Ratios and Proportion Write each fraction in simplest form. 1.2.3. Simplify each product. 4.5.6....

$: Ratios and Proportion Write each fraction in simplest form. 1.2.3. Simplify each product. 4.5.6. ALGEBRA 1 LESSON 4-1 49 84 24 42 135 180 35 25 99 144.$
Ratios and ProportionRatios and Proportion

Write each fraction in simplest form.

1. 2. 3.

Simplify each product.

4. 5. 6.

ALGEBRA 1 LESSON 4-1ALGEBRA 1 LESSON 4-1

4984

2442

135180

3525

99144

2181 40

149688

10856

(For help, go to Skills Handbook pages 724 and 727.)

4-1


1.

2.

3.

4.

5.

6.

49 7 • 7 7 84 7 • 12 12

= =

24 6 • 4 442 6 • 7 7

= =

135 45 • 3 3180 45 • 4 4

= =

35 40 5 • 7 5 • 8 5 • 7 • 5 • 8 825 14 5 • 5 7 • 2 5 • 5 • 7 • 2 2

= = = = 4

99 96 9 • 11 8 • 12 9 • 11 • 8 • 12 9 3144 88 12 • 12 8 • 11 12 • 12 • 8 • 11 12 4

= = = =

21 108 3 • 7 3 • 3 • 3 • 4 3 • 7 • 4 4 181 56 3 • 3 • 3 • 3 7 • 8 3 • 7 • 8 8 2

= = = =


4

4

Solutions

4-1


Another brand of apple juice costs $1.56 for 48 oz. Find the unit rate.


The unit rate is 3.25¢/oz.

4-1

cost $1.56ounces 48 oz = $.0325/oz


The fastest recorded speed for an eastern gray kangaroo is

40 mi per hour. What is the kangaroo’s speed in feet per second?


40 mi1 h

5280 ft1 mi

1 h 60 min

1 min60 s

• • • Use appropriate conversion factors.

The kangaroo’s speed is about 58.7 ft/s.

40 mi1 h

5280 ft1 mi

1 h 60 min

1 min60 s

• • • Divide the common units.

= 58.6 ft/s Simplify.

4-1


Solve = .y3

34

y3 • 12 =

34

Multiply each side by the least common multiple of 3 and 4, which is 12.

• 12

4y = 9 Simplify.

4y4 =

94 Divide each side by 4.

y = 2.25 Simplify.


4-1


Use cross products to solve the proportion = – .


w 4.5

65

w4.5

= –65

w(5) = (4.5)(–6) Write cross products.

5w = –27 Simplify.

5w5

= –27 5 Divide each side by 5.

w = –5.4 Simplify.

4-1

Define: Let t = time needed to ride 295 km.

Relate: Tour de Franceaverage speed

Write:

equals

=

295-km tripaverage speed

363092.5

295

tkilometershours


In 2000, Lance Armstrong completed the 3630-km Tour de France course in 92.5 hours. Traveling at his average speed, how long would it take Lance Armstrong to ride 295 km?


363092.5 =

295t

3630t = 92.5(295) Write cross products.

t = Divide each side by 3630.92.5(295)

3630 t 7.5 Simplify. Round to the nearest tenth.Traveling at his average speed, it would take Lance approximately 7.5 hours to cycle 295 km.

4-1


Solve the proportion = .


z + 34

z – 46

z + 34

z – 46=

(z + 3)(6) = 4(z – 4) Write cross products.

6z + 18 = 4z – 16 Use the Distributive Property.

2z + 18 = –16 Subtract 4z from each side.

2z = –34 Subtract 18 from each side.

Divide each side by 2.=2z2

–342

z = –17 Simplify.

4-1

Ratios and ProportionRatios and ProportionALGEBRA 1 LESSON 4-1ALGEBRA 1 LESSON 4-1

pages 185–188 Exercises

1. $9.50/h

2. $.40/lb

3. 131 cars/week

4. 400 cal/h

5. $.24/oz

6. $.09/oz

7. A

8. A

9. B

10. A

11. 480

12. 1.2

13. 10,800

14. 7.5

15. 11.25

16. 5

17. 25.2

18. 7.5

19. 6

20. –20

21. 14.4

22. 9

23. –16.5

24. 6

25. –5.25

26. 90

27. 28

28. 17.6

29. 67.5

30. 700

31. 105.6 km

32. 0.5

33. 8

34. 7

35. –3

23

111213

12

4-1


36. 8

37. 165

38. 12.5

39. 18.75

40. 14.60

41. 18.25

42. 504

43. 2520

44. 20 mi/h

45. 15 mi/h

46. 12 mi/h

47. 1 mi/h

48. 1 mi/h

49. about 0.28 mi/h

50. 50.4 min

51. 10.5 mm

52. 246.4 ft/s

53. 3

54. 5.3

55. –16

56. 115.2

57. 45

58. 4.4

59. –17

60. 59

61. –8.4

62. about 646 students



65. Answers may vary. Sample: Multiply the numerator of each side bythe denominator of the other side. Set the products equal to each other and solve the equation.

= , (7)(15) = 5x, x = 21

66. $.05/mi

67. 4 people/mi2, 2485 people/mi2, 78 people/mi2

75

x 15

4-1


68. Check students’ work.

69. a. 7, 14b. 21c. 21d. x = 7a

70. Bonnie: $56.00, Tim: $32.00

71. 48 V

72. 9

73. –7.5

74. 9

75. –32

76. a. 5.47 min/mib. 5.37 min/mi

4-1

77. D

78. G

79. C

80. G

81. [2] = ;

x = 53 rings, which is 53 years OR equivalent explanation

[1] incorrect proportion solved correctly OR correct proportion solved incorrectly

82.

83.

84.

12 in. 32 rings

20 in. x rings

13

13


97. –5

98. –5.5

99. –90

100. –6

85.

86.

87. no solution

88. t = height (in.), t 72

89. s = students, s 235

90. m = miles, m 344

91. w = weight (lb), w > 20

92.

93.

94.

95. 6

96. 136

12

4-1

>–

<–

<–


Solve.

1. Find the unit rate of a 12-oz bottle of orange juice that sells for $1.29.

2. If you are driving 65 mi/h, how many feet per second are you driving?

Solve each proportion.

3. 4.

5. 6.

c6

1215

= 2112

7y

=

3 + x7

48

= 2 + xx – 4

2535

=

10.75¢/oz.

about 95.3 ft/s

4.8 4

12

–17

4-1

Proportions and Similar FiguresProportions and Similar Figures

Simplify

1. 2. 3.

Solve each proportion.

4. 5. 6.

7. 8. 9.


3642

81 108

2652

x 12

7 30

= y 12

8 45

= w 15

1227

=

9a

8110

=25 75

= z 30

n 9

=n + 1

24

(For help, go to Skills Handbook and Lesson 4-1.)

4-2


1. 5. 7. 9.

2.

3.

4.

6. 8.


3642

6 • 6 66 • 7 7

= =

81108

27 • 3 327 • 4 4= =

2652

26 • 1 126 • 2 2= =

x12

730=

30x = 12(7)

30x = 84

x = 8430

x = 245

y 12

=8

4545y = 12(8)

45y = 96

y = 9645

y = 22

15

w 15

=1227

27w = 15(12)27w = 180

w = 18027

w = 623

9a

=8110

81a = 9(10)

81a = 90

a = 9081

a = 119

25 75

=z

3075z = 25(30)75z = 750

z = 75075

z = 10

n9

=n + 1

2424n = 9(n + 1)

24n = 9n + 9

15n = 9

n = 9

15

n = 35

Solutions

4-2


In the figure below, ABC ~ DEF. Find AB.


Write: =

Relate:

=

EFBC

DEAB

Define: Let x = AB.

69

8x

Write a proportion comparing the lengths of the corresponding sides.

Substitute 6 for EF, 9 for BC, 8 for DE, and x for AB.

6x = 9(8) Write cross products.

= Divide each side by 6.6x6

726

x = 12 Simplify.

AB is 12 mm.

4-2


A flagpole casts a shadow 102 feet long. A 6 ft tall man casts

a shadow 17 feet long. How tall is the flagpole?


=10217

x6 Write a proportion.

17x = 102 • 6 Write cross products.

17x = 612 Simplify.

Divide each side by 17.=17x17

61217

x = 36 Simplify.

The flagpole is 36 ft tall.

4-2


The scale of a map is 1 inch : 10 miles. The map distance from Valkaria to Gifford is 2.25 inches. Approximately how far is the actual distance?


=mapactual

1 10

2.25x

mapactual

Write a proportion.

1 • x = 10 • 2.25 Write cross products.

x = 22.5 Simplify.

The actual distance from Valkaria to Gifford is approximately 22.5 mi.

4-2

Proportions and Similar FiguresProportions and Similar FiguresALGEBRA 1 LESSON 4-2ALGEBRA 1 LESSON 4-2

10. 12 in.

11. 87.5 mi

12. 145.25 mi

13. 325.5 mi

14. 350 mi

15. a. Lincoln to San Paulo = 16 miLincoln to Duncanville = 26 miSan Paulo to Duncanville = 18 mi

b. 26 mi roundtrip

16. 1 cm : 8 km

17. 4 in. by 6 in.

18. 2 in. by 4 in.

19. 2 in. by 3 in.


1. AB PQ, BC QR, CA RP, A P, B Q, C R

2. ED JH, DF HK, FE KJ, D H, E J, F K

3. 3.125 ft

4. 13.33 cm

5. 80 in.

6. 40 m

7. 20.25 cm

8. 7.2 ft

9. 4.8 ft

23

4-2

–


20. 3.2 in. by 4.8 in.

21. 33.75 in.

22. 22.5 ft by 27 ft

23. a. Answers may vary. Sample: GK and RQ are not corresponding sides.

b. =

24. 1 in. : 12 ft

25. 9 ft by 12 ft

26. 3 ft

27. 216 ft2

28. yes; because it is 6 ft wide and 9 ft long

29. 48 cm long by 20 cm wide

GHPQ

HLRQ

30. a. 6 mb. 6 m, 18 mc. Yes, the ratio of the sides

is equal to the ratio of perimeters in similar figures.

d. 2 m2, 18 m2

e. Answers may vary. Sample: The area ratio is the square of the side ratio.

31. Answers may vary. Sample: doll house to regular house, model car to real car

32. a. Yes; the sides are proportional.b. The volume ratio is the cube

of the side ratio.c. 27 : 1

33. a = 8, b = 6, c = 10

4-2


34. about 1 in. : 30.5 mi

35. 400,400 km

36. a. =

b. 3.2

c. 11.2 in.

d. 39.2 in.2

37. B

38. I

8 8 + x

57

39. [2] smaller area: 6 • 5 = 30; 30 ft2; larger area: 10 • 12 = 120; 120 ft2;

=

0.5(120) = 30x

60 = 30x

2 = xTwo gallons of paint should cover

a 10 ft 20 ft wall.

[1] incorrect calculation for one area and proportion solved correctly OR correct area calculations but proportion set up incorrectly

0.530

x 120

4-2

40. [4] a. =

77.5(12) = 16(55) 930 = 880Since the cross

products are not equal, the

proportion is not true. So the postcard is not

similar to the painting.

b. = OR =

The postcard should be

12 cm 16.9 cm OR

11.4 cm 16 cm.


77.516

5512

[3] appropriate methods, but with one computational error OR found only one possible postcard size[2] incorrect proportions solved correctly[1] correct answer with no work shown

41. 4.5

42. 16

43. –22

44. 40

45. b < –4

46. x ≥ 7

47. m < –4

48. h > –

2367

32

4-2

5512

77.5x

55y

77.516

/

/

1. In the figure below, ABC ~ DEF. Find DF.

2. A boy who is 5.5 feet tall casts a shadow that is 8.25 feet long. The tree next to him casts a shadow that is 18 feet long. How tall is the tree?

3. The scale on a map is 1 in.: 20 mi. What is the actual distance between two towns that are 3.5 inches apart on the map?


About 19.7 cm

70 mi

4-2

12 ft

Proportions and Percent EquationsProportions and Percent Equations

Find each product.

1. 0.6 • 9 2. 3.8 • 6.8 3. 4.

Write each fraction as a decimal and as a percent.

5. 6. 7. 8.

9. 10. 11. 12.

(For help, go to skills handbook pages 727 and 728.)


2360

2046

• 17135

534

•

710

23100

25

1320

3540

716

425

170200

4-3

Proportions and Percent EquationsProportions and Percent EquationsALGEBRA 1 LESSON 4-3ALGEBRA 1 LESSON 4-3

1. 0.6 • 9 = 5.4

2. 3.8 • 6.8 = 25.84

3.

4.

5.

6.

7.

2360

2046

• = 23 • 20 20 • 3 • 23 • 2

= 1 3 • 2

= 16

17135

534

• = 17 • 5 5 • 27 • 17 • 2

= 1 27 • 2

= 154

710

= 7 ÷ 10 = 0.7; 0.7(100%) = 70%

23100

= 23 ÷ 10 = 0.23; 0.23(100%) = 23%

25

= 2 ÷ 5 = 0.4; 0.4(100%) = 40%

Solutions

4-3


Solutions (continued)

4-3

1320

= 13 ÷ 20 =0.65; 0.65(100%) = 65%

3540

= 35 ÷ 40 = 0.875; 0.875(100%) = 87.5%

716

= 7 ÷ 16 = 0.4375; 0.4375(100%) = 43.75%

425

= 4 ÷ 25 = 0.16; 0.16(100%) = 16%

170200

= 170 ÷ 200 = 0.85; 0.85(100%) = 85%

8.

9.

10.

11.

12.


What percent of 90 is 27?


90n = 2700 Find the cross products.

n = 30 Divide each side by 90.

30% of 90 is 27.

4-3

percent = n

1002790

partwhole


Find 25% of 480.

= 25 100

n 480

partwhole

25% of 480 is 120.

12,000 = 100n Find the cross products.

120 = n Divide each side by 100.

4-3


Water covers about 361,736,000 km2, or about 70.8% of the

earth’s surface area. Approximately what is the total surface area of

the earth?


361,736,000t

Relate: 70.8% of the total surface area is 361,736,000 km2.

Define: Let t the total surface area.

Write: = 70.8100

partwhole

70.8t = 361,736,000,000 Find cross products.

t = 510,926,553.7 Divide each side by 70.8.

The total surface area of the earth is approximately 510,926,554 km2.

4-3




Relate: What percent of 140 is 84?

Define: Let p = the decimal form of the percent.

Write: p • 140 = 84

140p = 84

60% of 140 is 84.

p = 0.6 Divide each side by 140.

p = 60% Write the decimal as a percent.

4-3




Relate: What percent of 60 is 114?

Define: Let n = the decimal form of the percent.

Write: n • 60 = 114

60n = 114

190% of 60 is 114.

n = 1.90 Divide each side by 60.

n = 190% Write the decimal as a percent.

4-3


a. Estimate the number that is 19% of 323.


• 325 = 651565 is approximately 19% of 323.

b. What is 73% of 125? Use fractions to estimate the answer.

• 124 = 9334

125 124 124 and 4 are compatible numbers.

93 is approximately 73% of 125.

323 325 325 and 5 are compatible numbers.

= 20%. So is a good approximation of 19%.19% 15

15

15

= 75%. So is a good approximation of 73%.73% 34

34

34

4-3


A candidate for mayor sent out surveys to 8056 people in his

city. After two weeks, about 18% of the surveys were returned. How

many surveys were returned?


Relate: What is 18% of 8056?

Define: Let n = the unknown number.

Write: n = 0.18 • 8056

n = 0.18 • 8056

About 1450 surveys were returned.

n = 1450.08 Simplify.

4-3



1. 50%

2. 25%

3. 33 %

4. 20%

5. 25%

6. 20%

7. 8

8. 16

9. 21

10. 28

11. 10

12. 20

13. = , 50

14. = , 25

15. = , 160

16. = , 240

17. = , 70

18. = , 140

19. 30 h

20. 50 = 0.25x; 200

21. 25 = 0.50x; 50

22. 96 = n • 150; 64%

40 100 80 100 15 100 20 100 60 100 30 100

20x

20x24x

48x42x

42x

23. 45 = n • 60; 75%

24. x = 0.05(300); 15

25. x = 0.05(200); 10

26. 21

27. 200%

28. 4

29. 300%

30. 0.42

31. 0.3%

32. 200

33. 100

34. 22

4-3

13


35. 150

36. 68

37. 32


39–40. Proportions or equations may vary. Samples are given.

39. = , 4

40. = , 400%

41. x = 0.002(900), 1.8

42. 0.02x = 1.8, 90

43. 1000x = 988, 98.8%

44. 1.4(84) = x, 117.6

75 100

3x

x 100

30075

45. 62; 50% is 61 and 51.3% > 50%.

46. 20; 25% is 21 and 23.9% < 25%.

47. 73; 10% is 74 and 9.79% < 10%.

48. 185; 75% is 180 and 76.02% > 75%.

49. $297.00

50. $32.70

51. $3896.00

52. Answers may vary. Sample: $1.20; take 10% and 5% of $8 and add them.

53. a. $74.25b. 3.75%c. 6 yr

54. $61.20

4-3

55. $1250.00

56. 7%

57. 2 yr


66. 20

67. 13 mi

68. 60 mi

69. 90 mi

70. 106 mi

71. b < –4;

72. x 7;

73. h > –21;

74. p < – ; 13

4-3

13

23

>–

58. Answers may vary. Sample: 33%;

I sleep 8 h a day, and there are

24 h in a day, so = , x = 33 .

59. a.

b. 200

c. 400%

60. Yes; 16.99(1.15)(0.66) < 13.99(1.06), 12.90 < 14.83.

61. 27237

62. 11712

63. 425

64. 35

65. 667

x 100

8 24

13

3 50


1. What is 35% of 160?

2. What percent of 450 is 36?

3. 32 is 80% of what number?

4. What is 0.03% of 260,000?

5. What percent of 50 is 75?

6. Estimate 62% of 83?

56

8%

40

78

150%

51

4-3

Percent of ChangePercent of ChangeALGEBRA 1 LESSON 4-4ALGEBRA 1 LESSON 4-4

Write an equation for each problem and solve.

1. What is 20% of 20? 2. 8 is what percent of 20?

3. 18 is 90% of what number? 4. 27 is 90% of what number?

Estimate each answer.

5. 67.3% of 24 6. 65% of 48

(For help, go to Lesson 4-3.)

4-4


1. What is 20% of 20?n = 0.2(20)n = 4

2. 8 is what percent of 20?8 = n • 20

n = = 8 ÷ 20 = 0.4 = 40%

3. 18 is 90% of what number?18 = 0.9n

n = = 18 ÷ 0.9 = 20

820

180.9

4. 27 is 90% of what number?27 = 0.9n

n = = 27 ÷ 0.9 = 30

5. 67.3% of 24 of 24 = • 24 = 16

6. 65% of 48 = • 48 = 32

270.9

23

23

23

23

Solutions

4-4

23

Percent of ChangePercent of Change

The price of a skirt decreased from $32.95 to $28.95. Find

the percent of decrease.


percent of decrease =amount of change

original amount

32.95 – 28.9532.95=

Subtract to find the amount of change.Substitute the original amount.

The price of the skirt decreased by about 12%.

0.12 or 12% Write as a decimal and then as a percent.

4-4

4 32.95= Simplify the numerator.


Between 1940 and 1980, the federal budget increased from

$9.5 billion to $725.3 billion. What was the percent of increase in the

federal budget?


percent of increase =amount of change

original amount

=725.3 – 9.5

9.5 Substitute.

=715.8

9.5 Simplify the numerator.

= 75.35 or 7535% Write as a decimal and then as a percent.

The federal budget increased nearly 7535%.

4-4


You read the bathroom scale as 122 lb. What is your greatest possible error?

The scale is read to the nearest 1 lb, so the greatest possible error is one half of 1 lb, or 0.5 lb.


4-4


When a garden plot was measured, the dimensions were

156 in. 84 in. Use the greatest possible error to find the minimum and maximum possible areas.


Both measurements were made to the nearest whole inch, so the greatest possible error is 0.5 in.

Minimum Area Maximum Area

155.5 in. 83.5 in. = 12,984.25 in.2

156.5 in. 84.5 in. = 13,224.25 in.2

The minimum area is 12,984.25 in.2, and the maximum area is 13,224.25 in.2.

The length could be as little as 155.5 in. or as great as 156.5 in.

The width could be as little as 83.5 in. or as great as 84.5 in. Find the minimum and maximum areas.

4-4


Suppose you measure a library book and record its width as

17.6 cm. Find the percent of error in your measurement.

Since the measurement is to the nearest 0.1 cm, the greatest possible error is 0.05 cm.


The percent error is about 0.3%.

percent error =greatest possible error

measurement Use the percent error formula.

=0.0517.6 Substitute.

0.0028409091 Divide.

= 0.3% Round and write as a percent.

4-4


A small jewelry box measures 7.4 cm by 12.2 cm by 4.2 cm.

Find the percent error in calculating its volume.


The measurements are to the nearest 0.1 cm. The greatest possible error is 0.05 cm.

as measured maximum value minimum value

V = • w • h V = • w • h V = • w • h

= 7.4 • 12.2 • 4.2= 379.18 cm3

= 7.45 • 12.25 • 4.25= 387.87

= 7.35 • 12.15 • 4.15= 370.61

Possible Error: maximum – measured measured – minimum387.87 – 379.18 = 8.69 379 – 370.61 = 8.57

4-4


Use the difference that shows the greatest possible error to find the percent error.

The percent error is about 2%.

percent error =greatest possible error

measurement

=387.87 – 379.18

379.18

Use the percent error formula.

Substitute.

4-4

(continued)

= 0.0229178754

= 2%

Write as a decimal.

Round and write as a percent.

=8.69

379.18 Simplify the numerator.



1. 50%; increase

2. 33 %; decrease

3. 25%; increase

4. 20%; decrease

5. 33 %; increase

6. 25%; decrease

7. 25%; increase

8. 20%; increase

9. 84.4%; increase

10. 71.1%; increase

11. 60.7%; decrease

12. 14.4%; increase

13. 39%

14. 60%

15. 0.5 ft

16. 0.05 cm

17. 0.005 g

18. 0.5 in.

19. 19.25 cm2, 29.25 cm2

20. 48.75 mi2, 63.75 mi2

21. 46.75 in.2, 61.75 in.2

22. 51.75 km2, 68.75 km2

23. 253.75 in.2, 286.75 in.2

24. 303.75 km2; 340.75 km2

25. 25%

26. 25%

27. 12.5%

28. 12.5%

29. a. 48 cm3

b. 74.375 cm3

c. 28.125 cm3

d. 26.375 cm3

e. 55%

30. 23%; decrease

31. 22%; decrease

4-4

13

13


32. 157%; increase

33. 175%; increase

34. 4%; increase

35. 3%; decrease

36. 56%; decrease

37. 9%; decrease

38. 17%; increase

39. 2%

40. 19%

41. 1 mm

42. no; 16% increase but a 14% decrease

43. no; increases to $70.40 but decreases to $63.36

44. Answers may vary. Sample: Joan bought shoes for $10. Sarah bought the same shoes 3 days later for $7. What was the percent change? 30% decrease

45. 24.5 cm2, 25.5 cm2

46. 58 mi2, 59.6 mi2

47. 54.1 in.2, 54.3 in.2

48. a. 100%b. 100%c. 50%d. 50%

49. 11%

50. 34%

51. Answers may vary. Sample: Use the greatest possible error to calculate the maximum, minimum, and measured areas. Find the amounts by which the maximum and minimum differ from the measured area. Divide the greater difference by the measured area.

4-4


52. Jorge found the change of $5 but divided by the final price instead of the original price.

53. a. 9%, 3%b. Answers may vary. Sample: The larger a measure, the smaller is the percent error.

54. Yes; 148.3 > 3 (48.7) = 146.1,

and 205%.55. a. 21%

b. 21%c. 21%; answers may vary.

Sample: 1.1a • 1.1a = 1.21a2, which is 21% greater than a • a = a2.Relationship between % increase of side and area of the square doesn’t depend on the side length.

148.3 – 48.748.7

56. C57. I58. A59. [2] perimeter of softball diamond:

4(60) = 240, perimeter 240 ft, side of baseball diamond: 1.5(60) = 90, side 90 ft, perimeter of baseball diamond: 4(90) = 360, perimeter 360 ft, % of increase

= = 50%; area of softball

diamond: 60(60) = 3600, area 3600 ft2, area of baseball diamond: 90(90) = 8100, area 8100 ft2, percent of increase

= = 125%

OR computation that gives same results

360 – 240240

8100 – 36003600

4-4


[1] appropriate methods, but with one computational error OR finds only one % of increase

60–65. Equations may vary.

60. = , 7%

61. = , 87%

62. = , 179.5

63. = , 300%

64. = , 1.7

65. = , 37.2

66. n <

x 100 x 100

x 100

44 100

0.2 100266100

5 67131579x9632 x 834 x 14

1 15

67. q –17

68. x –1

4-4

>–

>–


Find each percent of change. Describe the percent of change as an increase or decrease.

1. $6 to $9 2. 15 cm to 12 cm

Find the greatest possible error.

3. 13.2 m 4. 34.62 g

5. Find the percent error for the measurement 6 cm.

6. Find the minimum and maximum possible areas for a rectangle measured as 3 m x 7 m.

50% increase 20% decrease

0.05 m 0.005 g

about 8.3%

min: 16.25 m2; max: 26.25 m2

4-4

Applying Ratios to ProbabilityApplying Ratios to ProbabilityALGEBRA 1 LESSON 4-5ALGEBRA 1 LESSON 4-5

Rewrite each decimal or fraction as a percent.

1. 0.32 2. 0.09 3. 4.45

2009

50

(For help, go to Skills Handbook page 728.)

4-5


1. 0.32 = = 32%

2. 0.09 = = 9%

3. = = = 22.5%

4. = = = 18%

32100

9100

45200

2 • 22.52 • 100

22.5100

950

9 • 250 • 2

18100

Solutions

4-5

A bowl contains 12 slips of paper, each with a different name

of a month on it. Find the theoretical probability that a slip selected at

random from the bowl has the name of a month that ends with “ber.”


Applying Ratios to ProbabilityApplying Ratios to Probability

P(event) =number of favorable outcomesnumber of possible outcomes

The probability of picking a month that ends with “ber” is .13

=4

12There are 4 months out of 12 that end with “ber”:September, October, November, and December

=13 Simplify.

4-5


For a number cube, find the probability of not rolling a

number divisible by 3.

The probability of not rolling a number divisible by 3 is .23

P(÷ 3) =number of favorable outcomesnumber of possible outcomes =

26 =

13

P(not ÷ 3) = 1 – P(÷ 3) Use the complement formula.

= 1 – = Simplify.13

23

4-5


Quality control inspected 500 belts at random. They found no

defects in 485 belts. What is the probability that a belt selected at

random will pass quality control?

P(no defects) =number of times an event occurs

number of times the experiment is done

The probability that a belt has no defects is 97%.

= Substitute.485500

= 0.97 = 97% Simplify. Write as a percent.

4-5

Applying Ratios to ProbabilityApplying Ratios to Probability

If the belt manufacturer from Additional Example 3 has 6258

belts, predict how many belts are likely to have no defects.

number with no defects = P(no defects) • number of belts


= 0.97 • 6258 Substitute. Use 0.97 for 97%.

= 6070.26 Simplify.

Approximately 6070 belts are likely to have no defects.

4-5



1.

2.

3.

4.

5.

6. 0

7.

8. 1

9.

10.

1213161223

13

1356

11.

12.

13. 1

14. 80%

15. 24%

16. 43%

17. 15%

18. 85%

19. 39%

20. 67%

21. a. about 40%b. about 200 oak trees

22. a. 40%b. about 23 families

23.

24. 0

25.

26.

27.

28.

29.

30. 0

31.

5612 1

6

13561289 1 450

1 30

4-5

32.

33. 1%

34. 1.6%

35. a. Answers may vary. Sample: 20 students, 12 girls and 8 boys: 5%

b. 60%c. Answers may vary.

Sample: Subtract P(picking a boy) from 1.

36. a. 15%

b. 15%

37. a. b. c.


49

3 20

38. Answers may vary. Sample: For theoretical probability, all possible outcomes are equally likely to happen, but experimental probability is based on observed outcomes.

39.

40.

41.

42.

43.

44.25

34

3 1638 7 16581353

4-5


45. Answers may vary. Sample: You can add the numerator and denominator and make the sum the denominator, keeping the numerator the same.

46.

47.

48.

49. a. Check students’ work.b.

14 3 10 3 10

c. , , ,

d. no

e. Answers may vary. Sample: Yes; the more you roll, the closer you get to the theoretical probability.

50. A

51. G

52. D

1 18

19

19

1 36

4-5


53. [4] a. theoretical P(red) = OR 20%

b. experimental P(red) = OR about 19.6%

c. For the red beads, the manufacturer’s claim seems to be true. However, the experimental probabilities of the other colors are not as close to 20%, so Rasheeda’s experiment does not support the manufacturer’s claim.

[3] one computational error with complete explanation OR correct computation with weak explanation

[2] correct computation but no conclusion[1] error(s) in computation and no conclusion

54. [2] P(defective stapler) = 5.1%; production should be

stopped because 5.1% > 4%.

[1] correct calculation with no conclusion OR incorrect calculation but correct reasoning based on incorrect calculation

15

55 280

18 350

4-5


55. 25%; increase

56. 50%; increase

57. 40%; increase

58. 50%; decrease

59. 25%; decrease

60. 12.5%; decrease

61. –3 ≤ t ≤ 4;

62. 5 < b < 7;

63. h < 2 or h > 5;

64. –2 ≤ w < 1;

65. x < 2 or x ≥ 4;

66. 1 ≤ k ≤ 3;

67. 6.17, 5, 5

68. 17.29, 11, 35

69. 3 4 7 94 1 96 57 1

70. 1 2 4 6 2 3 7 4 76 87 9

4-5


Find each probability for the roll of a number cube.

1. P(4) 2. P(not 4) 3.P(odd)

4. You harvest 50 cherry tomatoes from your garden. You randomly inspect 15 tomatoes and find that 2 have bad spots on them.

a. What is the experimental probability that a tomato has a bad spot?

b. Predict how many of the tomatoes you picked will have bad spots.

16

56

12

about 13%

about 7 tomatoes

4-5

Probability of Compound EventsProbability of Compound EventsALGEBRA 1 LESSON 4-6ALGEBRA 1 LESSON 4-6

(For help, go to Lessons 4-5.)

Find each probability for one roll of a number cube.

1. P(multiple of 3) 2. P(greater than 4)

3. P(greater than 5) 4. P(greater than 6)

Simplify.

5. 6. 7.2

1476•

1524

1230•

655

443•

4-6


1. P(multiple of 3) = P(3 or 6) = =

2. P(greater than 4) = P(5 or 6) = =

3. P(greater than 5) = P(6) =

4. P(greater than 6) = 0

5.

6.

7.

26

13

26

13

16

214 • =

76

2 • 72 • 7 • 6 =

16

1524 • =

1230

15 • 12 12 • 2 • 15 • 2 =

16=

1 2 • 2

655 • =

443

3 • 2 • 4 • 11 5 • 11 • 3 =

85=

2 • 45 = 1

35

Solutions

4-6


Suppose you roll two number cubes. What is the probability that you will roll an odd number on the first cube and a multiple of 3 on the second cube?

P(odd and multiple of 3) = P(odd) • P(multiple of 3)

The probability that you will roll an odd number on the first cube and a multiple

of 3 on the second cube is . 16

P(odd) = =36

12

There are 3 odd numbers out of six numbers.

P(multiple of 3) = =26

13

There are 2 multiples of 3 out of 6 numbers.

= •12

13

= 16

Substitute.

Simplify.

4-6

Probability of Compound EventsProbability of Compound Events

Suppose you have 3 quarters and 5 dimes in your pocket.

You take out one coin, and then put it back. Then you take out

another coin. What is the probability that you take out a dime and then

a quarter?Since you replace the first coin, the events are independent.


P(dime and quarter) = P(dime) • P(quarter)

= •

=

58

38

1564

Multiply.

The probability that you take out a dime and then a quarter is .1564

P(dime) = There are 5 out of 8 coins that are dimes.58

P(quarter) = There are 3 out of 8 coins that are quarters.38

4-6


Suppose you have 3 quarters and 5 dimes in your pocket.

You take out one coin, but you do not put it back. Then you take out

another coin. What is the probability of first taking out a dime and then

a quarter?

P(dime then quarter) = P(dime) • P(quarter after dime)

= •

=

58

37

1556

Multiply.

The probability that you take out a dime and then a quarter is .1556

P(dime) = There are 5 out of 8 coins that are dimes.58

P(quarter after dime) = There are 3 out of 8 coins that are quarters.37

4-6


A teacher must select 2 students for a conference. The

teacher randomly picks names from among 3 freshmen,

2 sophomores, 4 juniors, and 4 seniors. What is the probability that a

junior and then a senior are chosen?

P(junior then senior) = P(junior) • P(senior after junior)

The probability that the teacher will choose a junior then a senior is .4

39

P(junior) = There are 4 juniors among 13 students.4

13

P(senior after junior) = There are 4 seniors among 12 remaining students.

412

= =

= • 4

134

12

16156

Substitute.

439 Simplify.

4-6



1.

2.

3.

4.

5.

6.

7. 1

8. 0

9.

1 36 1 18 1 1819142536

4 81

10.

11.

12.

13.

14.

15.

16.

17.

18.

19. 0

2 27191681 4 27 4 27 2 11 3 11 1 55 3 11

20. 1

21.

22.

23.

24.

25.

26.

27. 0

28.

29. Indep.; you still have 2 choices for each coin with or without the other coin.

27 3 22162919 1 15

1 45

30. Dep.; with one name gone the data set changes.

31. Indep.; the data set hasn’t changed.

32. Answers may vary. Sample: For dep. events, the outcome of the first event affects the outcome of the second (example: picking a marble out of a bag, and then picking a second marble without replacing the first one). For independent events, the outcomes do not affect each other (example: picking the second marble after replacing the first).

4-6

33. a. 0.58 b. 0.003248

34. a-c. Check students’ work.

35. 0.0036

36.

37.

38.

39.

40.

41.

42. a.

b.


16 1 10 1 1215 1 15

c.

d.

e. Answers may vary.

Sample: 1; + + + = 1

43. a. b. c. 5

44. a. b. c.

45. a. 12 b. c.

46. C

47. F

48. B

1 18

27

1577

20772077

27

1577

2077

2077

1 3125

1 15,625

1 36

1 36

16

56

13

4-6


49. [2] P(green, green) = • = = ,

P(red, red) = • = = ,

P(r, r) is twice as likely as P(g, g).[1] correct calculations for both

probabilities but incorrect statement OR correct calculations for one probability and correct statement based on that answer

50.

51.

52.

53.

54. 4, –4

39

28

6 72

1 12

49

38

1272

16

11214 2127 8 21

55. 2

56. all real numbers

57. No solution; abs. value can’t be negative.

58. No solution; abs. value can’t be negative.

59. t < 17 or t > 35

4-6


You roll two number cubes. Find each probably.

1. P(odd and even) 2. P(1 or 2 and less than 5)

You select letters from the following: A A B B B C D D E F G G G and do not replace them. Find each probability.

3. P(A then B) 4. P(vowel then G)

14

29

126

352

4-6

Solving and Applying ProportionsSolving and Applying ProportionsALGEBRA 1 CHAPTER 4ALGEBRA 1 CHAPTER 4

1. 15

2. 7.5

3. 2.4

4. 20

5. 40

6. 64%

7. 20

8. 12 cm

9. 4%

10. $7.80

11. 11.1%; increase

12. 25%; decrease

13. 10%; decrease

14. 33.3%; increase

15.

16.

17. 2.24

18.

19. 3080

20. 1

21. 162.5 mi

22. 12.5 ft

3515

16

23. a.

b.

c. 0

24. 12 carnations for $6.99

25. Answers may vary. Sample: Four cards have one letter each: A, B, C, or D. What is the probability that the first card you select is A and the second is B, if you don’t replace the first card before selecting the second card?

9 1412

1 12

4-A

26. a. about 1143%b. Sample: Use the second row. Subtract the amount in the first column

from the amount in the second column. Divide the result by the amount in the first column and multiply by 100.

27. a.

b.

c.

Solving and Applying ProportionsSolving and Applying Proportions

14 4 1514

ALGEBRA 1 CHAPTER 4ALGEBRA 1 CHAPTER 4

4-A

Ratios and Proportion Write each fraction in simplest form. 1.2.3. Simplify each product. 4.5.6....

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