1. (For help, go to Lessons 1-4, 1-5 and 1-6) ALGEBRA 1 LESSON 2-1 Simplify each expression. 1. x – 2 + 22. n + 2 – 23. 54. For each number, state its opposite and its reciprocal. 5. 26. –27. 8. – 7m 7 c5c5 3535 3535 Solving One-Step Equations 2-1

2. Solutions 1. x – 2 + 2 = x + 0 = x 2. n + 2 – 2 = n + 0 = n 3. 5 = c = c 1 = c 4. = m = 1 m = m c5c55 7m77m77 5. opposite of 2 is –2; reciprocal of 2 is 1212 6. opposite of –2 is 2; reciprocal of –2 is – 1212 7. opposite of is – ; reciprocal of is 3535 3535 3535 5353 8. opposite of – is ; reciprocal of – is – 3535 3535 3535 5353 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

3. Solve b – 5 = 7. b – 5 + 5=7 + 5Add 5 to each side to get the variable alone on the left side of the equal sign. b=12Simplify. Check:b – 5=12Check your solution in the original equation. 12 – 57Substitute 12 for b. 7=77=7 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

4. Find the value of a. PQ=PR 11=6 + aThe lengths of congruent sides are equal. 5=aSimplify. The value of a is 5. 11 – 6=6 + a – 6Subtract 6 from each side. Check:11=6 + aCheck your solution in the original equation. 116 + 5Substitute 5 for a. 11=11 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

5. Together, you and your puppy weigh 128 lb. If you alone weigh 115 lb, how much does your puppy weigh? 115 + p=128 115 + p – 115=128 – 115Subtract 115 from each side. p=13Simplify. The puppy weighs 13 lb. Check: Is the solution reasonable? The puppy weighs about 15 lb. The puppy’s weight plus your weight is about 130 lb, which is close to 128 lb. The answer is reasonable. Relate: your weight plus puppy’s weight equals combined weight Define: Let p = the puppy’s weight. Write: 115 + p = 128 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

6. Solve – = –9.3 t6t6 t=55.8Simplify. Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1 – 6 ( – ) =– 6(9.3)Multiply by –6 on each side to get the variable alone on the left side of the equal sign. t6t6

7. Solve y = –10 2525 y=–25Simplify. Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1 ( ) y= (–10)Multiply each side by, the reciprocal of. 2525 5252 5252 5252 2525 Check:y=–10 2525 (–25)–10Substitute –25 for y. 2525 –10=–10

8. Solve –6m = 42. m=–7Simplify. =Divide each side by –6. –6m –6 42 –6 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

9. Solve each equation. 1. b – 8 = –22. –12 = x + 93. – = 14 4. 28 = x5. 12r = –72 4545 y7y7 6–21–98 35–6 Solving One-Step Equations ALGEBRA 1 LESSON 2-1 2-1

10. (For help, go to Lessons 2-1.) ALGEBRA 1 LESSON 2-2 Solve each equation and tell which property of equality you used. 1. x – 5 = 142. x + 3.8 = 93. –7 + x = 7 4. x – 13 = 205. = 86. 9 = 3x Solve each equation. 7. 10x = 28. x = –69. x + 2 = 6 2323 x4x4 3434 1212 Solving Two-Step Equations 2-2

11. Solutions Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 1515 1. x – 5=14 x – 5 + 5=14 + 5Addition Property x=19of Equality 2. x + 3.8=9 x + 3.8 – 3.8=9 – 3.8Subtraction Property x=5.2of Equality 3. –7 + x=7 –7 + x + 7=7 + 7Addition Property x=14of Equality 4. x – 13=20 x – 13 + 13=20 + 13Addition Property x=33of Equation 5. =8 4=8 4Multiplication Property x=32of Equality x4x4 x4x4 6. 9=3x =Division Property 3=xof Equality 3x33x3 9393 7. 10x=2 = x= 2 10 10x 10 2-2

12. Solutions (continued) 8. x=–6 x= –6 x=–9 2323 2323 3232 3232 9. x + 2 =6 x + 2 – 2 =6 – 2 x=3 3434 1212 3434 3434 3434 1212 3434 Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2

13. Solve 13 = + 5. y3y3 24 = y Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2 13 – 5 = + 5 – 5Subtract 5 from each side. y3y3 8 = Simplify. y3y3 3 8 = 3 Multiply each side by 3. y3y3 Check: 13 = + 5 y3y3 13 + 5Substitute 24 for y. 24 3 13 8 + 5 13 = 13

14. You order iris bulbs from a catalog. Iris bulbs cost $.90 each. The shipping charge is $2.50. If you have $18.50 to spend, how many iris bulbs can you order? Relate: cost per iris times number of iris bulbs plus shipping equals amount to spend. Define: Let b = the number of bulbs you order. Write: 0.90 b + 2.50 = 18.50 0.90b + 2.50 = 18.50 0.90b + 2.50 – 2.50 = 18.50 – 2.50Subtract 2.50 from each side. 0.90b = 16Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2

15. You can order 17 bulbs. b = 17.7 0.90b 0.90 = 16 0.90 Divide each side by 0.90. Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2 (continued) Check: Is the solution reasonable? You can only order whole iris bulbs. Since 18 bulbs would cost 18 0.90 = 16.20 plus $2.50 for shipping, which is more than $18.50, you can only order 17 bulbs.

16. Solve 21 = –p + 8 21 – 8=–p + 8 – 8Subtract 8 from each side. 13=–pSimplify. –1(13)=–1(–p)Use the Multiplication Property of Equality. Multiply each side by –1. –13=pSimplify. Check: 21=–p + 8 21–(–13) + 8Substitute –13 for p. 21=21Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2

17. Solve 8 = – + 4. Justify each step. –96=cSimplify. c 24 Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2 8 – 4=– + 4 – 4Subtraction Property of Equality c 24 4=–Simplify. c 24 –24(4)=(–24)(– )Multiplication Property of Equality c 24

18. Solve 3 – 5z = 18. Justify each step. 3 – 5z – 3=18 – 3Subtraction Property of Equality –5z=15Simplify. z=–3Simplify. Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2 =Division Property of Equality –5z –5 15 –5

19. Solve each equation. 1. 3b + 8 = –102. –12 = –3x – 9 3. – + 7 = 144. –x – 13 = 35 5. What is the justification for the following step? 12 – 2y = 46 12 – 2y – 12 = 46 – 12 c4c4 –61 –28–48 Subtraction Property of Equality Solving Two-Step Equations ALGEBRA 1 LESSON 2-2 2-2

20. (For help, go to Lessons 1-2 and 1-7.) ALGEBRA 1 LESSON 2-3 Simplify each expression. 1.2n – 3n2.–4 + 3b + 2 + 5b 3.9(w – 5)4.–10(b – 12) 5.3(–x + 4)6.5(6 – w) Evaluate each expression. 7.28 – a + 4a for a = 58.8 + x – 7x for x = –3 9.(8n + 1)3 for n = –210.–(17 + 3y) for y = 6 Solving Multi-Step Equations 2-3

21. Solutions 1. 2n – 3n = (2 – 3)n = –1n = –n 2.–4 + 3b + 2 + 5b = (3 + 5)b + (–4 + 2) = 8b – 2 3. 9(w – 5) = 9w – 9(5) = 9w – 45 4. –10(b – 12) = –10b – (–10)(12) = –10b + 120 5. 3(–x + 4) = 3(–x) + 3(4) = –3x + 12 6. 5(6 – w) = 5(6) – 5w = 30 – 5w 7. 28 – a + 4a for a = 5: 28 – 5 + 4(5) = 28 – 5 + 20 = 23 + 20 = 43 8. 8 + x – 7x for x = –3: 8 + (–3) – 7(–3) = 8 + (–3) + 21 = 5 + 21 = 26 9. (8n + 1)3 for n = –2: (8(–2) + 1)3 = (–16 + 1)3 = (–15)3 = –45 10. –(17 + 3y) for y = 6: –(17 + 3(6)) = –(17 + 18) = –35 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

22. Solve 3a + 6 + a = 90 4a + 6=90Combine like terms. 4a + 6 – 6=90 – 6Subtract 6 from each side. 4a=84Simplify. 3a + 6 + a=90 Check: 3(21) + 6 + 2190Substitute 21 for a. 63 + 6 + 2190 90=90 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 4. a=21Simplify. 4a44a4 84 4

23. You need to build a rectangular pen in your back yard for your dog. One side of the pen will be against the house. Two sides of the pen have a length of x ft and the width will be 25 ft. What is the greatest length the pen can be if you have 63 ft of fencing? Relate: length plus 25 ft plus length equals amount of side of side of fencing Define: Let x = length of a side adjacent to the house. Write: x + 25 + x = 63 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

24. x + 25 + x=63 The pen can be 19 ft long. 2x + 25=63Combine like terms. 2x + 25 – 25=63 – 25Subtract 25 from each side. 2x=38Simplify. x=19 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 (continued) =Divide each side by 2. 2x22x2 38 2

25. Solve 2(x – 3) = 8 2x – 6=8Use the Distributive Property. 2x – 6 + 6=8 + 6Add 6 to each side. 2x=14Simplify. x=7Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 2. 2x22x2 14 2

26. Solve + = 17 3x23x2 x5x5 x=10Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 Method 1: Finding common denominators + =17 3x23x2 x5x5 x + x=17Rewrite the equation. 3232 1515 x=17Combine like terms. 17 10 ( x ) = (17)Multiply each each by the reciprocal of, which is. 17 10 10 17 17 10 10 17 x + x=17A common denominator of and is 10. 15 10 2 10 3232 1515

27. Solve + = 17 3x23x2 x5x5 15x + 2x=170Multiply. 17x=170Combine like terms. x=10Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 Method 2: Multiplying to clear fractions + =17 3x23x2 x5x5 10 ( + ) =10(17)Multiply each side by 10, a common multiple of 2 and 5. 3x23x2 x5x5 10 ( ) + 10 ( ) =10(17)Use the Distributive Property. 3x23x2 x5x5 =Divide each side by 17. 17x 17 170 17

28. Solve 0.6a + 18.65 = 22.85. 100(0.6a + 18.65)=100(22.85)The greatest of decimal places is two places. Multiply each side by 100. 100(0.6a) + 100(18.65)=100(22.85)Use the Distributive Property. 60a + 1865=2285Simplify. 60a + 1865 – 1865=2285 – 1865Subtract 1865 from each side. 60a=420Simplify. Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3 =Divide each side by 60. 60a 60 420 60 a=7Simplify.

29. Solve each equation. 1. 4a + 3 – a = 242. –3(x – 5) = 66 3. + = 74. 0.05x + 24.65 = 27.5 n3n3 n4n4 7–17 1257 Solving Multi-Step Equations ALGEBRA 1 LESSON 2-3 2-3

30. (For help, go to Lessons 1-7 and 2-3.) Simplify. 1.6x – 2x2.2x – 6x3.5x – 5x4.–5x + 5x Solve each equation. 5.4x + 3 = –56.–x + 7 = 12 7.2t – 8t + 1 = 438.0 = –7n + 4 – 5n Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

31. Solutions 1.6x – 2x=(6 – 2)x = 4x2.2x – 6x= (2 – 6)x = –4x 3.5x – 5x=(5 – 5)x = 0x = 04.–5x + 5x=(–5 + 5)x = 0x = 0 5.4x + 3=–56.–x + 7=12 4x=–8–x=5 x=–2x=–5 7.2t – 8t + 1=438.0=–7n + 4 – 5n –6t + 1=430=–12n + 4 –6t=4212n=4 t=–7n= 1313 Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

32. The measure of an angle is (5x – 3)°. Its vertical angle has a measure of (2x + 12)°. Find the value of x. 5x – 3=2x + 12Vertical angles are congruent. 5x – 3 – 2x=2x + 12 – 2xSubtract 2x from each side. 3x – 3=12Combine like terms. 3x – 3 + 3=12 + 3Add 3 to each side. 3x=15Simplify. Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 =Divide each side by 3. 3x33x3 15 3 x=5Simplify.

33. You can buy a skateboard for $60 from a friend and rent the safety equipment for $1.50 per hour. Or you can rent all items you need for $5.50 per hour. How many hours must you use a skateboard to justify buying your friend’s skateboard? Relate: cost of plus equipment equals skateboard and equipment friend’s rental rental skateboard Define: let h = the number of hours you must skateboard Write: 60 + 1.5 h = 5.5 h Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4

34. 60 + 1.5h=5.5h 60 + 1.5h – 1.5h=5.5h – 1.5hSubtract 1.5h from each side. 60=4hCombine like terms. You must use your skateboard for more than 15 hours to justify buying the skateboard. Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 (continued) 60 4 4h44h4 =Divide each side by 4. 15=hSimplify.

35. Solve each equation. a.–6z + 8=z + 10 – 7z –6z + 8=z + 10 – 7z –6z + 8=–6z + 10Combine like terms. –6z + 8 + 6z=–6z + 10 + 6zAdd 6z to each side. 8=10Not true for any value of z! This equation has no solution Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 b.4 – 4y=–2(2y – 2) The equation is true for every value of y, so the equation is an identity. 4 – 4y=–2(2y – 2) 4 – 4y=–4y + 4Use the Distributive Property. 4 – 4y + 4y=–4y + 4 + 4yAdd 4y to each side. 4=4Always true! 2-4

36. Solve each equation. 1.3 – 2t = 7t + 42. 4n = 2(n + 1) + 3(n – 1) 3.3(1 – 2x) = 4 – 6x 4.You work for a delivery service. With Plan A, you can earn $5 per hour plus $.75 per delivery. With Plan B, you can earn $7 per hour plus $.25 per delivery. How many deliveries must you make per hour with Plan A to earn as much as with Plan B? 4 deliveries 1 no solution Equations with Variables on Both Sides ALGEBRA 1 LESSON 2-4 2-4 – 1919

37. (For help, go to Lesson 1-1.) ALGEBRA 1 LESSON 2-5 Write a variable expression for each situation. 1.value in cents of q quarters 2.twice the length 3.number of miles traveled at 34 mi/h in h hours 4.weight of 5 crates if each crate weighs x kilograms 5.cost of n items at $3.99 per item Equations and Problem Solving 2-5

38. Solutions 1.value in cents of q quarters: 25q 2.twice the length : 2 3.number of miles traveled at 34 mi/h in h hours: 34h 4.weight of 5 crates if each crate weighs x kilograms: 5x 5.cost of n items at $3.99 per item: 3.99n Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

39. The width of a rectangle is 3 in. less than its length. The perimeter of the rectangle is 26 in. What is the width of the rectangle? Relate: The width is 3 in. less than the length. Then x – 3 = the width. Define: Let x = the length. The width is described in terms of the length. So define a variable for the length first. Write: P = 2 + 2wUse the perimeter formula. 26=2 x + 2( x – 3 ) Substitute 26 for P, x for, and x – 3 for w. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

40. 26= 2x + 2x – 6Use the Distributive Property. 26= 4x – 6Combine like terms. 26 + 6= 4x – 6 + 6Add 6 to each side. 32= 4xSimplify. The width of the rectangle is 3 in. less than the length, which is 8 in. So the width of the rectangle is 5 in. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5 (continued) = Divide each side by 4. 8= xSimplify. 32 4 4x44x4

41. The sum of three consecutive integers is 72. Find the integers. Relate: first plus second plus third is 72 integer integer integer Define: Let x = the first integer. Then x + 1 = the second integer, and x + 2 = the third integer. Write: x + x + 1 + x + 2 = 72 Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

42. If x = 23, then x + 1 = 24, and x + 2 = 25. The three integers are 23, 24, and 25. x + x + 1 + x + 2 = 72 3x + 3 = 72Combine like terms. 3x + 3 – 3 = 72 – 3Subtract 3 from each side. 3x = 69Simplify. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5 (continued) = Divide each side by 3. 3x33x3 69 3 x = 23Simplify.

43. An airplane left an airport flying at 180 mi/h. A jet that flies at 330 mi/h left 1 hour later. The jet follows the same route as the airplane on parallel altitudes. How many hours will it take the jet to catch up with the airplane? AircraftRateTimeDistance Traveled Airplane180t180t Jet330t – 1330(t – 1) Define:Let t = the time the airplane travels. Then t – 1 = the time the jet travels. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

44. Relate:distance traveledequalsdistance traveled by airplaneby jet Write:180 t=330( t – 1 ) 180t=330(t – 1) 180t=330t – 330Use the Distributive Property. 180t – 330t=330t – 330 – 330tSubtract 330t from each side. –150t=–330Combine like terms. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5 (continued) =Divide each side by –150. –150t –150 –330 –150 t =2Simplify. 1515 t – 1 =1 1515 The jet will catch up with the airplane in 1 h. 1515

45. Suppose you hike up a hill at 4 km/h. You hike back down at 6 km/h. Your hiking trip took 3 hours. How long was your trip up the hill? Define:Let x = time of trip uphill. Then 3 – x = time of trip downhill. Relate: distance uphillequals distance downhill Part of hikeRateTimeDistance hiked Uphill4x4x Downhill63 – x6(3 – x) Write:4 x= 6( 3 – x ) Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

46. 4x=6(3 – x) 4x=18 – 6xUse the Distributive Property. 4x + 6x=18 – 6x + 6xAdd 6x to each side. 10x=18Combine like terms. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5 (continued) =Divide each side by 10. 10x 10 18 10 x=1Simplify. 4545 Your trip uphill was 1 h long. 4545

47. Two jets leave Dallas at the same time and fly in opposite directions. One is flying west 50 mi/h faster than the other. After 2 hours, they are 2500 miles apart. Find the speed of each jet. Define:Let x = the speed of the jet flying east. Write: 2 x + 2( x + 50 ) = 2500 Then x + 50 = the speed of the jet flying west. Relate:eastbound jet’s plus westbound jet’s equals the total distance distance distance JetRateTimeDistance Traveled Eastboundx22x Westboundx + 5022(x + 50) Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

48. 2x + 2(x + 50)=2500 2x + 2x + 100=2500Use the Distributive Property. 4x + 100=2500Combine like terms. 4x + 100 – 100=2500 – 100Subtract 100 from each side. 4x=2400Simplify. x=600 x + 50=650 The jet flying east is flying at 600 mi/h. The jet flying west is flying at 650 mi/h. Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5 (continued) =Divide each side by 4. 4x44x4 2400 4

49. 1.The sum of three consecutive integers is 117. Find the integers. 2.You and your brother started biking at noon from places that are 52 mi apart. You rode toward each other and met at 2:00 p.m. Your brother’s average speed was 4 mi/h faster than your average speed. Find both speeds. 3.Joan ran from her home to the lake at 8 mi/h. She ran back home at 6 mi/h. Her total running time was 32 minutes. How much time did it take Joan to run from her home to the lake? 38, 39, 40 your speed: 11 mi/h; brother’s speed: 15 mi/h about 13.7 minutes Equations and Problem Solving ALGEBRA 1 LESSON 2-5 2-5

50. (For help, go to Lessons 1-2, 1-4 and 1-6.) Evaluate each formula for the values given. 1.distance: d = rt, when r = 60 mi/h and t = 3 h 2.perimeter of a rectangle: P = 2 + 2w, when = 11 cm, and w = 5 cm 3.area of a triangle: A = bh, when b = 8 m and h = 7 m 1212 Formulas ALGEBRA 1 LESSON 2-6 2-6

51. Solutions 1.d = rt = 60(3) = 180 mi 2.P = 2 + 2w = 2(11) + 2(5) = 22 + 10 = 32 cm 3.A = bh = (8)(7) = 4(7) = 28 m 2 1212 1212 Formulas ALGEBRA 1 LESSON 2-6 2-6

52. Solve the formula for the volume of a rectangular prism V = wh for width w in terms of its volume V, length, and its height h. V= wh =wh Simplify. V Formulas ALGEBRA 1 LESSON 2-6 2-6 = Divide each side by, = 0. V wh / = Divide each side by h, h = 0, to get w alone on one side of the equation. V h wh h / =w Simplify. V h

53. Solve y = 4x – 3 for x. y + 3=4x – 3 + 3Add 3 to each side. y + 3=4xSimplify. Formulas ALGEBRA 1 LESSON 2-6 2-6 =Divide each side by 4. y + 3 4 4x44x4 =xSimplify. y + 3 4

54. Solve z – br = p for b in terms of z, r, and p. z – br – z=p – zSubtract z from each side. –br=p – zCombine like terms. Formulas ALGEBRA 1 LESSON 2-6 2-6 =Divide each side by –r. –br –r p – z –r b=–Simplify. p – z –r

55. The formula K = C + 273.15 gives the Kelvin temperature K in terms of the Celsius temperature C. Transform the formula to find Celsius temperature in terms of Kelvin temperature. Then, find the Celsius temperature when the Kelvin temperature is 400°. Solve for C. K=C + 273.15 K – 273.15=C + 273.15 – 273.15Subtract 273.15 from each side. K – 273.15=C Simplify. Find C when K = 400. 400 – 273.15=C Substitute 400 for K. 126.85=C 400º K is 126.85º C. Formulas ALGEBRA 1 LESSON 2-6 2-6

56. Solve each equation for the given variable. 1. 3x + 2y = z; y2. = ; a 3. 3x + 4 = 2(3 – y); y 4.The formula v 2 = can be used to find the constant speed v of a satellite revolving around Earth in a circular orbit of radius r. G is a number known as the gravitational constant and M is the mass of Earth. Transform this formula to find the mass of Earth. a – b c d3d3 GM r M = v2rGv2rG Formulas ALGEBRA 1 LESSON 2-6 2-6 a = + b cd 3 y = z – x 1212 3232 y = – x + 1 3232

57. (For help, go to Lessons 1-4 and 1-6.) Write the numbers in each group in order from least to greatest. 1.2.4, 9.8, 3.6, 7.5, 1.92.144, 235, 98, 72, 58, 195 3.–12, 14, –3, –8, 7, 04.2, –3, –4, 6, –2, 4 Use mental math to simplify. 5.6. 3 + 4 + 5 + 6 + 7 5 5 + 6 + 8 + 9 4 1212 2323 3838 1414 5858 1212 Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

58. Solutions 1. 1.9, 2.4, 3.6, 7.5, 9.8 2. 58, 72, 98, 144, 195, 235 3. –12, –8, –3, 0, 7, 14 4. –4, –3, –2, 2, 4, 6 5. = = 5 6. = = 7 3838 2323 5858 1212 1212 1414 3 + 4 + 5 + 6 + 7 5 25 5 5 + 6 + 8 + 9 4 28 4 Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

59. Find the mean, median, and mode of the data below. Determine which measure of central tendency best describes the data. 14 10 2 13 16 3 12 11 Mode: noneThe data item that occurs most often. The mean is 10.125, the median is 11.5, and there is no mode. The mean is less than 5 of the 8 data items because of the outliers 2 and 3. The median best describes the data. Mean: 14 + 10 + 2 + 13 + 16 + 3 + 12 + 11 8 = 10.125 total number of data items Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7 Median: 2 3 10 11 12 13 14 16List data in order. = 11.5For an even number of data items, find the mean of the two middle terms. 11 + 12 2

60. 82 + 94 + 89 + x 4 Mean (average): = 90 Let x = the grade of the fourth exam. Suppose your grades on three science exams are 82, 94, and 89. What grade do you need on your next exam to have an average of 90? 265 + x 4 = 90Simplify the numerator. 4 ( ) = 4(90)Multiply each side by 4. 265 + x 4 265 + x= 360Simplify. Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

61. (continued) 265 + x – 265= 360 – 265Subtract 265 from each side. x= 95Simplify. Your grade on the next exam must be 95 for you to have an average of 90. Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

62. Find the range and mean of each set of data. Use the range to compare the spread of the two sets of data. 45 47 34 36 3856 35 27 47 35 Range: 47 – 34=13Range: 56 – 27=29 Both sets have a mean of 40. The range of the first set of data is 13 and the range of the second set of data is 29. The first set of data clusters nearer the mean. Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7 Mean: 45 + 47 + 34 + 36 + 38 5 = 200 5 =40 Mean: 56+ 35 + 27 + 47 + 35 5 = 200 5 =40

63. Make a stem-and-leaf plot for the data. 56 44 63 58 51 59 47 51 67 50 65 49 66 63 5|1 means 51 Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

64. Find the mean of the city mileage and the mean of the highway mileage. 4|2 means 42 Mean City Mileage: 17 + 18 + 20 + 21 + 24 + 26 + 36 + 38 8 = 25 mi/gal Mean Highway Mileage: 22 + 24 + 30 +30 + 32 + 33 + 38 + 43 8 = 31.5 mi/gal Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7

65. 1.Find the mean, median, and mode. Which measure of central tendency best describes the data? 49 52 53 56 62 61 55 52 2.Make a stem-and-leaf plot for the data above. 3.Your test grades are 83, 94, 86, and 91. What grade do you need to earn on your next test to have an average of 90? mean 55; median 54; mode 52; The mean best describes the data. 4|9 means 49 96 Using Measures of Central Tendency ALGEBRA 1 LESSON 2-7 2-7