Yass

5 B Fluency 6 a Yes. James and Jenny’s vertices are connected by an edge which indicates that they played together. b James, Jenny and Belinda 7 a Yes. Xanthe and Tina’s vertices are connected by an edge which indicates that they have worked together.

14 No, it cannot be redrawn as a simple graph. A simple graph is undirected with no loops or multiple edges. If we changed edges from directed to undirected or removed edges or loops to make it simple, then it would be a different network. So. it is not possible without changing the network. 15 No. If there is only one vertex, then the only edge would be a loop which is not allowed for simple graphs. 16 There are four simple graphs with three vertices.

b • Tina and Xavier • Tina and Derek • Xanthe and Xavier 8 a Aaron has worked with Rochelle and Bianca b • Aaron and Mario • Rochelle and Bianca 9 a Teribithia b Oz c A directed edge from Oz to Teribithia 10 a Based on the graph, the edge with the highest value represents the highest conversion fee. The value of 7 between New Zealand Dollars and Euros is the largest so, the conversion between New Zealand Dollars and Euros has the highest fee percentage.

We can rotate and move around the vertices, but they would represent the same information. 17 a lt is an app where people are friends. Otherwise, it would need to be a directed graph. b It would not be possible to make this a non-simple graph. To do so, you would need to do be a friend with yourself to make a loop or be a friend with a person multiple times, neither of which are possible. c It would be possible to make a non-simple graph using a different app with a followerfollowing set-up as it would be a directed graph.

Answers mathspace.co

783

Problem-solving 18

19 a

Fluency 4 a Example answer:

b Example answer:

c Example answer:

Expo Centre

Hillsboro Beaverton

Rose Quarter Pioneer Square

5 a i Example answer: b i

Airport Gateway

Clackmas Town Centre

b Airport - Gateway - Rose Quarter - Pioneer Square - PSU b

Kingfisher

Small Fish

Foxes

Tadpole

Water Beetle

ii f = 4

ii f = 3

c i Example answer: A

Squirrels

Snail Plants

Algae

ii f = 2

Killer Whale

ii E = 7

6 a i Graph A Tuna

iii D = 14 iii D = 16

Zooplankton

11.03 Planar graphs Understanding 1 a f=5

iv D = 2E

7 a i 6

ii 5

iii 9

iv Yes

b i 5

ii 5

iii 8

iv Yes

c i 8

ii 4

iii 10

iv Yes

d i 6

ii 4

iii 8

iv Yes

8 a i v=6

b e=8

c v=5

iv D = 2E ii E = 8

b i Graph B

Mackerel

d 5+5−8=2

ii e = 6

iii

e Yes

2 B

iv f = 2 3 a e=7

784

D C

Rabbits

Hawks

Frog

E F

Gresham

PSU

20 a

Example answer:

b f=7

c v=4

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

v 2

b i v=5 iii

ii e = 5

Problem-solving

13 a f = 2

b Example answer:

D B

14 a e = 4

iv f = 2

v 2

9 a Yes

b No

c No

d Yes

10 a f = 9

b e=8

c v = 11

d e=7

e v=9

f f=3

A B

15 a v = 2

b Example answer:

Reasoning 11 a

b Example answer:

A B C D

Vertices

Faces

Edges

6 13

v+f−e

2 2 2 2 Y Y Y Y

11.04 Walks, paths, and trails

Understanding

Planar? (Y/N) No. of vertices with odd degree

b The number of vertices with odd degree is always even. Each edge has two ends so each edge is connected to two vertices. Adding an edge means adding 1 degree each to two vertices, or a total of 2 degrees. This means the total degree of all the vertices is always even. If we have an even number of vertices with odd degrees and any number of vertices with even degrees, the total degree is even, since even times any integer is even and the sum of even is always even. If we have an odd number of odd degree vertices and any number of vertices with an even degree, the total degree will be odd, since odd times odd is also odd and the sum of odd and even is odd. This is not possible, as the total needs to be even. 12 The cube has f = 6, e = 12, v = 8, therefore v + f − e = 8 + 6 − 12 = 2. Euler’s rule holds true.

16 a 1

b 1

17 a e = 5

b f=3

1 a i No

ii Yes

v Yes

vi No

b i Yes

ii No

2 a ii: Path c i: Walk

iii Yes

iv No

iii Yes b iii: Cycle d iv: Closed walk

3 A path is a non-repeating sequence of vertices and edges between two distinct vertices, while a cycle is a closed loop that starts and ends at the same vertex. Both are concepts in graph theory, but paths have distinct endpoints and no repeated vertices, while cycles form closed loops. Fluency 4 a Example answers: • E, D, C, B, A • A, B, C, E, D • D, A, B, C, E b Example answers: • P, R, Q, S, T • T, S, P, R, Q • Q, P, R, T, S 5 a W, X, W, X, W b B, W, Q, B or B, Q, W, B

Answers mathspace.co

785

6 Example answer: A, B, C, A, E

17 a F, A, G, E = 19

b E, C, B = 8

c E, C, D, B = 19

d X, Y, V = 20

7 Example answer: E, F, D, B, E, C, A

e T, X, Y, Z = 17

8 a Example answers: • A, B, D, H, G, C, A • E, F, G, H, D, B, E b Example answer: F, C, D, B, E, G, F 9 a A, C, E, B, A b T, S, U, T, P, R, T or T, P, R, T, S, U, T 10 a F, D, C, B, D, A, E, F or F, D, C, B, A, D, E, F b S, T, R, P, U, Q, S 11 a S, Q, S or Q, S, Q b R, T, P, R or T, P, R, T 12 a Example answers: • E, D, C • A, B, D b Example answers: • D, C • A, B, C c Example answers: • B, A • A, B, C d Example answers: • E, C, B • A, D, C e Example answers: • A, B, D, G, F • D, B, A, C, F, G • E, B, D, C, A

Reasoning 18 A path does not contain repeated vertices (other than the start and the end), while a walk may contain repeated vertices. In other words, a path is a walk with distinct vertices. 19 Example answer: A real-life example would be GPS navigation systems that help to determine the most efficient route between two locations, considering factors like distance, travel time, and current traffic conditions. The significance of finding the shortest path in this context lies in its ability to assist drivers in saving time, fuel, and navigating through congested or unfamiliar routes with greater efficiency. Problem-solving 20 E, B, H = 35 21 A, D, F, G, H 22 L, J, I, H, F, E, B, A or L, J, I, H, D, A 23 D, C, E, G, F, I 24 a

3 5

c A, G, E

b 10 mins

13 3 paths

b P, Q, U, W, X

25 a 5.5 hrs

b Q, S, T, P or Q, R, T, P

26 A, C, B, E, F, H, J

c X, H, Y, Q, B or X, Y, Q, B, W

27 a

d P, U, T, S, Q or P, R, T, S, Q

9 3

15 Many answers including: A, B, C, D, E, A or A, B, D, E, C, A or A, D, C, B, E, A 16 a Closed walk b X, H, Y, X or X, Y, H, X

786

11 G

f Example answers: • E, F, G, H, D, B • A, B, D, H, G, C

14 a B, Y, Q, X or B, H, W, X

9 5

Mathspace Victorian Curriculum – Year 10 Pre-General mathspace.co

5 11

b 14 km

c A, F, E

Chapter 11 review

11 a e = 7

1 a Non-valid

b Valid

2 a i 7

ii 10

iii A = 6, B = 1, C = 2, D = 3, E = 3, F = 3, G = 2 b i 6

ii 7

b f=4

c We can use Euler’s formula and show that v + f − e = 2. In this case: 5 + 4 − 7 = 9 − 7 = 2, so Euler’s formula holds true. 12 a

iii A = 2, B = 2, C = 1, D = 2, E = 3, F = 4

4 11

3 a

D 8

A 5 B

b 12 km

9 A

c A, B, D

b The degree c • Astrid and Christina • Bernice and Christina • Christina and Edwina • Bernice and Dora d A complete graph 4 a Walk

b Neither

c Walk and path

d Walk and path

5 a S, T, S

b R, T, S, P, R

6 E, D, A, C 7 A, B, D, F, G = 8 8 C, B, E, F, I 9 a i Y B

ii f = 2

H X W

b i P

ii f = 4

U S T

10 a 6

b 9

c No

Answers mathspace.co

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