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A n s w e r s

341

answ

ers

CHAPTER 18 Number systems: the Real Number System

Exercise 18A — Classification of numbers

1

2

Exercise 18B — Set notation

1 a

{3, 9, 15, 21}

b

{3, 6, 9, 15, 18, 21}

c

{6, 12, 18, 24}

d

{12, 24}

e

{3, 9, 12, 15, 21, 24}

f

{ } or

∅

g

{3, 6, 9, 12, 15, 18, 21, 24}

h

{6, 18}

i

{3, 6, 9, 15, 18, 21}

j

{3, 9, 15, 21}

2 a

{11, 12, 13, 14, 15, 16, 17, 18, 19}

b

{10, 11, 12, 18, 19, 20}

c

{10, 11, 12, 13, 14, 20}

d

{10, 13, 14, 15, 16, 17, 20}

e

{10, 15, 16, 17, 18, 19, 20}

f

{13, 14, 15, 16, 17, 18, 19}

g

{ } or

∅

h

{ } or

∅

i

{15, 16, 17}

j

{10, 13, 14, 15, 16, 17, 18, 19, 20}

k

{11, 12}

l

{10, 15, 16, 17, 20}

3 a

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

b

{1, 6, 7, 8, 9, 10, 12}

c

{1, 2, 3, 4, 7, 9, 11}

d

{1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

e

{1, 7, 9}

f

{5, 13, 14}

g

{6, 8, 10, 12}

4

Exercise 18C — Recurring decimals

1 c f g j k l n o q r t w x y z

2

7

Irene. It can also be written as .

Exercise 18D — The Real Number System

1

2

3

9 a

2

i

b

3

i

c

−

5

i

d

−

10

i

Exercise 18E — Surds

1 b d f g h i l m o q r s t w z

Exercise 18F — Simplifying surds

1

Answers

a

Q

b

Q

c

Q

d

Q

′

e

Q

′

f

Q

g

Q

h

Q

′

i

Q

j

Q

k

Q

l

Q

m

Q

′

n

Q

o

Q

′

p

Q

q

Q

r

Q

′

s

Q

′

t

Q

′

u

Q

v

Q

′

w

Q

′

x

Q

y

Q

′

z

Q

a

Q

b

Q

c

Q

d

Q

e

Q

f

Q

′

g

Q

′

h

Q

i

Q

′

j

Undefined

k

Q

′

l

Q

′

m

Q

′

n

Q

o

Q

p

Q

q

Q

′

r

Q

′

s

Q

t

Q

u

Q

′

v

Q

w

Q

x

Undefined

y

Q

′

z

Q

3

B

4

E

5

C

6

D

a

T

b

T

c

F

d

F

e

T

f

T

g

T

h T i T j T5 B 6 A 7 D 8 E

a b c d

e f g h

i j k 2 l

m n 1 o 3 p

q r s t 3

u v w x

y 1 z

3 E 4 D 5 C 6 E

a J + b Q c Q d Qe J + f Q g J + h J +

i J + j J − k J − l Qm J − n J + o Q p Qq J + r J − s J + t Qu J − v Q w J + x J +

y J − z Qa J + b Q′ c Q′ d Qe J − f Q′ g J − h Q′i Q j Q k J + l J −

m Q n Q o J + p Q′q J − r Q′ s Q′ t J −

u Q′ v J + w Q x Q′y J + z Q′a Q b Q′ c Q d J −

e J + f J g J + h Qi J + j Q′ k J + l Jm Q′ n Q o Q′ p Qq J + r J − s Q t Qu J − v Q′ w J − x J+

y J z Q′4 C 5 C 6 B 7 D

2 A 3 E 4 B 5 C6 Any perfect square 7 m = 4

a b c d

e f g h

i j k l

29--- 7

9--- 8

9--- 5

9---

49--- 1

6--- 17

45------ 19

45------

3145------ 32

45------ 28

45------ 53

99------

433------ 34

99------ 367

495--------- 361

999---------

427999--------- 868

1665------------ 323

999--------- 152

333---------

1318------ 157

300--------- 1237

1980------------ 5611

9000------------

268999--------- 2

13------

0.02

2 3 3 2 2 6 2 14

3 3 5 3 5 5 3 11

3 6 2 15 4 7 7 2 18A➔

18F

Answers Page 341 Thursday, December 30, 1999 2:03 PM

342 A n s w e r san

swer

s

2

3

Exercise 18G — Addition and subtraction of surds1

2

3

Exercise 18H — Multiplication of surds1

2

Exercise 18I — Division of surds1

m n o p

q r s t

u v w x

y z

a b c d

e f g h

i j k l

m n o p 5

q r s t

u v w x

y z

a 4a b 9ab c

d e f

g h i

j k l

m n o

p q r

s t u

v w x

y z

4 E 5 C 6 D 7 C

a b c

d e f

g h i

j k l

m n o

p

a b

c d

e f

g h

i j

k l

m 15 − 10 + 10 n 0

o p

q

2 17 5 6 6 5 13 2

2 22 3 15 9 2 10 2

7 5 8 5 8 7 3 15

9 5 7 6

4 2 15 2 24 10 24 7

36 5 10 17 21 6 40 2

30 3– 18 7 28 5– 18 30

64 3 10 2 2

2 2 6 2 3 2 213--- 15 20 5 3

2--- 7 7

2--- 11

8 3 32--- 5–

6a 2

3ab 6 3a 10b 4a 3ab

13a2 2 5a2b 6 13ab 2ab

2a2b3 3ab 2ab2 17ab 4x3 5y

5x3y2 5 24x y 20xy 5x

14xy 7xy 54c3d2 2cd 9c2d2 14d

18c3d4 5cd 28c5d5 6 22ef23---e2 f 3 30 7e5 f 5 2ef 3

4---e6 f 2 7 f

19---xy4 6xy 1

3---x5y6 3

7 5 17 2 8 3

19 7 15 5 5 3+ 15 2 7 6+

4 11 5 13 13 2

10 7 11 5– 3 6– 7 2– 5 6+

17 3 18 7– 5 xy 8 x 3 y+

x 5 y 7 xy+–

10 2 3–( ) 2 2

5 5 6+( ) 6 6– 2 3+

7 3 10 2

4 5 5 5

14 3 3 2+ 11 4 11–

3 6 6 3+ 17 2

10 15

8 11 22+– 39 3

12 30 16 15–

r

s t 0

u v

w x

a b

c d

e f

g h

i j

k l4 D 5 E 6 A 7 E 8 B

9 a cm b cm

c cm d m

e m f m

a b c d

e f g 10 h

i j k 27 l

m n o p 126

q 120 r 144 s t

u v w 2 x

y z

a b

c d

e f

g h

i j

3 a 98 cm2 b 75π cm2

c m2 d m2

e m2 f m2

4 E 5 C 6 D 7 A 8

a b c 2 d

e f g 4 h

i j k l

m n 1 o 1 p

q 1 r s t 2

2 5– 5 2– 2 30– 2 15+

12 ab 7 3ab+72--- 2 2 3+ 3 2–

15 2 58--- 3

34 a 6 2a– 52 a 29 3a–

6 6ab 32a 2 6a 8a 2+ +

a 2a a 2 2a+

3a a a2 3a+ a2 a+( ) ab

4ab ab 3a2b b+ 3 ab 2a 1+( )

6ab 2a– 4a2b3 3a+ 2a b–

12 2 6 6 8 3+( )

18 2 3– 2 5+( ) 3π 5

18 2 2 5+( ) 21 11

14 55 42 2 6

4 3 6 2 5 15

3 7 4 10 30 3

10 33 96 6 180 5

120 3 360 3

2 6 623--- 4

3--- 5

25--- 6 3 3

x2y y x2y3 x

3a4b2 2ab 5abc2 2abc

6a5b2 2b 6a3b4

3x2y2 10xy 15x6y2 292---a2b4 5ab 1

2---a3b2 2ab

20 11 6 6

45π 96 10+( ) 72 15

15360 2

5 7 2 3

6 1535---

34

------- 4 23

---------- 52

------- 5 6

2 315--- 2 6

45--- 3 3 2 17

25---


A n s w e r s 343

answ

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2

7

8

9

Exercise 18J — The Distributive Law

1

2 a

b

c

d

e

f

g

h

i

j

3

4

9

Exercise 18K — Rationalising denominators

1

2

7

Exercise 18L — Rationalising denominators using conjugate surds1 a

b

c

d

e

f

g

h

u v w x

a b c d

e f

3 B 4 E 5 A 6 C

a m b cm c m

d m e cm f cm

a b c

a b 126 L

a b

c d

e f

g h

i j

a b

c d

e f

g h

i j

a −46 b 18 c 11d 50 e 6 f −2g 10 h −5 i 7j 51 k 26 l −1m 7 n 17 o 63p 44 q 53 r 343s 76 t 17 u x − yv 2x − 3y w 9x − 16y x 4x3 − 25yy xy(49x − 9y) z xy(81x − 25y)

5 A 6 C 7 E 8 D

xy-- 1

x3y2----------- 2

x3y4----------- 6x xy

2xy 3y2x2

3y-------------

4 a3

----------3b2 2b

2a a--------------------

2 2

3m3n m----------------------

15

2m2n2-----------------

4 13 4 6 7 11

3 7 5 13 152

------ 5

4 55

3------- 2 2

35 2

21 6 3+ 3 10 7 5–

2 5 10– 6 10+

126 2 14 3– 10 21 4 6–

72 14 30+ 30 15– 80 6+

24 2– 12+ 60 3 24 5+

3 10 9 2 5 5 15––+

35– 11–

4– 40 3–

24 3 18 30– 8 10– 60+

112 140 3– 24 6 90 2–+

2 55 2 22– 4 15– 4 6+

10 35 14 14 15 10– 42–+

180 30 3– 18 6– 9 2+

15x 26 xy 8y+ +

4x 2 5xy 10y–+

27 10 2+ 16 4 15+

18 6 5+ 53 10 6+

35 12 6+ 53 12 10+

104 60 3+ 14 6 5–

10 2 21– 37 8 10–

a b

c

a b c d

e f g h

i j k l

m n o

a b

c d

e f

g h

i j

k

3 B 4 D 5 C 6 A

a b c

57 12 15–5410 1368 15–

57 12 15–----------------------------------------

59 12 15– 9 5 6 3–+

5 22

---------- 7 33

---------- 4 1111

------------- 4 63

----------

2 217

------------- 102

---------- 2 155

------------- 3 355

-------------

5 66

---------- 4 1515

------------- 5 714

---------- 8 1515

-------------

8 2149

------------- 8 1057

---------------- 103

----------

2 2+ 3 10 2 33–6

---------------------------------

12 5 5 6–10

------------------------------ 9 105

-------------

3 10 6 14+4

--------------------------------- 5 63

----------

3 22 4 10–6

--------------------------------- 21 15–3

--------------------------

14 5 2–6

---------------------- 12 10–16

----------------------

6 15 25–70

-------------------------

217

----------± 153

----------± 2 63

----------±

5 2–

3 6+3

----------------

2 2 5+3

------------------------

2 6 7+17

------------------------

8 11 4 13+31

---------------------------------

2 21 35–14

-----------------------------

15 15 20 6–13

------------------------------------

9 11 9+20

---------------------- 18G➔

18L

Answers 43 Thursday, December 30, 1999 2:03 PM

344 A n s w e r san

swer

s

i

j

k

l

m

n

o

p

q

r

s

t

u

v

w

x

y

z

2 a

b

c

d

e

f

g

h

i

j

3

4 a

b

c

d

e 44

f

g

h

5 Yes

6 a

b

Exercise 18M — Modulus1

6 a

b c R and y ≥ 0

5 14 2 10 25 7– 10 5–+155

------------------------------------------------------------------------

12 2 17–

19 4 21–5

-------------------------

9 2 154+4

------------------------------

20 2 9 10 4 30+ +– 9 6–2

--------------------------------------------------------------------------

312-------

3 3 2 6+18

---------------------------

10 3 15 6 9 2 27+ + +( )–42

----------------------------------------------------------------------

12 3 4– 3 6 2–+52

----------------------------------------------------

60 2 10 30 6 10– 5 6–+35

------------------------------------------------------------------------

115 31 21+148

-------------------------------

71 12 33–17

----------------------------

18 2 10 6 9 3– 15–+

102 48 6+95

----------------------------

9 154– 132 42 2 8 77–+ +50

---------------------------------------------------------------------------

7 3 9+3

-------------------

21 5 6 14– 5 70 20––27

------------------------------------------------------------------

6– 6 2 10 2 5–+ +2

---------------------------------------------------------

9 2 8+14

-------------------

9 7 13 3–120

------------------------------

36 14– 48 210+231

----------------------------------------------

6 7 2–

45 15 14 9 10 6 35+ + +( )–5

----------------------------------------------------------------------------

66 24 6+5

-------------------------

a −2 b

7 D 8 B 9 C 10 A

a 19 b c 0.75 d

e 8 f 2a g 12 h

i 3.21 j 0 k − l 4

m 10 n 10 o a2b2 p −16q 27 r 15 s −72 t −54

u v −8 w −11 x 30

y −3a z −6cd2 C 3 E 4 D 5 B

x –2 –1 0 1 2 3 4 5 6

y –8 –6 –4 –2 0 2 4 6 8

|x | 8 6 4 2 0 2 4 6 8

5 4 14–

959 281 77 182 7 6 11+ + +629

-----------------------------------------------------------------------------

3 7 65 16 11–+28

---------------------------------------------

41 6 30+( )–12

---------------------------------

1419------

6 3519

-------------

210 2 120–41

-------------------------------

200 2 126–41

-------------------------------

99 238 50 400 2–1681

---------------------------------------------

99 120 50 460 2–1681

---------------------------------------------

103 90 2–

7 2 4+

295 2 382–49

-------------------------------

11 5+6

-----------------------

42 5 28 7+17

---------------------------------

14--- 15

12---

23---

12---

y = |2x – 4| y = 2x – 4y

x2

–4

4


A n s w e r s 345

answ

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Chapter review

Multiple choice

Short answer1 a Irrational, since equal to non-recurring and non-

terminating decimal.b Rational, since can be expressed as a whole

number.c Rational, since given in a rational form.d Rational, since it is a recurring decimal.e Irrational, since equal to non-recurring and non-

terminating decimal.2

a {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}b {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}c {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}d {∅}e {2}f {3, 5, 7, 11, 13, 17, 19}g {∅}h {1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19}i {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 17, 18,

19, 20}j {3, 5, 7, 11, 13, 17, 19}k {2, 3, 5, 7, 11, 13, 17, 19}

3

4

5 a , , ,

b , ,

6

7

8

9

10

11 a

b

c

12

13

14

15 m2

16

Analysis

1

2

3

CHAPTER 19 MatricesExercise 19A — Operations with matrices

1

2

3

4

5

6

1 A 2 D 3 A 4 D 5 E6 B 7 E 8 B 9 C 10 A

11 D 12 A 13 E 14 A 15 D16 B 17 C 18 E 19 C 20 A

a b c

a J − b J + c Q d Q′

a b

a b

a m b cm

c m d 22 cm

a 27 b

εB

C

4, 6, 8, 10, 12, 14, 16,

18, 20

9,15, 1

23, 5, 7, 11, 13,17, 19

A

6299------ 337

900--------- 157

165---------

2m20m------ m3 8m3

25mm16------ 20

m------

72x3y4 2xy14---x2y5 xy–

25 3 3ab ab

5 17 4 6–( )

26 4 2–( )720 2

23 6 48–

2323 594 14–50

-------------------------------------

2277 606 14–50

-------------------------------------

51 12 14– 18 7– 27 2+

5 74

----------

a b c 3a

a −11 b −3

a amps b amps

c amps d amps

a cm b cm3

c cm3 d cm3

a cm3 b cm c cm

Matrix Order 2, 1 element 1, 3 element

A 2 × 2 8 —

B 3 × 1 5 —

C 1 × 4 — 10

D 2 × 3 4 4

E 3 × 3 1 2

a b c d

a b c

d e

a C b D c E d A e B

a b c

d e f

a b c d

x 5y2

------------- x2y 2

3 7 3–40

-----------------------

32

-------

5 103

------------- 5 1026

----------------

135 3819

------------------- 5 342

-------------

2 37 18π 37

54π 3 18π 37 3 3+( )

360 103 10π

π----------------- 2 15π

π-----------------

3 3

0 9

7– 3

8 5

6 3–

2– 8

3 6–

2– 1–

4– 6

8 14

9– 6

12 12

11 6

4– 20

9 9

0 27

7 18

12– 4–

2 0 14

4 0 0

6 0 18

0 8 0

0 10 16

0 12 0

2 8 14

4 10 16

6 12 18

3 4 21

6 5 8

9 6 27

4 0 28

8 0 0

12 0 36

1– 4 7–

2– 5 8

3– 6 9–

2 0

3 1–

2– 0

6– 2

1– 0 1

2 3 1–

2 6–

1– 12–

2– 1– 18M➔

19A


346 A n s w e r san

swer

s7 Different orders

8

9

10

11

12

Exercise 19B — Multiplying matrices1 a A (2 × 2), B (2 × 2), C (3 × 2), D (1 × 2),

E (2 × 3), I (2 × 2)b CA, DB, AE, AI, IA, IB, A2, ECc (3 × 2), (1 × 2), (2 × 3), (2 × 2), (2 × 2), (2 × 2),

(2 × 2), (2 × 2)

d

2

3

4 a i ii iii

b All are I c Multiplicative inverses5

6

7 a b

c Carlton 120, Essendon 99, West Coast 74, Sydney 70

8 a b Shop A = $820, Shop B = $345

History of Maths1 Matrix theory and number theory2 Computer development3 Cross of Honour4 Caltech

Exercise 19C — Multiplicative inverse and solving matrix equations

1 a b

2

3

4

56 Answers will vary.

7

8 a D − det = 0 b E − det = 0c F − Not a square matrix

9

10

a b

a b

a True b True c False d True

a b No

a b c d

e f g h

i j

a A b C c D d B

a

b Sharks have a total of 32 points. Dolphins have a total of 31 points.

0 1 3 1

1 0 2 2

3 2 0 1

1 2 1 0

0 0 1 1

0 0 1 2

1 1 0 3

1 2 3 0

82 54

76 68

91 82

15 14 104

7 10 52

13 7 5 1 31 18 26

12 4 4 4 17 15 16

14 8 5 1 35 19 29

13 4 4 5 18 19 16

20 14

44 22

4 5

2 2–4– 18 8–

8– 8– 6

2 3–

4 5

2 3–

4 5

1 1

1 0

8– 21–

28 13

14 15

24– 30–

10 20

5– 10

8 26

4– 12

4 3

0 9–

2 7–

24 9

0 0

0 0

3 2

8– 5

2 0

0 3–

10 4–

24– 9–31 0

0 31

10 11–

16 1

0 0

0 0

3 2

8– 5

1 0

0 1

1 0

0 1

1 0

0 1

3

1

0

32

31

a 5 b 12 c −2 d −8 e 7 f 14

a b c

d e f

a C b E c D d A

a b c

d e f

a b

a b c

d e f

g h

18 12

14 15

10 14

9 16

6

1

10

25

12

AB 6 1 0

0 1= A 1– 1

6---B= B 1– 1

6--- A=

MN 2 1 0

0 1M 1–,–

12---N N 1–,–

12---M–= = =

15--- 10 3–

5– 2112------ 0 3

4– 2–

12--- 1 6

0 2–

18--- 1– 3

4 4–

17--- 5– 1

3 2–

114------ 4 1

6– 2

12--- 1 6

0 2–

14--- 1 2

2– 012 2

2– 1–

18--- 1 2

2– 12–

18--- 11– 2

4 0

18--- 1 2

2– 12–

0 8

1– 2–

18--- 2– 8–

1 0

12--- 31– 22–

24 18

12--- 5– 5

14 8–

16--- 6– 2

6– 4

12--- 18 23

12– 16–

130------ 78 103

24– 34–1 0

0 1

115------ 1 5–

2 5

115------ 132– 114–

186 162


A n s w e r s 347

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11

12

Exercise 19D — Applications of matrices1

2 a and b Answers will vary. c det = 0d i ii

e In i there are parallel lines; in ii there is only one line.

3 a E b B 4 a C b D5 16, 4 6 15, 10 7 Anh 8 $51 070

Investigation — Matrix multiplication using a graphics calculator

1

2

3

Exercise 19E — Matrices and transformations

1

23

4 (0, −1), (−1, 0)

5

6

78

b (2, 4), (8, 12), (12, −1)c Same as a.d i (0, 0), (−6, 8), (−10, −5)

ii Lengths, angles, area same; gradients − , , .

e i (0, 0), (−8, 6), (−10, −5)ii Lengths, angles, area same; gradients − , 2, .

f

9 a

c (3, 0), (0, 1) d e

f Answers will vary. g (4, 0), (0, 4)h Answers will vary.

Investigation — Commutativity of transformationsRotation and dilation are commutative.Reflection and dilation are commutative.All other combinations are generally not commutative.

a b

a x = −2, y = 1 b x = 1, y = 2c x = −2, y = 3 d x = 7, y = 4

a (5, −1) b (3, 0) c (10, 2)d (0, 0) e (4, 4) f (−2, −3)

a b c d

a (7, 0) b (0, 3) c (2, −8) d (0, 0)a i (3, −1) ii (4, 2) iii (−2, 4) iv (−1, −3)b i (−3, 1) ii (−4, −2) iii (2, −4) iv (1, 3)c i (1, 3) ii (−2, 4) iii (−4, −2) iv (3, −1)

2

1–

2

3

Fig. 19x

y

O 2 3

–3

–2

–14–3

Fig. 20

Both lines

x

y

O 2

–3

30

55

98

40.36 36.99 36.92

35.57 5.45 87.14

27.99 6.77 63.58

37.15 28.62 44.9

74 67

66– 4–

62 40–

69 42–

56 48

4–

0

4–

8–

3

3–

3–

4–

0 1–

1– 0

a b c d

e f g h

i j

a i (0, 2) ii (0, −2) iii (−2, 0) iv (1, − )

b i (−1, 3) ii (1, −3) iii (−3, −1) iv

c i (−2, –2)ii (2, 2) iii (2, −2) iv (−1 + , + 1)d i (−4, 0) ii (4, 0) iii (0, −4) iv (2 , 2)a D b B c Ea i 10, 11.2, 13.6 ii , , −

iii 80°, 46°, 54° iv 54.78

Transformation Matrix Length Gradient Angle Area

Translation No change

No change

No change

No change

Reflection No change Change No

changeNo

change

Rotation No change Change No

changeNo

change

b i (0, 0), (6, 5), (10, 1) ii (0, 0), (9, 5), (15, 1)iii (0, 0), (3, 10), (5, 2) iv (0, 0), (3, −10), (5, −2)v (0, 0), (9, 10), (15, 2)vi (0, 0), (9, 15), (15, 3)

0 1–

1 0

1– 0

0 1–

0 1

1– 0

1 0

0 1

0 1

1– 0

1

2------- 1–

2-------

1

2------- 1

2-------

12---– 3

2-------

32

-------– 12---–

32

-------– 12---–

12--- 3

2-------–

12--- 3

2-------–

32

------- 12---

12---– 3

2-------–

32

------- 12---–

3

3 3+2

---------------- 3 3–2

------------- 12---+,

3 33

43--- 1

2--- 13

4------

43--- 1

2---

134

------

34--- 4

13------

2

4

1– 0

0 1

0 1–

1 0

Figure 22

x

y

O

21

1 2 3 4 5

345

(5, 1)

P(3, 5)

3 0

0 1

k 0

0 1

19B➔

19E


348 A n s w e r san

swer

sChapter review

Multiple choice

Short answer

4 a b (5, −1)

5 a (2, 3), (−3, 2) b (3, 2), (3, −2)6 a No change

b Rotation of 180°c Reflection in the x-axisd Reflection in the y-axise Reflection in the line y = xf Rotation of 270°g Reflection in the line y = −xh Dilation of 3 about Oi Dilation of −3 about Oj Rotation of 90°

Analysisa 1: Despatch for Deluxe model takes 1 hour.b 14: Packaging at Plant 1 has a wage rate of $14 per

hour.c (3 × 3), (3 × 2), (3 × 2)

d

e The total costs for the Standard model at Plants 1 and 2.f The assembly costs for each model at Plant 1.g i $529.50, ii $514.00

CHAPTER 20 Algebra and logicExercise 20A — Statements, connectives and truth tables

4 B

5 D

1 D 2 C 3 B 4 D 5 C6 C 7 A 8 B 9 A 10 B

11 E 12 C 13 C

1 b 2 a, f, g, i 3

1 a Opinion b T/F c Td Question e T f Opiniong Instruction h T/F i Fj Near-statement

2 a The car has 4 seats.The car has air-conditioning.

b The Department of Finance was over budget in 1998.The Department of Defence was over budget in 1998.

c Bob went to an hotel.Carol went to an hotel.Ted went to an hotel.Alice went to an hotel.

9 3

6– 0110------ 1 2

3– 4

433.25 420.50

529.50 514.00

605.25 587.50

d To be a best-seller a novel must be interesting to the reader.To be a best-seller a novel must be relevant to the reader.

e Sam will win the trophy.Nancy will win the trophy

f You can choose vanilla ice-cream for dessert.You can choose strawberry ice-cream for dessert.You can choose fruit for dessert.

g There are some statements which cannot be proved to be true.There are some statements which cannot be proved to be false.

h Most of my friends studied Mathematics.Most of my friends studied Physics.Most of my friends studied Engineering.Most of my friends studied Law.Most of my friends studied Arts.

3 a John and Mary rode their bicycles to school.b The book you want is in row 3 or 4.c The weather is cold and cloudy.d Many people read novels or history.

e In a recent poll 80% preferred jazz or classical music.

f Two is an even prime number.Or, Two is the only even prime number.

6 a p q r

T T T

T T F

T F T

T F F

F T T

F T F

F F T

F F F

b p q r s p q r s

T T T T F T T T

T T T F F T T F

T T F T F T F T

T T F F F T F F

T F T T F F T T

T F T F F F T F

T F F T F F F T

T F F F F F F F


A n s w e r s 349

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6 c p q r s t p s r s t p q r s t p q r s t

T T T T T T F T T T F T T T T F F T T T

T T T T F T F T T F F T T T F F F T T F

T T T F T T F T F T F T T F T F F T F T

T T T F F T F T F F F T T F F F F T F F

T T F T T T F F T T F T F T T F F F T T

T T F T F T F F T F F T F T F F F F T F

T T F F T T F F F T F T F F T F F F F T

T T F F F T F F F F F T F F F F F F F F

7 a p q p ∧ q

p = Sydney on time T T T

q = Perth fully booked T F F

p ∧ q F T F

F F F

b p q r p ∧ q ∧ r

p = John passed T T T T

q = Zia passed T T F F

r = David passed T F T F

p ∧ q ∧ r T F F F

F T T F

F T F F

F F T F

F F F F

c p q r (p ∧ q) ∨ r

p = Alice does dishes T T T T

q = Renzo does dishes T T F T

r = Carla does dishes T F T T

(p ∧ q) ∨ r T F F F

F T T T

F T F F

F F T T

F F F F

d p q r p ∧ (q ∨ r)

p = female member T T T T

q = student T T F T

r = professor T F T T

p ∧ (q ∨ r) T F F F

F T T F

F T F F

F F T F

F F F F

8 p ~p p ∧ ~p

T F F

F T F

9 a p q p ∧ ~q b p q ~p ∧ ~q

T T F T T F

T F T T F F

F T F F T F

F F F F F T

c p q r (p ∧ q) ∧ r p q r (p ∧ q) ∧ r

T T T T F T T F

T T F F F T F F

T F T F F F T F

T F F F F F F F 20A➔

20A


350 A n s w e r san

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s

Not equivalent

Not equivalent

Equivalent

Not equivalent

Equivalent

Equivalentc Brackets have no effect on expressions with a

single ∨ or ∧ operator, but have an effect if they are mixed up together.

d p q p ∨ ~q

T T T

T F T

F T F

F F T

f p q r (p ∨ q) ∨ r p q r (p ∨ q) ∨ r

T T T T F T T T

T T F T F T F T

T F T T F F T T

T F F T F F F F

10 a It is raining and I bring my umbrella.b It is raining or I bring my umbrella.c It is not raining and I bring my umbrella.

11 a Peter and Quentin like football.b Peter or Quentin like football.c Peter likes football or Quentin does not like

football.

12 p q ~(p ∨ q) (~p ∧ ~q)

T T F F

T F F F

F T F F

F F T T

13 p q ~(p ∨ q) ~p ∨ ~q

T T F F

T F F T

F T F T

F F T T

14 a p q (p ∧ q) ~p (p ∧ q) ∨ ~p (p ∨ q) (p ∨ q) ∧ ~p

T T T F T T F

T F F F F T F

F T F T T T T

F F F T T F F

e p q ~p ∨ ~q

T T F

T F T

F T T

F F T

b p q (p ∨ q) ~p (p ∨ q) ∨ ~p (p ∨ ~p)

T T T F T T

T F T F T T

F T T T T T

F F F T T T

15 p q r (p ∧ q) ∨ r p ∧ (q ∨ r)

T T T T T

T T F T T

T F T T T

T F F F F

F T T T F

F T F F F

F F T T F

F F F F F

16 a p q r (p ∧ q) ∧ r p ∧ (q ∧ r)

T T T T T

T T F F F

T F T F F

T F F F F

F T T F F

F T F F F

F F T F F

F F F F F

b p q r (p ∨ q) ∨ r p ∨ (q ∧ r)

T T T T T

T T F T T

T F T T T

T F F T T

F T T T T

F T F T T

F F T T T

F F F F F


A n s w e r s 351

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Exercise 20B — Valid and invalid arguments

Conclusion is true whenever all premises are true(3rd row), thus a valid argument.

Conclusion is true whenever all premises are true (1st, 5th, 7th and 8th rows).

Conclusion is true whenever all premises are true (4th row), thus a valid argument.

1 p q ~p ~q (p → q) (~q → ~p)

T T F F T T

T F F T F F

F T T F T T

F F T T T T

2 p q ~p ~q (p → q) (~p → ~q)

T T F F T T

T F F T F T

F T T F T F

F F T T T T

3 If it is bread then it is made with flour; If it is made with flour then it is bread; If it is not made with flour then it is not bread; If it is not bread then it is not made with flour.

4 C 5 A

6 a Conclusion: My pet is fluffy.b Conclusion: Two is the only prime number.c Conclusion: Growing apples depends on good

irrigation.

7 a and c are valid.

8 b All men are liarsMary is not a liar.Mary is not a man.

d Cannot be made into a valid argument.e All musicians can read music.

Louise is a musician.Louise can read music.

9 D

10 p q p ∨ q ~p

T T T F

T F T F

F T T T

F F F T

11 a p q r p → q q → r p → r

T T T T T T

T T F T F F

T F T F T T

T F F F T F

F T T T T T

F T F T F T

F F T T T T

F F F T T T

b p q p → q ~q ~p

T T T F F

T F F T F

F T T F T

F F T T T

c p q r s p → q r → s p → q ∧ r → s p ∧ r q ∧ s

T T T T T T T T T

T T T F T F F T F

T T F T T T T F T

T T F F T T T F F

T F T T F T F T F

T F T F F F F T F

T F F T F T F F F

T F F F F T F F F

F T T T T T T F T

F T T F T F F F F

F T F T T T T F T

F T F F T T T F F

F F T T T T T F F

F F T F T F F F F

F F F T T T T F F

F F F F T T T F FConclusion is true whenever all premises are true (1st row). 20A➔

20B


352 A n s w e r san

swer

s

13 C

p = If elected with a majorityq = My government will introduce new tax laws

b Conclusion is false whenever all premises are true (3rd and 4th rows), thus an invalid argument.

15 a p → q~p ~q

Conclusion false when premises true (3rd row), therefore invalid.

Valid argument

Invalid argument

Invalid argument18 2nd statement is usually implied, 1st and 3rd are

stated in advertisement. Invalid argument — affirming the consequent.

Exercise 20C — Techniques of proof1 p = She plays well

q = She wins

Therefore, this is a tautology.

Therefore, this is a tautology.

12 a Disjunctive syllogismb Modus tollensc Modus ponens

14 a p q p → q ~p ~q

T T T F F

T F F F T

F T T T F

F F T T F

b p q ~p ~q p → q

T T F F T

T F F T F

F T T F T

F F T T T

16 a p q r p → q r → ~q p → ~r

T T T T F F

T T F T T T

T F T F T F

T F F F T T

F T T T F T

F T F T T T

F F T T T T

F F F T T T

b p q r ~p ∧ ~q r → p

T T T F T

T T F F T

T F T F T

T F F F T

F T T F F

F T F F T

F F T T F

F F F T T

Invalidargument

c p q ~p → ~q

T T T

T F T

F T F

F F T

17 a Valid — hypothetical syllogismb p → q; r → q; p → r, invalidc Valid – modus tollensd Valid – constructive dilemmae p = The team plays well

q = The offence was goodr = The defence was goodp → (q ∨ r)~q ∧ ~p~r

p q r (q ∨ r) p → (q ∨ r) ~q ∧ ~p ~r

T T T T T F F

T T F T T F T

T F T T T F F

T F F F F F T

F T T T T F F

F T F T T F T

F F T T T T F

F F F F T T T

p q p → q (p → q) ∨ ~q

T T T T

T F F T

F T T T

F F T T

2 p q p → q ~p → ~q (p → q) ∨ ~p → ~q

T T T T T

T F F T T

F T T F T

F F T T T


A n s w e r s 353

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Therefore, this is not a tautology.

Hence, this is a valid argument.

Hence, this is an invalid argument (row 3).

Hence, this is a valid argument.

3 p q p ∧ q (p ∧ q) ∨ ~q

T T T T

T F F T

F T F F

F F F T

4 p q r p → q q → r (p → q) ∧ (q → r) p → r (p → q) ∧ (q → r) → (p → r)

T T T T T T T T

T T F T F F F T

T F T F T F T T

T F F F T F F T

F T T T T T T T

F T F T F F T T

F F T T T T T T

F F F T T T T T

5 a p q r ~p → ~q ~q → r (~p → ~q) ∧ (~q → r) ~p → r (~p → ~q) ∧ (~q → r) → (~p → r)

T T T T T T T T

T T F T T T T T

T F T T T T T T

T F F T F F T T

F T T F T F T T

F T F F T F F T

F F T T T T T T

F F F T F F F T

b p q r p → ~q q → ~r (~p → ~q) ∧ (q → ~r) p → ~r (~p → ~q) ∧ (q → ~r) → (p → ~r)

T T T F F F F T

T T F F T F T T

T F T T T T F F

T F F T T T T T

F T T T F F T T

F T F T T T T T

F F T T T T T T

F F F T T T T T

c p q ~p → ~q (~p → ~q) ∧ q (~p → ~q) ∧ q → p

T T T T T

T F T F T

F T F F T

F F T F T20C➔

20C


354 A n s w e r san

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s

Hence, this is an invalid argument (row 1)

6 a ~p → ~q b ~p → q~q → r ~p~p → r ~q

c ~p → ~q d ~p → ~q~q ~q → ~r~p ~p → ~r

7 D

16 Let x = 2, then x 2 = 4, let x = −2 , then x2 = 4.17 Demonstrate that 2 is both a prime number and is

even.18 Prove by counter-example.

19 Consider the contrapositive statement.20 Consider what would happen if x > y + z. This

would imply that the shortest distance from A to C is not a straight line!

Exercise 20D — Sets and Boolean algebra

3 B

5 A, C, D, E6

7

d p q ~p → q (~p → ~q) ∧ p (~p → ~q) ∧ p → ~q

T T T T F

T F T T T

F T T F T

F F F F T

8 If x2 is even, then write it as 2n.(2n)2 = 4n2

If a number is multiplied by 4 then it is even.Therefore 4n2 is even.

9 If a number, x, is even, then x2 is even.24 is even, therefore 242 is even.

10 Assume n is not odd, show that n2 is even.11 a Assume a ≠ b, multiply both sides by x.

b Assume n < 2, write it as 2 – x and square it.c Assume n is not divisible by 2, therefore it is

odd, therefore write it as (2x + 1), then square it.12 Assume it is positive and compare it with the

product of 2 positive numbers of the same magnitude.

13 Assume that a is the smallest positive real number.Let x = Since a > 0, then x > 0 and x < a (property of division).This contradicts the assumption that a is the smallest positive number.

14 Assume is rational, so that = , where a

and b are integers that have no common factors.

Therefore 2 = , or a2 = 2b2.

Therefore a2 is a multiple of 2 and therefore a is a multiple of 2 (from a2 = a × a)Since it is a multiple of 2, write a = 2x.

Therefore a2 = 4x2 = 2b2.

Therefore b2 = 2x2 and is thus a multiple of 2.Therefore both a and b are multiples of 2 and have a common factor of 2.This contradicts our initial statement, so it must be false.

15 Assume n is the largest possible integer.Let x = n + 1. Therefore x > n, which contradicts our initial statement.

a2---

2 2ab---

a2

b2

-----

1

2

4 a A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}b B = {4, 8, 12, 16, . . .}c C = {2}d D = {Jack, Queen, King}e E = {∅}f F = {9, 8, 7, 6, …}

8 a b

c

9 a {4}

A B

C

A B

C

A B

C

A B

C

A B254981

ε

A B22446688

ε

A B

ε

A B

ε

A B

C


A n s w e r s 355

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11 Part 1: Show that (A . B) + (A′ + B′) = I(A . B) + (A′ + B′) = (A + A′ + B′) . (B + A′ + B)

= (I + B′) . (A′ + I)= (I) . (I) = I= I QED

Part 2: Show that (A . B) . (A′ + B′) = O(A . B) . (A′ + B′) = A . B . A′ + A . B . B′

= O . B + A . O= O + O= O QED

13 Answers will vary.14 (p ∧ q) ∧ ~p = (p . q) . p′ = (p . p′) . q = O . q = O

p ∧ ~p = p . p′ = O QED

Exercise 20E — Digital logic

2 C

b Used where there are 2 people who can activate the light separately.

5 D

6

9 E

b Same truth table as question 8.

b {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}c {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}d {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}e {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

10 a (B ∪ C) = {1, 2, 4, 6, 8, 9, 10}b A ∩ (B ∪ C) = {1, 2, 4, 6, 8, 9, 10}c (A ∩ C) = {1, 4, 9}d (A ∩ B) = {2, 4, 6, 8, 10}e (A ∩ B) ∪ (A ∩ C) = {1, 2, 4, 6, 8, 9, 10}

12 a A + B b I c A + B d A . B

15 a A + B + A′ + B′ = A + A′ + B + B′ = I + I = Ib (A + B) . A′ . B′ = A . A′ . B′ + B . A′ . B′

= OB′ + OA′ = O + O = Oc (A + B) . (A + B′) = (A + B) . A + (A + B) . B′

= A + A = Ad A . B + C . (A′ + B′) = A . B + C . (A . B)′

= A . B + C

16 Rule Dual

1st commutative 2nd commutative

1st identity 2nd identity

1st complement 2nd complement

1st associative 2nd associative

1st distributive 2nd distributive

1 x y Q

0 0 0

0 1 0

1 0 0

1 1 1

3 a x y z Q

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

b Q = (x . y) + z

4 a x y Q

0 0 1

0 1 0

1 0 0

1 1 1

7 a b c Output

0 0 0 0

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

8 a b c Output

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

10 a a b c Output

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

P x

y

Q

20D➔

20E


356 A n s w e r san

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s

b When S1 = 0, the system is disabled; the safe can’t be opened.When S1 = 1 and S2 = 0, the alarm rings.When S1 = 1 and S2 = 1, the safe can be opened.

14

15

16

17

Chapter reviewMultiple choice

Short answer1 a m ∧ j b c ∨ (w ∧ s) c l ∧ e ∧ p

3 Converse = If she sends her children to good schools, the politician is intelligent.Contrapositive = If she doesn’t send her children to good schools, the politician is not intelligent.Inverse = If a politician isn’t intelligent she doesn’t send her children to good schools.

The conclusion (column 7) is false when the premises (columns 4, 5, 6) are all true (*). Invalid argument

11 a b c d Output

0 0 0 0 0

0 0 0 1 0

0 0 1 0 1

0 0 1 1 0

0 1 0 0 0

0 1 0 1 0

0 1 1 0 1

0 1 1 1 0

1 0 0 0 1

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 0

12 a S1 S2 a b c Q R

0 0 0 0 1 0 0

0 1 0 1 0 0 0

1 0 1 0 1 0 1

1 1 1 1 0 1 0

13 a a . b = [(a . b)′]′a . b = (a′ + b′)′

b

18 a iii a′ + bb

a

cb

a

Q

ab Qc

ba

a'c

Q

Qb

a

c

d a . be (a . b)′ + bf

a b (a . b)′ (a . b)′ + b

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 1

1 C 6 C2 D 7 C3 A 8 A4 D 9 D5 B 10 B

2 p q (p ∧ q) (~p ∧ ~q) (p ∧ q) ∨ (~p ∧ ~q)

T T T F T

T F F F F

F T F F F

F F F T T

4 r I g ~r → I ~I → g r ~I

T T T T T T F *

T T F T T T F *

T F T T T T T

T F F T F T T

F T T T T F F

F T F T T F F

F F T F T F T

F F F F F F T

Qa

b

Qb

a


A n s w e r s 357

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5 Let: a = ; multiply both sides by x.ax = 1a(x1) = 1 = x0 (Definition of x0)

a = = x –1 (Definition of division of indices)

6 Assume < xx < x2 (Square both sides)1 < x (Divide both sides by x)This contradicts the initial condition, so the assumption is wrong.

8 (A + A′ . B) . (B + B . C) = (A + A′ . B) . (B . I + B . C) {B . I = B}= (A + A′ . B) . B(I + C) {Distributive law}= (A + A′ . B) . B . I {I + C = I}= A . B . I + A′ . B . B . I {Distributive law}= A . B + A′ . B {B . I = B, B . B = B}= (A + A′) . B {Distributive law}= I . B {A + A′ = I}= B QED

9

10 Q = A . [(B + C)′ + B . C]

Analysis

Premises (S1, S2) are all true in rows 4,12, 13, 15, 16Conclusion (S3) is true in rows 3, 4, 7, 8, 9, 10, 11,12, 13, 14, 15, 16Conclusion is true when premises are true, so this is a valid argument.

7 a {11, 13, 17, 19, 31, 41, 61, 71, 91}b {2, 3, 7, 11, 13, 16, 17, 23, 29, 31, 37, 41, 43,

47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97}c {61}

1x---

x0

x1

-----

x

1 a S1 = (g → n) ∧ (a → c)S2 = ~n ∨ ~cS3 = ~g ∨ ~a

Q

c

ba

Q

bc

a

b, c g n a c g → n a → c S1 S2 = ~n ∨ ~c S3 = ~g ∨ ~a

T T T T T T T F F

T T T F T F F T F

T T F T T T T F T

T T F F T T T T T

T F T T F T F T F

T F T F F F F T F

T F F T F T F T T

T F F F F T F T T

F T T T T T T F T

F T T F T F F T T

F T F T T T T F T

F T F F T T T T T

F F T T T T T T T

F F T F T F F T T

F F F T T T T T T

F F F F T T T T T


358 A n s w e r san

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Argument is valid because conclusion is true when all premises are true (row 15).

e Answers will vary.

b a → b and b → ac Yesd Yes, because a ↔ b implies both a → b and

b → a.e Equivalences: i, iii, iv, vi, viii

e This circuit ‘decodes’ or distinguishes the inputs, which are 00, 01, 10 and 11. Only one of the outputs = 1 for each of the 4 possible inputs.

f 3-bit: 3 NOT, 8 AND4-bit: 4 NOT and 16 ANDn-bit: n NOT and 2n AND

CHAPTER 21 Undirected graphs and networksExercise 21A — Vertices and edges

1

2 a, b, d, f

3 c e

4 d and e

5 There are 18 edges.deg(A) = 3 deg(E) = 2deg(B) = 5 deg(F) = 6deg(C) = 4 deg(G) = 5deg(D) = 5

Note: This is only one possibility.

6 a b

c d

e A connected graph cannot be drawn.

d a b c d a → b b → c ~d → a ~c

T T T T T T T F

T T T F T T T F

T T F T T F T T

T T F F T F T T

T F T T F T T F

T F T F F T T F

T F F T F T T T

T F F F F T T T

F T T T T T T F

F T T F T T F F

F T F T T F T T

F T F F T F F T

F F T T T T T F

F F T F T T F F

F F F T T T T T

F F F F T T F T

2 a a b a → b a ↔ b

T T T T

T F F F

F T T F

F F T T

3 a–d b a Q R S T

0 0 0 0 0 0

0 1 0 1 0 0

1 0 0 0 1 0

1 1 0 0 0 1

a i V = 4 ii E = 6 iii deg(A) = 2deg(B) = 4deg(C) = 3deg(D) = 3

b i V = 4 ii E = 3 iii deg(A) = 2deg(B) = 1deg(C) = 2deg(D) = 1

c i V = 4 ii E = 3 iii deg(A) = 2deg(B) = 2deg(C) = 2deg(D) = 0

d i V = 4 ii E = 8 iii deg(A) = 4deg(B) = 4deg(C) = 3deg(D) = 3

e i V = 5 ii E = 8 iii deg(A) = 2deg(B) = 2deg(C) = 3deg(D) = 3deg(E) = 4

f i V = 6 ii E = 7 iii deg(1) = 3deg(2) = 3deg(3) = 4deg(4) = 1deg(5) = 2deg(6) = 1

A B

CD

A CD

EB

AB

G CF

E D

A

B

C

DE

A

BC

D

E

A B

E

DC

B

E D

CA


A n s w e r s 359

answ

ers

7 a deg(A) = 3 b deg(A) = 3deg(B) = 4 deg(B) = 2deg(C) = 3 deg(C) = 3deg(D) = 4 deg(D) = 2deg(E) = 2 deg(E) = 2

c deg(A) = 4 d deg(A) = 2deg(B) = 6 deg(B) = 2deg(C) = 6 deg(C) = 2deg(D) = 6 deg(D) = 2deg(E) = 4 deg(E) = 2

e A connected graph could not be drawn.8 49

10

11 A 12 A 13 E

Exercise 21B — Planar graphs1

2 a

b Connected, 2, Euler’s

3 a Euler’s formula holds, and the given graph is a connected planar graph.

b Sum of degrees is 14. There are 7 edges. The sum of degrees is twice the number of edges

c Two odd-degree vertices

4

5

6 True

7 a, b, c, g, h, i, j

8

9 B 10 D 11 E

Investigation — Schlegel diagrams

a b

a 8 b 12c deg(A) = 2 deg(B) = 4 deg(C) = 5

deg(D) = 4 deg(E) = 3 deg(F) = 3deg(G) = 1 deg(H) = 0

d No, since vertex H is not connected to any other vertex.

e Draw an edge from vertex H to any of the other vertices.

a b

c d

V R E V + R – E

i 5 5 8 2

ii 5 4 7 2

iii 6 6 10 2

iv 6 5 9 2

v 5 2 4 3

vi 7 8 13 2

vii 4 2 4 2

viii 4 1 2 3

A B

C D

H

G

F

E

a i 4 ii 2 iii deg(A) = 2deg(B) = 2deg(C) = 2deg(D) = 2

iv 8 v 4

b i 5 ii 5 iii deg(A) = 3deg(B) = 3deg(C) = 4deg(D) = 3deg(E) = 3

iv 16 v 8

c i 5 ii 4 iii deg(A) = 2deg(B) = 3deg(C) = 4deg(D) = 3deg(E) = 2

iv 14 v 7

d i 6 ii 6 iii deg(A) = 4deg(B) = 3deg(C) = 2deg(D) = 4deg(E) = 5deg(F) = 2

iv 20 v 10

e i 6 ii 5 iii deg(A) = 2deg(B) = 4deg(C) = 3deg(D) = 2deg(E) = 3deg(F) = 4

iv 18 v 9

f i 7 ii 8 iii deg(A) = 3deg(B) = 3deg(C) = 4deg(D) = 5deg(E) = 3deg(F) = 5deg(G) = 3

iv 26 v 13

vi 2

a 0 b 4 c 2 d 2 e 2 f 6

a E = 9 b E = 9 c R = 3 d R = 4e V = 5 f V = 4 g E = 11 h R = 7i V = 6 j V = 4

CubeOctahedron

Icosahedron Dodecahedron 21A➔

21B


360 A n s w e r san

swer

sExercise 21C — Eulerian paths and circuits12

34

56 a Yes

b A–B–J–F–G–J–I–G–E–I–H–E–F–B–C–E–D–C–A–D–H–A

7 D 8 B 9 B 10 E

Exercise 21D — Hamiltonian paths and circuits12 O–2–4–1–6–7–8–5–3–O (This is one of many

possible answers)

3

4

5 i, ii, iii, vi6 O–D–E–F–C–B–A–O, 69 km7 D–E–F–G–H–I–D, 53 km8 a 22 b 31 c 60 d 329 a

b D–A–B–C–G–F–E–D, 96 km10 a 19 km b 27 km c 35 km11 a 24 b 2912 934 km13 a b 1348 km c Yes d 1986 km

14 D15 A

Exercise 21E — Trees1 a, c and f are trees.

b is not a tree as the network contains a circuit.d is not a tree as the network is not connected.e is not a tree as the network contains multiple edges and a circuit.f is not a tree as the network contains multiple edges and a circuit.g is not a tree as the network contains multiple edges and a circuit.h is not a tree as the network contains multiple edges and a circuit.

2

a i, ii, iii, v b i, v

Vertex

Degree of

vertex in

graph i

Degree of

vertex in

graphii

Degree of

vertex in

graphiii

Degree of

vertex in

graphiv

Degree of

vertex in

graphv

Degree of

vertex in

graphvi

A 2 3 2 3 4 1

B 2 2 2 3 2 4

C 2 3 3 3 4 3

D 2 2 3 3 4 2

E 4 2 3

F 1

No. of odd

vertices

0 2 2 4 0 4

Eulerian path

(Yes/No)

Y Y Y N Y N

Eulerian circuit

(Yes/No)

Y N N N Y N

a 0 or 2 b Even

a i deg(A) = 4deg(B) = 4deg(C) = 2deg(D) = 5deg(E) = 2deg(F) = 5deg(G) = 2deg(H) = 4

ii Yes iii No

b i deg(A) = 5deg(B) = 5deg(C) = 5deg(D) = 5deg(E) = 4

ii No iii No

c i deg(A) = 5deg(B) = 4deg(C) = 4deg(D) = 5

ii Yes iii No

d i deg(A) = 3deg(B) = 4deg(C) = 3deg(D) = 5deg(E) = 4deg(F) = 4deg(G) = 5deg(H) = 2

ii No iii No

a There are 2 odd vertices b Edge FI

a i, ii, iii, iv, vi b i, iii, iv

b d e

g h

5 1

3

24

A

F

B CD E

A

D E

F

G

C

B

15

12 12

15

16

2914

22

19

15

10

8

A

B

CD

E

280300

219

291

258

453

402

353

345

417


A n s w e r s 361

answ

ers

3

45 Possible answers:

6

7 321 km 8 C 9 D


Short answer1

2

3

4

5 a V = 6, E = 10b

c There are 2 odd-degree vertices. For an Eulerian circuit to exist there must be 0 odd-degree vertices.

d A–B–A–D–B–C–F–D–E–F–D6 a i Yes

ii A–D–I–J–E–B–C–F–K–J–L–H–M–B–Aiii No, M–B–C–F–K–J–E–B–A–D–I–J–L–H–M

b i Yesii Eulerian

iii L–H–M–B–A–D–I–J–E–B–C–F–K–J–LL–J–K–F–C–B–E–J–I–D–A–B–M–H–L

7 a 1–5–3–4–6–2–11–2–6–3–4–5–1

b 48c Postal deliveries, paper deliveries, garbage and

recycling collection, etc.d Yes

8

9 a

10 43

Analysisa A planar network is a graph without crossing edges.b There are 3 roads to/from F.c deg(A) = 2 deg(B) = 4 deg(C) = 3

deg(D) = 2 deg(E) = 4 deg(F) = 3d 18 e 9 f S = 2Eg F–A–B–F–E–D–C–B–E–C h iii 51 kmj A minimal spanning tree would connect each of the

towns for the least cost.k 83 l B–A–F–E–D–C–B m iii

a i ii 4 edges

b i ii 10 edges

c V − 1

a 9 b 10 c 17 d 18 e 21

a b

c d

a 10 b 13 c 22d 30 e 24 f 33

1 D 2 C 3 E 4 A 5 B6 C 7 C 8 D 9 B 10 C

11 C 12 D 13 D 14 E 15 D

a 9 b 16 c deg(A) = 4deg(B) = 3deg(C) = 4deg(D) = 4deg(E) = 2deg(F) = 5deg(G) = 2deg(H) = 2deg(I) = 0

d No

e Connect vertices G and I to any of the other vertices, that is, to A, B, C, D, E or F.

a i Yes ii No iii Yes iv Yesb i 17 ii 13 iii 1 iv 3

6

D

A

B

JH

CE

F

GI

A D

E F

CB

deg(A) = 3 deg(B) = 4 deg(C) = 2deg(D) = 5 deg(E) = 2 deg(F) = 4

a b 1620 km

c Yes d 2239 km

b i 5 ii 10 iii 13

c i ii

iii

E B

CD

A

489

420470 495 290

320

300

310 400462

A D

B

E

C

F

2C D

A

E

B

F

C

D

A

E

B

F

GH

IJK

C D

A

E

F

G

H

I

J

KL

M

N

B

21C


362 A n s w e r san

swer

sCHAPTER 22 Sampling for attributesExercise 22A — Populations and samples1 0.255

2

5

6 D

7

8

9

10

11

12

13

14 As this is a random experiment, answers will differ from sample to sample.

Exercise 22B — Distribution of sample proportions

4

5

8 a 0.319; true proportion = 0.417b


10 As this is a random experiment, answer will differ from sample to sample; hence a dot plot of the distribution will also differ.Theoretical population proportion = 0.125

11

12 a Multiply the sample size by for each clinic — this is the number of patients. Then add, and divide by the total number in the entire survey.

b (or approximately 32%)

c No, because sample sizes are not the same, each is not really significant; only the total count

matters.

Exercise 22C — The 95% spread of sample proportions1

a Population b Sample c Populationd Sample e Sample f Population

3 0.25 spoiled; 0.75 unspoiled 4 0.6a 12.9% b Samplec Probability not a representative sample — few

girls study Engineering.

a 0.0833b Very little — it is a very small sample; perhaps

5–10% defective rate in the population.

a 0.3 and 0.33b Statistician B used a much larger sample and is

therefore more likely to be closer.

a 0.74 b 148

a 0.65b Not very representative, population proportion = 0.5

a 0.3b No, because only Frankston was sampled, not the

entire state.

a 0.3b Better than question 11 — a more representative

sample.

a 0.375 b Theoretical result = 0.388

15 Theoretical result = 0.22 16 Theoretical result = 0.25

1 0.46 2 0.58 (real proportion 0.56) 3 0.746

6 D 7 E

4

3

2

1

0

0.6

0.65 0.7

0.75 0.8

0.85 0.9

0.95 1

4

3

2

1

6

5

0

0.20

00.

267

0.33

30.

400

0.46

70.

533

0.60

00.

667

0.73

30.

800

0.86

70.

933

a 0.65 b

ClinicSample

size Patients

Abbotsford 0.429 7 3

Brunswick 0.385 13 5

Carlton 0.300 10 3

Dandenong 0.333 15 5

Eltham 0.400 10 4

Frankston 0.250 8 2

Geelong 0.381 21 8

Hawthorn 0.273 11 3

Inner Melbourne 0.133 15 2

N. Melbourne 0.231 13 3

S. Melbourne 0.294 17 5

E. Melbourne 0.333 9 3

W. Melbourne 0.357 14 5

St. Kilda 0.375 8 3

a b [0.0, 0.4]

43210

0.00

0

0.12

5

0.25

0

0.37

5

0.62

5

0.75

0

0.50

0

4567

3210

0.3

0.4

0.5

0.6

0.8

0.9

0.7

p̂

54171--------- 0.316=

p̂

p̂

81012

6420

0.3

0.4

0.5

0.6

0.8

0.2

0.10

0.7


A n s w e r s 363

answ

ers

2

345


7

8

9

1011

They roughly have the same middle (mean), but the interval is half as ‘wide’.

Exercise 22D — Confidence intervals12 0.05853

b Maximum standard error (0.05) at = 0.5, 0.00 at either end and symmetric about 0.5.

4 B5

b When the sample size is multiplied by 4, the standard error is divided by 2.When the sample size is multiplied by 2, the

standard error is divided by .6 See the results to question 5. Also consider values of

.7

8 C9

12 [0.145, 0.175]13 Liberals [0.232, 0.270] Labor [0.209, 0.247]

Democrats [0.085, 0.111] Undecided [0.402, 0.446]14 Practice [0.252, 0.596] or [25.2%, 59.6%],

Real exam 57.6%; Within the confidence interval, so studying did not significantly improve his result.It could be due to random chance.

15 Practice [16.9%, 49.7%],Real exam 57.6%; Outside the interval, so studying did improve her result.

16

17

Exercise 22E — Using confidence intervals

4 a

b

Since n is much smaller, the width of each interval is larger.

5 a p = 0.2; [164, 0.236]p = 0.4; [356, 0.444]p = 0.6; [556, 0.644]p = 0.8; [764, 0.836]

a b [0.1, 0.7]

a 0.2025 b 0.195a 0.413 b 0.414a i ii [0.000, 0.200]

b 0.096 c 0.086

a [0.4, 1.0] b [0.4, 1.0]c 0.7333, 0.736; Both results compare well and are

fairly similar.a As this is a random experiment, answers will

differ from sample to sample.b p = 0.42, = 0.41 c = 0.415a [0.15, 0.8] b [0.25, 0.7]c A significant change in the interval due to the

large number (400) of samples.a 0.492 b [0.36, 0.64]a 0.505 b [0.42, 0.59]

a 0.0733 b 0.0933 c 0.0367 d 0.0467

a 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

s.e. 0.00 0.03 0.04 0.046 0.049 0.05 0.049 0.046 0.04 0.03 0.00

a n 1 2 4 8 16 32 64 128 256 512 1024

s.e. 0.5 0.354 0.25 0.177 0.125 .0884 0.0625 0.0442 0.03125 0.0221 0.0156

a [0.013, 0.307] b [0.133, 0.507]c [0.087, 0.233] d [0.227, 0.413]

0 0.1 0.2 0.3 0.4 0.595% C.I.

[0.0, 0.0]

[0.04, 0.16]

[0.12, 0.28]

[0.208, 0.392]

[0.302, 0.498]

[0.4, 0.6]

0.6 0.7 0.8 0.9 1.095% C.I.

[0.502, 0.698]

[0.608, 0.792]

[0.72, 0.88]

[0.84, 0.96]

[1.0, 1.0]

2025303540

151050

0.3

0.4

0.5

0.6

0.8

0.2

0.10

0.7

81012

6420

1 2 3 4 5 6 7 80

p̂ p̂

p̂

p̂( )p̂

p̂( )

2

p̂ 1 p̂–( )

p̂

p̂

10 [0.006, 0.094] 11 [0.367, 0.737]

a 0.059 b [0.040, 0.277]c 0.23 is within the interval, so the assertion cannot

be denied.d 0.23 ± 2(0.059) = [0.112, 0.348]e The sample lies within the interval, so the

assertion cannot be denied.a 0.014, [25.3%, 30.9%]b No, because 35% is outside the interval.

1 [0.129, 0.291] 2 [0.131, 0.353] 3 [0.1, 0.2]

p Confidence interval

0 [0, 0]

0.1 [0, 0.2]

0.2 [0.067, 0.333]

0.3 [0.147, 0.453]

0.4 [0.237, 0.563]

0.5 [0.333, 0.667]

0.6 [0.437, 0.764]

0.7 [0.547, 0.853]

0.8 [0.667, 0.933]

0.9 [0.8, 1.0]

1 [1, 1]

0.70.80.9

0.60.50.40.30.20.1

Confidence interval

0

00.

10.

20.

30.

40.

50.

60.

70.

80.

9 1

p

22A➔

22E


364 A n s w e r san

swer

sb The intervals are much smaller since n = 500.

13 a No, because the confidence interval would be 0.75 ± 0.087 and 0.54 is outside this range.

b Yes, because 0.82 is in the range of the confidence interval.

14 3615 [0.81, 1.0] (Upper limit exceeded 1)


Short answer1 0.32823 [0.25, 0.50]45

6 [0.18, 0.53]7 Yes, because the confidence interval = [0.17, 0.25] and

a sample of which is within this range.

8 a No.b The confidence interval for good sets is

[0.92, 1.0] and a sample proportion of

is outside this range.

Analysis1 a

b

c 0.52 d [0.4, 0.7] e 0.112f [0.297, 0.743] g 0.128

h [0.264, 0.776]; Very close to C.I. in f, (s.e. almost equal to s.d.).Not close to the 95% spread because n was small (10) and there were only 20 samples taken.

2 Answers will vary.

CHAPTER 23 CombinatoricsExercise 23A — The addition and multiplication principles1 a AB BA CA

AC BC CBb 6

2 BG GB YB RBBY GY YG RGBR GR YR RY

3 ACB BAC CABABC BCA CBA

4567 a i 90 ii 90

b 1808 1269 C

10 C11 D12 10013 614 4815 25616 129617 108018 13319

20 a 200 b 40 c 50 d 29021 a 13 230

b 17 640Jack may wear 13 230 outfits with a jacket or 4410 outfits without a jacket. Therefore he has a total of 17 640 outfits to choose from. The assumption made with this problem is that no item of clothing is exactly the same; that is, none of the 7 shirts is exactly the same.

Exercise 23B — Factorials and permutations1 a 4 × 3 × 2 × 1

b 5 × 4 × 3 × 2 × 1c 6 × 5 × 4 × 3 × 2 × 1d 7 × 6 × 5 × 4 × 3 × 2 × 1

6 96 7 292 8 504 9 6710 364 11 267 12 17

1 A 2 D 3 C 4 D 5 C6 D 7 B 8 C 9 D 10 E

a 30 b 0.355

a 0.05 b 0.0128 c 0.067 d 0.1a [0.4, 0.6] b [0.864, 0.916]c [0.767, 1.00] d [0.3, 0.7]

9 1584 10 [0.086, 0.314]

Sample 1 2 3 4 5 6 7 8 9 10

Proportion, 0.3 0.4 0.8 0.5 0.5 0.7 0.6 0.5 0.4 0.4

Sample 11 12 13 14 15 16 17 18 19 20

Proportion, 0.4 0.4 0.6 0.7 0.5 0.7 0.5 0.5 0.5 0.5

0.8

0.6

0.4

0.20

00.

10.

20.

30.

40.

50.

60.

70.

80.

9

71400--------- 0.178=

91100--------- 0.91=

p̂

p̂

56

9

78

43210

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

a 42 b 210 c 840 d 2520a 24 b 6 c 12 d 24 e 1320a 49 b 252 c 16

a 1000 b 27c 271

272273281282283291292293

371372373381382383391392393

471472473481482483491492493


A n s w e r s 365

answ

ers

2

34 a n(n − 1)(n − 2)(n − 3)(n − 4)

b (n + 3)(n + 2)

c

d

5 a 8 × 7 = 56b 7 × 6 × 5 × 4 × 3 = 2520c 8 × 7 × 6 × 5 × 4 × 3 × 2 = 40 320

6

78 569 3024

10 218411 358 80012 12013 819014 336015 362 88016 479 001 60017 D18 E19

20

21

Exercise 23C — Arrangements involving restrictions and like objects1 3602 83 1603 104 12605 27 7206 1 307 504789

10111213 B14 D1516

Exercise 23D — Combinations

1

23

4 13655 2526

78 5609 100

10 59 4001112 201 37613 D14 C1516171819

Exercise 23E — Applications of permutations and combinations1

23 244 245 376 9926

7

89 D

10 D11 a 28

b 9 17 27 41 12 14, 9 17 27 41 12 37,9 17 27 41 12 34

c 612 a 84

b 7 15 25 32 10 12, 7 15 25 32 10 35,7 15 25 32 10 37

c 101314

Exercise 23F — Pascal’s triangle and the binomial theorem1

a 24 b 120 c 720d 3 628 800 e 8.717 829 12 × 1010 f 362 880g 5040 h 6a 3024 b 151 200 c 840 d 720

a b c

a 27 907 200 b 639 200 c 1 028 160

a 5P3 = 60 b 5P4 = 120 c 5P5 = 120

a 7P3 = 210 b 7P4 = 840 c 7P7 = 5040

a 6P6 = 720 b

a 5.4 × 1010 b 3.6 × 109 c 4.0 × 1010

a 120 b 20 c 60a 30 240 b 3024 c 672a 120 b 48 c 72a 1680 b 420 c 360a 1320 b 110

a 720 b 24 c OYSTER

a 36530 b 365P30

a b c d

a 8C2 b 9C3 c 8C0 d 10C4

a 1 b 20 c 120 d 220

1n n 1–( ) n 2–( )-------------------------------------

1n 2+( ) n 1+( )n n 1–( )

-------------------------------------------------------

9!3!----- 60 480= 5!

3!----- 20= 18!

13!-------- 1 028 160=

6P62

---------- 360=

8P33!

----------19P2

2!-------------

1P11!

----------5P00!

----------

a 495 b 11 c 1 d 1e 54 264 f 120 g 100 h 680a 2 598 960 b 65 780 c 65 780 d 2 467 400

a 28 b 120

a 120 b 10 daysa 57 b 4 days 6 hoursa 15 b 1 day 4 hoursa 8 145060 b 11 480 c 820a 220, 220 b 6435, 6435 c 10, 10 d 56, 56e 1, 1 f The value of nCr is the same as nCn − r.

a 45 b 120 c 120 d 210 e 105a 720 b 252 c 10 d 120 e 2.4 × 1018

a 210 b 126 c 84 d 140

a 126 b 56 c 21 d 70a 8008 b 5005 c 4004 d 4004

a 5005 b 455 c 6545 d 2 977 975a 455 b i 5005 ii 1 623 160iii 8 123 915 800 iv 834 451 800

Row0th 11st 1 12nd 1 2 13rd 1 3 3 14th 1 4 6 4 15th 1 5 10 10 5 16th 1 6 15 20 15 6 17th 1 7 21 35 35 21 7 18th 1 8 28 56 70 56 28 8 1 23A➔

23F


366 A n s w e r san

swer

s2 a 70

b 1 9 36 84 126 126 84 36 9 1c 45

3 a 8b 220c 10C0

10C1 10C2

10C3 10C4

10C5 10C6

10C7 10C8

10C9 10C101 10 45 120 210 252 210 120 45 10 1

4 a x2 + 2xy + y2

b n3 + 3n2m + 3nm2 + m3

c a4 + 12a3 + 54a2 + 108a + 8156 B 7 D8

b i The sum of the elements in each row of Pascal’s triangle is a power of 2:Row Sum0 20 = 11 21 = 22 22 = 43 23 = 84 24 = 165 25 = 32

ii ‘The sum of the elements in the nth row of Pascal’s triangle is 2n.’

9 26 = 64

Chapter review

Multiple choice

Short answer12 2000P2,

19P6, 12P9 3 12

4 19C6, 22C15,

2000C2 5 175

67 756 7568 a 210 b 10C0

10C1 10C2

10C3 10C4 . . . 10C10 c 1024

9

Analysis1 50 878 2 2 944 656

3

CHAPTER 24 Introductory probabilityExercise 24A — Events, event spaces and random events1 ε = {1, 2, 3, 4, 5, 6}, Y = {5, 6}2 ε = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),

(2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Z = {(1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} (30 events in all)

3, 4 ε = {AS, 2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AC, 2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C, JC, QC, KC, AH, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AD, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD, QD, KD}, P = {AS, 2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, JC, JH, JD}, Q = {JS}

5

6

7 {M, M, M, M, M, M, M, F, F, F,}

8 E9 C

10

11

12

Exercise 24B — Random numbers and simulations1 Answers will vary.2 Answers will vary.3 Answers will vary.4 a One way is to use randInt(0,1,10) to generate

10 values that are either equal to 0 or 1, and let 0s represent Heads, and 1s represent Tails.

b Answers will vary.c Answers will vary.

5 Generally, the histogram for 100 tosses will be more even than that for 10 tosses.

6 a, b The player can expect to win about once every 15 games, spending $15 to win $10 (a loss of $5).

7 Answers will vary.8 Answers will vary.

Exercise 24C — Long-run proportion1 0.12 0.4993 a, b Note the steady improvement from about 0.25

(25% success) to about 0.36 (36%) success.

4 D5 D6 0.708 This result supports the suspicion that the coin

is biased because the expected result would be 0.5.7 0.788 Yes

a 80x b 56p5q3 c 4608x2

a i 1 ii 2 iii 4iv 8 v 16 vi 32

1 C 2 B 3 A 4 D 5 A6 C 7 A 8 E 9 B 10 C

11 B 12 D

a 180 b 648

a 325 b 6 c 676

a 405x8 b 3240x7 c 196 830x

a 2 598 960 b 1287 c 65 780 d 1584

a b c

Dice total 2 3 4 5 6 7 8 9 10 11 12

Probability

a b c d e f

Sales 7 11 17 25 41 53 60 72 84 97

Houses 28 47 68 93 135 164 186 217 244 270

Proportion 0.25 0.234 0.25 0.269 0.304 0.323 0.323 0.332 0.344 0.359

13--- 5

6--- 4

13------

152------

710------

136------ 2

36------ 3

36------ 4

36------ 5

36------ 6

36------ 5

36------ 4

36------ 3

36------ 2

36------ 1

36------

313------

2345------ 11

45------ 13

45------ 1

45------ 20

45------ 0

45------


A n s w e r s 367

answ

ers

9

There is a huge bias for 2s and 4s, against 3s and 5s, 1s and 6s OK.

10 Baker 3, proportions were 0.75, 0.84 and 0.86 respectively. However, Baker 1 wasted only 22 loaves, versus 32 for Baker 2 and 42 for Baker 3. This could reduce Baker 3’s true effectiveness.

11 a, b Answers will vary.12 a Left-handers caught 6 out of 14 (0.43),

right-handers caught 24 out of 56 (0.43)— no difference.

b Caught 30 out of 70, bowled 30 out of 70 — no difference.

c Left-handers — not out 4 out of 14 (0.29), right-handers — not out 6 out of 56 (0.11) — more likely to be not out against left-handers.

d The relatively small number of left-hander observations means comparisons are not very accurate. However, there seems little difference between left- and right-hander effectiveness.

13 1329

Exercise 24D — Simple and compound events — independent events

1

2

3 See the answer to Exercise 24A question 2,

Pr(9) = =

4 7 is mostly likely, Pr(7) = = 56 {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H),

(T, H, T), (T, T, H), (T, T, T)}

7 C8 B

9 × × =

10

0.7 × 0.45 × 0.5 = 0.157511 0.3 × 0.55 × 0.5 = 0.082512

Pr(2 Tails) = 0.2025

13

a There are 8 paths.b Compare answers with Exercise 22A, question 3,

by adding up similar outcomes’ probabilities.14 E15

a 0.12 = 12% b 0.32 = 32%16 b, d, e17

Pr(P − F) = 0.4 × 0.25 = 0.1

18 a Pr(no drug) = = 0.35

Pr(small dose) = = 0.40

Pr(large dose) = = 0.25

b, c, d

e Although few patients were tested, it appearsthat a greater percentage (80%) of those givena large dose of the drug recovered, whereasa much smaller percentage (29%) of those notgiven the drug recovered. 20% of people testedwere given a large dose of the drug andrecovered, 15% of people tested were given asmall dose and recovered, whereas only 10%of people were not given the drug and recovered.So it could be said that a patient is more likelyto recover if the drug is taken.

Exercise 24E — Compound events — mutually exclusive events1 a, d, e, g

2

1 2 3 4 5 6

0.180 0.233 0.074 0.246 0.092 0.175

a b

a 0.2646 b 0.0204 c 0.1764

a b c d

14---

12--- 1

2---

436------ 1

9---

636------ 1

6---

18--- 3

8--- 7

8--- 3

8---

16--- 1

6--- 1

6--- 1

216---------

New York up

New York down

Tokyo up

Tokyo up

Tokyo down

Tokyo down

Australia up

Australia up

Australia up

Australia up

Australia down

Australia down

Australia down

Australia down

0.7 x 0.45 x 0.5 = 0.16

0.7 x 0.45 x 0.5 = 0.16

0.7 x 0.55 x 0.5 = 0.19

0.7 x 0.55 x 0.5 = 0.19

0.3 x 0.45 x 0.5 = 0.07

0.3 x 0.45 x 0.5 = 0.07

0.3 x 0.55 x 0.5 = 0.08

0.3 x 0.55 x 0.5 = 0.08

0.55 x 0.55 = 0.3025

0.55 x 0.45 = 0.2475

0.45 x 0.55 = 0.2475

0.45 x 0.45 = 0.2025

H

H

T

H

T

HH

HT

TH

TT

0.550.450.55

0.45

0.55

0.45T

a = b c

H

H

T

H

T

H

TH

T

TH

T

T

H

Pr(H, H, H) =

Pr(H, H, T) =

Pr(H, T, H) =

Pr(H, T, T) =

Pr(T, H, H) =

Pr(T, H, T) =

Pr(T, T, H) =

Pr(T, T, T) =

1—81—81—81—81—81—81—81—8

0.6 x 0.2 = 0.12

0.6 x 0.8 = 0.48

0.4 x 0.2 = 0.08

0.4 x 0.8 = 0.32

W

B

B'

B

B'

WB

WB'

MB

M'B'

0.60.80.2

0.8

0.2

0.4M

P

T

S

0.4 x 0.3 = 0.120.4 x 0.25 = 0.100.4 x 0.45 = 0.18

PBPFPL

B

LF

0.25 x 0.3 = 0.0750.25 x 0.25 = 0.0630.25 x 0.45 = 0.112

SBSFSL

B

LF

0.35 x 0.3 = 0.1050.35 x 0.25 = 0.0880.35 x 0.45 = 0.157

TBTFTL

B

LF

1440------

1640------

1040------

0.35 x 0.286 = 0.10

0.35 x 0.714 = 0.250.40 x 0.375 = 0.15

0.40 x 0.625 = 0.250.25 x 0.80 = 0.20

0.25 x 0.20 = 0.05

N

NR

NR'SR

SR'LR

LR'

R

R'

L

SR

R'

R

R'

452------ 1

13------ 6

13------ 15

52------

24A➔

24E


368 A n s w e r san

swer

s3

4 E

5 = 0.517

6 a = = b

7

89

1011

Exercise 24F — Compound events — Venn diagrams1

2 a = 0.275 b 0.4

3

The events are mutually exclusive and the Venn diagram could have been drawn as two circles which did not overlap.

4 0.18

5 0.9009

6 C78 0.27, much higher probabilities of winning with

roulette.9 36

10 9

Exercise 24G — Compound events — conditional probability

1 × = = 0.2

2

3 Note that Pr(factory A) = =

4 D5 B6 Pr(support and Labor) = 0.42 × 0.2 = 0.084,

Pr(support and Lib/Nat) = 0.58 × 0.56 = 0.3248, thus Pr(support) = 0.081 + 0.3248 = 0.4088

7 a Pr(passArts) = 0.6 × 0.56 = 0.336Pr(passScience) = 0.4 × 0.78 = 0.312, thereforePr(pass) = 0.336 + 0.312 = 0.648

b

8 a

b Pr(R1) = 0.5 × = 0.278

Pr(R2) = 0.5 × = 0.222; Pr(R) = 0.5

c Pr(2R) = = 0.444

910 a

× = 0.0045 b × = 0.0724

c × = 0.8507

a b = c d e 1

a = b = c =

a 0.258 b 0.449 c 0.865a 0.037 b 0.296 c 0.667 d 0.333a 0.32 b 0.46 c 0.31a 0.4999 b 0.9997 c 649 773

a 4 b 5 c 8

a 37% b 63%

1645------ 35

45------ 7

9--- 29

45------ 19

45------

16 76+178

------------------

59 13+148

------------------ 72148--------- 18

37------ 19

37------

1236------ 1

3--- 24

36------ 2

3--- 15

36------ 5

12------

317 o'clock 11 o'clock

5580 34

ε55

200---------

IT 16MT

031 29

ε

S 0.51A

0.180.08 0.23

ε

A 0.9009B

0.00090.0291

0.0691

ε

A14

46

7

302

C

B

S 103

8

9

5

84

N

C

a = b =

a 0.27 b 0.71

1430------ 6

14------ 6

30------

W

B

B'

B

B'

14

8

4

12

6

16M

1030------ 1

3--- 12

30------ 2

5---

10003000------------ 1

3---

A

D

D'

D

D'

0.95

0.07

0.93

0.05

B

2—3

1—3

Pr(D and A) = × 0.05 = 0.01667Pr(D and B) = × 0.07 = 0.04667Thus Pr(D = 0.01667 + 0.04667 = 0.0633

13---23---

0.3120.648------------- 0.481=

10.5

0.5

R

W

R

W

2

15—27

12—27

12—27

15—27

1527------

1227------

0.2220.5

-------------

A

A

P

NA

P

NA

P

N

N

P

4—52

4—51

3—51 16—

5132—51

15—5132—51

16—5131—51

4—51

16—52

32—52

452------ 3

51------ 48

52------ 4

51------

4852------ 47

51------


A n s w e r s 369

answ

ers

d To get a blackjack, you need either(A = ace, P = picture card):

Pr(A and P) = × = 0.0241 or

Pr(P and A) = × = 0.0241, so

Pr(blackjack) = 0.0241 + 0.0241 = 0.0482

Exercise 24H — Recording and interpreting simulations1

8


Short answer1 0.39752 randInt(−20,20,10)

3 Player A in the ratio of 14:134 0.49285 0.999967 68 Pr(A ∩ B) = 0.259

10

Analysis1

23 This is a perfectly fair game.4 0.359

a 16 b 10 and 11 are equally likely2 D 3 Player B 4–7 Answers will vary.

0 heads 1 head 2 heads 3 heads 4 heads

Theoretical probability

1 B 2 A 3 A 4 C 5 E6 D 7 D 8 C 9 C 10 C

11 E 12 B 13 E 14 D 15 D

452------ 16

51------

1652------ 4

51------

116------ 4

16------ 6

16------ 4

16------ 1

16------

a 0.5 b c

a b = a

b

c 0.03

a 0.455 b 0.718

1536------ 32

36------

1622------ 5

15------ 1

3---

R0.3

0.9 0.27

0.1 0.03

0.45 0.315

0.55 0.3850.7

U

U'U

U'

R'

0.270.27 0.315+------------------------------- 0.462=

E 11CP

66 17

ε

24F➔

24H


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