CCG SA Ch10 - Edl€¦ · Selected Answers © 2014 CPM Educational Program. All rights reserved. 2...

Selected Answers © 2014 CPM Educational Program. All rights reserved. 1

Lesson 10.1.1 10-6. a: V = (2)(5)(6)− π(0.52 )(6) ≈ 55.3 cm3

b: Answers vary. One possibility: It could represent a pencil sharpener. 10-7. a: 70° b: 50° c: 2x 10-8. She is constructing an angle bisector. 10-9. Height = 6 3 , area = 15(6 3)+ 1

6 (144π) = 90 3 + 24π ≈ 231.3 in.2 ; perimeter = 15 +12 +12 +12 +15 + 1

6 (24π) = 54 + 4π ≈ 66 in. 10-10. a: 17, −17,17, −17,17 ; geometric b: 32,11, 12 , −4

34 , −7

38 ; neither

c: 81, 81, 81, 81, 81 ; arithmetic and geometric 10-11. They intersect only once at (3, 5). 10-12. D


Lesson 10.1.2 10-18. a: 64° b: 128° c: 64° d: 180° e: 128° f: 52° 10-19. Central angle = 3.6°; A ≈ 795.51 square units 10-20. a: 5m +1= 3m + 9, m = 4 b: 2(x + 4

) = 3x − 9, x = 17° c: (p − 2)2 + 62 = p2, p = 10 d: 18t = 360, t = 20° 10-21. a: D(0, 4) and E(4, 7) b: DE = 5 units, so AC should be 10 units long. c: 62 + 82 = 10 units 10-22. b: 108°, inscribed angle c: 72°, central angle d: 216°, yes 10-23. a: 9 cubic units b: 10 cubic units c: Possible response: There are no “floating” blocks and that there are no hidden blocks

behind the ones visible. d: Either a left or back view would reveal any hidden blocks. 10-24. D

Lesson 10.1.3 10-31. a: 3 b: 6 c: 2 d: 1 e: 4 f: 5 10-32. a: They are similar. CD is a dilation of AB from P. b: They have the same measure because they have the same central angle. However, CD

is longer; it is similar to a part of a circle with shorter diameter. c: 60

360 (28π ) ≈14.7 units 10-33. OY ≅ KY ≅ EY ≅ PY (all radii are congruent), ∠PYO ≅ ∠EKY (arc measures are equal),

so ΔPOY ≅ ΔEKY (SAS≅ ). Therefore, PO ≅ EK because ≅ Δs→≅ parts. 10-34. It must be a rhombus and it could be a square. 10-35. π (2)2(4) 45

360( ) ≈ 6.28 ft3

10-36. a: Yes (AA ~), ΔABC ~ ΔLKH b: Not enough information . c: Yes (AA ~), ΔABC ~ ΔEDC 10-37. C


Lesson 10.1.4 10-43. a: 50° b: 50° c: 67° d: 126° e: 54° f: 63° 10-44. a: x ≈ 31.9°, y ≈ 10.5 in. b: x ≈ 3.7 m c: x ≈ 34.7°, y = 250 ≈15.8 ft. 10-45. a: 4 times b: 360 c: 360 ÷ 5 = 72 10-46. area of the regular pentagon ≈ 61.9 ft2; total SA≈ 2(61.9)+ (5)(6)(12) ≈ 483.9 ft2 ;

volume ≈ (61.9)(12) ≈ 743.3 ft3 10-47. a: It is a square. Demonstrate that each side is the same length and that two adjacent

sides are perpendicular (slopes are opposite reciprocals). b: C′ is at (–5, –8) and D″ is at (–7, 4). 10-48. The angles, from smallest to largest, measure 64°, 90°, 116°, 130°, and 140°, so the

probability is 45 . 10-49. C

Lesson 10.1.5 10-54. MA = 14 + 17 + 6 = 39 feet, MB = 6 feet, so AB = ER and ER = 1485 ≈ 38.5 feet 10-55. a: The arc length is 30360 ⋅ 2π(3) = π2 , so the ratio 23 =

π6 .

b: The arc length is 5π6 , so the ratio is 65 =π6 .

c: Possible response: Because all circles are similar, circle D is similar to circle E. Therefore sector D is similar to sector E (the angles are the same and the radii and arc lengths are proportional by the same ratio of similarity). Therefore we can write a proportionality equation, arc length D

radius D = arc length Eradius E . In fact all sectors with 30º central

angles will be similar, so the ratio of arc lengthradius , that is, the radian, will be the same for

all 30º sectors. 10-56. 12 (9)(12)

12 (20)h, h – 5.4"

10-57. The ratio of the volumes must be 1

5( )3 , so the volume must be 15( )3 500π( ) = 4π cm3 .

10-58. a: 1.05 b: ≈ $130,588 c: f (t) = 130588(1.05)t 10-59. a: A ≈ 549.78 sq. cm b: A ≈ 74.21 sq. cm c: A ≈ 25.02 sq. cm 10-60. E


Lesson 10.2.1 10-66. a: No, there may be red marbles that she has not selected in her draws. b: No, it is less likely that there are red marbles, but no number of trials will ever assure

that there are no red ones. c: This is not possible, no number of draws will assure this. 10-67. a: 16 ; 12 b: 5

6 (60) = 50 times c: Answers vary, but two squares should have an even number, no numbers can be 3, and

all numbers must be less than 5. 10-68. a: 13 b: 6.5 c: 67.4º d: 134.8º 10-69. a: Have Ken’s analyzed for $30, and use the ratios of similarity to calculate the data for

Erica’s nugget themselves. b: Since r = 5, the ratio of the areas is 25. Thus, 20 ⋅25 = 500 cm2 . c: Since r = 5, the ratio of the volumes is 53 = 125 . Thus, Erica’s nugget weighs

125 ⋅5.6 = 700 g , which is about 25 ounces. 10-70. 2x + 3x + 4x + 5x , x ≈ 25.7� 10-71. a: True b: False c: False d: True 10-72. B


Lesson 10.2.2 10-78. a: x = y b: y = 2x or x = 1

2 y c: 3y = 5x d: x + y = 180�

10-79. a: Yes, because of the Triangle Sum Theorem, 180�− 64�− 26�= 90�. b: Yes, because 82 +152 = 172 . 10-80. a: See diagram at right. b: 1.4 ft3 10-81. a: See diagram at right. b: 56 ·

56 =

2536 ≈ 69.4%

c: 56 ⋅56 =

2536 ≈ 69.4%

d: 56 (49π ) ≈128.28 square cm e: 1− 1

4 −23 =

112 , 112 (360°) = 30°

10-82. 540° 10-83. a: Both are π4 . Note that a radian measure is a ratio and does not have units. b: If x is the unknown central angle of the circle, then the equation,

x360 (2π r)

r = π3 , can be

solved for x. x = 60º 10-84. B 10-85. a: P(on-campus given Engr) = 120

800+120 ≈13.0% b: P(on-campus) = 0.6 c: No. The probability of living on campus given that the student is an engineer is much

smaller than the probability of living on campus. 10-86. Region A is 14 of the circle, so it should result 14 (80) = 20 times. Regions B and C have

equal weight (which can be confirmed with arc measures), so they should each result (80 – 20) ÷ 2 = 30 times.

10-87. a:

360�9 = 40�

b: mAD = 2(97) = 194 ; m∠C = 0.5(1.94) = 97�

c: mAB = 125º and the length of AB

= 125º360º (16π ) ≈17.5 " ; area = 125º

360º (64π ) ≈ 69.8 in.2 10-88. Methods vary, a variety of relationships such as Parallel Line Angle Conjectures, the

Exterior Angle Theorem, or the Triangle Sum Theorem can be used. x = 109°, y = 71°, z = 99°.

10-89. a: x = 34 b: x = 4

3 c: x = −5 d: x = 325

10-90. She is incorrect, which can be tested by substituting both answers into the equation;

w = −6 or 4 . 10-91. A

1.5 ft

1 ft

1 ft 0.2 ft 0.2 ft 0.25 f

t

0.25 f

t

blue

red


Lesson 10.2.3 10-101. a: 236

236+274 = 46.3% b: P(laptop) ≠ P(laptop given business trip) so they are associated. 10-102. a: See table below. Entries not in bold are given in problem statement, and entries in

bold are computed from given information.

Daily Not Daily

Weekly local 25% 12% 37%

Not weekly local 40% 23% 63%

65% 35% 100%

b: 25% + 12% + 40% = 77% c: 25+4077 ≈ 84.4% 10-103. 57.75 = π (2.52 )h , h = 57.75

6.25π ≈ 2.94 in. 10-104. π ; Since the angle is doubled, the arc length must be doubled, and in turn, the ratio

of arc length to radius, or radian measure, must also double and thus be π. 10-105. a: See views at right. Depending on how they are oriented, the

top and bottom views are the same. (Or they could be reflections of each other.)

b: V = 7 units3 , SA = 28 units2; Methods vary. c: Since this solid has no “holes,” the surface area can be calculated by adding the

areas of each of the views. 10-106. b: ABAC c: BCAB d: BCAC e: ABAC f: BCAC 10-107. a: If x represents the length of chord AC , then x2 = 102 +102 − 2(10)(10) cos 80º ;

x ≈ 12.9 b: 18

F R T


Lesson 10.3.1 10-114. a: 24 b: The decision chart tells how many branches there are at each stage. c: 16 10-115. 5! = 120 10-116. a: 1260 = 20% b: 945 = 20% c: Yes 10-117. a: See table at right. b: 19 c:

19

19 +

118

≈ 66.7%

10-118. Edge = 1200/12 = 100 cm, SA = 6(1002 ) = 60, 000 cm2 ;V = 1003 = 1, 000, 000 cm3

10-119. Area of shaded region ≈ 93.53− 84.82 ≈ 8.71 in.2 10-120. C

Lesson 10.3.2 10-127. a: 7! = 5040 b: 1

5040 10-128. 12P3 = 12!

(12−3)! =12!9! = 12 ⋅11⋅10 = 1320

10-130. No, they should not charge a higher premium. P(ticket given red) = 9

348 = 0.02586 and P(ticket) = 507

20,000 = 0.02535 . Since they are approximately the same, they are most likely independent.

10-131. a: 30% b: 42% c: P(Green Fang) ⋅ P(alarm) ≠ P(Green Fang and alarm) , 0.64 ⋅0.28 ≠ 0.22 , so they

are weakly associated. 10-132. 158,184, 000 −17, 576, 000 = 140, 608, 000 ; 1

175760 10-133. a: cylinder b: triangle based prism c: cone

CIRCLES

red yellow

SQU

AR

ES yellow

red

blue

1

ticket no ticket red car 9 348 not red car 507 20000


Lesson 10.3.3 10-139. 9!= 362, 880 10-140. a: 1 b: 8 things taken 8 at a time. 8P8 should equal 8!. 8P8 = 8!

(8−8)! =8!0! . If 0!= 0 , you

would be dividing by 0. c: 3!3 = 2!, 2!

2 = 1!, 1!1 = 0!

10-141. a: 10P8 = 10!

2! = 1, 814, 400

b: 10C8 = 10 P88! = 10!

8!2! = 45

c: 6C1 = 6!1!5! = 6

10-142. Because they are independent, P(coffee and dairy) = P(coffee) � P(dairy) =

( 4263 )(2163) ≈ 22.2% . P(no caffeine and no dairy) = P(no caffeine) � P(no dairy) =

(63−4263 )(63−2163 ) ≈ 22.2%. Avoid the common mistake of computing P(no caffeine and no dairy) = 1 – P(coffee and dairy) because {coffee and dairy} includes all the drinks that have both coffee and dairy but does not include drinks that have only coffee or only dairy.

10-143. a: 16P6 = 16 ⋅15 ⋅14 ⋅13⋅12 ⋅11= 5, 765, 760 b: 1⋅ 15P5 = 1⋅15 ⋅14 ⋅13⋅12 ⋅11= 360, 360 c: 360,360

5,765,760 = 6.25% d: 15C5

16C6= 30038008 = 37.5%

10-144. a: 20 b: 30 c: 36 10-145. a: 5!= 5 ⋅4 ⋅3⋅2 ⋅1= 120 b: 5!2! =

5⋅4⋅3⋅2⋅12⋅1 = 60

c: 5!2!2! =

5⋅4⋅3⋅2⋅12⋅1⋅2⋅1 = 30

d: Because you cannot tell the repeated letters apart, there are fewer arrangements when there are repeated letters.


Lesson 10.3.4 10-153. 500C3 = 20, 708, 500 10-154. 3⋅7 ⋅2 ⋅1⋅4 = 168 choices 10-155. a: 22P3 = 9240 b: This is really a permutation lock. In this case the common use of the word

“combination” conflicts with the mathematical meaning. c: 22C3 = 1540 , but this does not make sense for a mechanical lock because it would

imply dialing the numbers in any order to open the lock. d: 22 ⋅21⋅21= 9702 10-156. a: 162º b: 16 sides c: ≈120.8 cm2 10-157. a: 12; (12 – 1), (12 – 2), (12 – 3) b: n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5) c: n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5) 10-158. a: There is one way to choose all five. 5!

5!0! = 1 . In order to have the formula give a reasonable result for all situations, it is necessary to define 0! as equal to 1.

b: There is one way to choose nothing. 5!0!5! = 1

c: This is a permutation in which some of the items occur multiple times, like in an anagram. 8!

2!5! = 168 arrangements of cars.


Lesson 10.3.5 10-173. 100C3 = 161, 700 10-174. The first because 3!> 2! 10-175. a: (n − 3)(n − 4)(n − 5)(n − 6)(n − 7) b: n + 2 , n +1 , n −1 , n − 2 c: n(n −1)(n − 2) d: (n + 2)(n +1)(n −1) 10-176. a: See diagram at right. Entries not

bolded are given in the problem statement, while bold entries are computed.

b: P(Senior⎢Ocean View) = 0.060.24 = 25%

c: They are associated because P(senior) ⋅ P(Ocean View) ≠ P(senior and Ocean View) , this is (0.10)(0.24) ≠ (0.06) .

10-177. ≈ 0.83 or a 17% decrease 10-178. a: See diagram at right. b: h2 = 172 −152 = x2 , x = 8 ,A(base) = 0.5(8)915)− 60,

600 = 60h , h = 10 cm c: TSA = 2(60) + 17(10) + 8(10) + 15(10) = 520 sq. cm d: 90º; sin a = 8

17 , a = 28º; sin b = 1517 ; b = 62º

10-179. 12C5, 12C4 , 12C3, 12C2, 12C1, 12C0 = 1586 10-180. This is permutation with some of the items to be arranged occurring multiple timers,

like an anagram. 10!3!2! = 302, 400 ways to layer the dip. 10-181. If t = hours, 2 + 1

4 t = 9 ; 28 hours 10-182. a: Similar b: Similar 10-183. a: b:

10-184. SA: A(Δ) = 1

2 (8)(4 3) ≈ 27.71; A(face) = (8)(10) = 80; TSA = 2(27.71) + 3(80) ≈ 295.43 sq. cm. V = 27.71(10) ≈ 277.13 cu. cm.

OceanView Not

OceanView

Senior (0.60)(0.10) = 0.06 0.04 0.10 Not senior (0.20)(0.90) = 0.18 0.72 0.90

0.24 0.76 1.00

17 cm

15 cm

FRONT RIGHT

FRONT RIGHT

CCG SA Ch10 - Edl€¦ · Selected Answers © 2014 CPM Educational Program. All rights reserved. 2...

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