31 Chapter 2 Modeling with Linear Functions Homework 2.1 1.a. b. c.In 1993, t = 13.Notice on the scattergram, when t = 13,p is approximately 43.5.Therefore, 43.5% of deaths due to car accidents in 1993 were due to drunk driving. d.When p = 51, t is approximately 8, which represents 1988.Therefore, it is estimated that 51% of deaths due to car accidents were due to drunk driving in 1988. 3.a. b. c.In 1997, t = 17.Notice on the scattergram, when t = 17, n is approximately 4.Therefore, the number of collisions in 1997 is estimated at 4000. d.When n = 1, t is approximately 27, which represents the year 2007.Therefore, it is estimated that in 2007, the number of collisions will be one thousand.Extrapolation was used to determine the number of collisions. e.The n-intercept for this linear model is (0, 8.8).This means that in 1980 (t = 0), there were 8800 collisions. f.The t-intercept for this model is (30.4, 0).This means that in the year 2010 (t = 30), it is predicted that there will be no collisions.This is extremely unlikely and gives an obviously false prediction.Model breakdown has occurred. 5.a. b. c.The n-intercept for this linear model is (0, 18.8).This means that in 1995 (t = 0), there were 18.8 million Internet users. d.The number of Internet users is increasing by approximately 20 million users per year. Homework 2.1 SSM: Intermediate Algebra 32 e.When n = 287, t is approximately 12, which represents 1995 + 12 = 2007.This means that in the year 2007, everyone is the United States will be an Internet user.This is highly unlikely.It is extremely improbable that everyone in the United States will ever become an Internet user.This shows that model breakdown has occurred. 7.a. b. c.In 1999, t = 14.Notice on the scattergram, when t = 14, n is approximately $175.This is an overestimate of $25 by the model. d.In 2001, t = 16.Notice on the scattergram, when t = 16, n is approximately $190.This is an overestimate of $30 by the model. e.Answers may vary. f. 9.a. b. According to the model, the value of a share on January 1, 2001 was approximately $150. c. d.The difference between the estimated value and the actual value of the stock on January 1, 2001 is$150 $43 = $107.The model did not take into account the investors sudden lack of confidence in dot-com stocks that occurred after January 2000. 11.a.To find the t-intercept, estimate the number of years since 1990 when D = 0.This estimate is approximately (7, 0).This point represents that in the year 1997 (t = 7), the percentages of sound recording sales for both pop and rap/hip-hop were the same. b.To find the D-intercept, estimate when t = 0.This estimate is approximately (0, 5.2).This point represents a 5.2% difference in the percentage of sales between pop and rap/hip-hop.In the year 1990, pop sound recordings outsold rap/hip-hop sound recordings by approximately 5.2%. SSM: Intermediate AlgebraHomework 2.2 33 13.Answers may vary.A linear model is a linear function that describes the relationship between two quantities for a true-to-life situation.Every linear model is a linear function.However, not every linear function is a linear model.Functions are used both to describe situations and to describe certain mathematical relationships between two variables. 15.Answers may vary. Homework 2.2 1. In this model, m will be the same as the original model.However, the value of b will increase. 3.Use (3, 5) and (7, 15) to write the equation of the line.First, find the slope. 15 5 10 57 3 4 2m= = = So, y = 2.5x + b.Substitute 3 for x and 5 for y since the line contains (3, 5) and then solve for b. ( )55 3210 152 252y mx bbbb= += += + = The equation of the line is y = 2.5x 2.5.(Your equation may be slightly different if you chose different points.)Use the graphing calculator to check your results. 5.Check the scatterplot obtained from the graphing calculator. Student C made the best choice of points. The line obtained by student A will have a slope that is not as steep as the actual slope.The line obtained by student B will have a slope that is steeper than the actual slope. 7.a. b.Use the points (75, 14.3) and (100, 33.2) to find the equation of the line. 33.2 14.3 18.90.77100 75 25m= = So, y = 0.77x + b.Substitute 75 for x and 14.3 for y since the line contains (75, 14.3) and then solve for b.( ) 14.3 0.77 7514.3 57.7543.45y mx bbbb= += += + = The equation of the line is y = 0.77x 43.45.(Your equation may be slightly different if you chose two other points.) c. 9.a. Homework 2.2 SSM: Intermediate Algebra 34 b.No, the data point (10, 184) does not fit the pattern of the rest of the data.This could have happened for a number of reasons including the illness of the student after having had 10 drinks. c.Use the points (0, 80) and (9, 135) to find the equation of the line. 135 80 556.19 0 9m= = So, y = 6.1x + b.Substitute 0 for x and 80 for y since the line contains (0, 80) and then solve for b.( ) 80 6.1 080y mx bbb= += += The equation of the line is y = 6.1x + 80.(Your equation may be slightly different if you chose different points.) 11.a. b.Use the points (6, 97) and (11, 78) to find the equation of the line. 97 78 193.86 11 5m= = = So, y = 3.8x + b.Substitute 6 for x and 97 for y since the line contains (6, 97) and then solve for b.( ) 97 3.8 697 22.8119.8y mx bbbb= += += += The equation of the line is y = 3.8x + 119.8.(Your equation may be slightly different if you chose different points.) c. 13.a. b.Use the points (69, 51.7) and (83, 47.99) to find the equation of the line. 47.99 51.7 3.710.2783 69 14m = = So, y = 0.27x + b.Substitute 69 for x and 51.7 for y since the line contains (69, 51.7) and then solve for b. ( ) 51.7 0.27 6951.7 18.6370.33y mx bbbb= += += += The equation of the line isy = 0.27x +70.33. (Your equation may be slightly different if you chose different points.) c. 15.a. b.Yes, the model predicts that the womens record time will equal the mens record time.This will happen in the year 2003 with a record time of approximately 42.6 seconds. c.Yes, the model predicts that the womens record time will be less than the mens record time in years 2004 and after. SSM: Intermediate AlgebraHomework 2.3 35 17.a. b.Use the points (6, 1.7) and (10, 0.5) to find the equation of the line. 1.7 0.5 1.20.36 10 4m= = = So, y = 0.3x + b.Substitute 6 for x and 1.7 for y since the line contains (6, 1.7) and then solve for b.( ) 1.7 0.3 61.7 1.83.5y mx bbbb= += += += The equation of the line is y = 0.3x + 3.5.(Your equation may be slightly different if you chose different points.) c. d. The regression equation from the graphing calculator is y = 0.28x + 3.32. e. The models are so similar that it is difficult to distinguish them on the graphing calculator screen. 19.a. b.Use the points (23, 34) and (43, 84) to find the equation of the line. 84 34 502.543 23 20m= = = So, y = 2.5x + b.Substitute 23 for x and 34 for y since the line contains (23, 34) and then solve for b.( ) 34 2.5 2334 57.523.5y mx bbbb= += += + = The equation of the line is y = 2.5x 23.5.(Your equation may be slightly different if you chose different points.) c. Homework 2.3 1.1.3 17 y x + =Substitute 2 for x and solve for y. 3 2 173 155yyy+ === 3.3 2 27 y x = Substitute 4 for x and solve for y. ( ) 3 2 4 273 8 273 19193yyyy = = = = 5.2 6 49 y x = Substitute 5 for x and solve for y. Homework 2.3 SSM: Intermediate Algebra 36 ( ) 2 6 5 492 30 492 79792yyyy = + = = = 7. ( ) 5 5 11 4 y x + = +Substitute 1 for x and solve for y. ( ) ( ) 5 5 11 4 15 25 11 45 25 155 102yyyyy + = + = = == 9.2.7 3.8 5.9 y x =Substitute 6.6 for x and solve for y. ( ) 2.7 3.8 6.6 5.92.7 25.08 5.92.7 30.9811.47yyyy = == 11. ( ) ( ) 1 3 1 2 3 2 1 f = + = + = 13. ( ) ( ) 9 3 9 2 27 2 25 f = + = + = 15. ( ) ( ) 2 3 2 2 6 2 8 f = + = + = 17. ( ) ( ) 2.3 6 2.3 4 13.8 4 9.8 g = = = 19. 1 16 4 1 4 36 6g = = = 21. 2 26 4 4 4 83 3g = = = 23. ( ) ( ) ( ) 4 1 4 1 f a a a = = 25. ( ) ( ) 1 4 1 1 4 4 1 4 5 f a a a a + = + = = 27. ( ) ( ) 4 1 4 4 1 f a h a h a h + = + = 29. To find x when( ) 6 f x = , substitute 6 for( ) f xand solve for x. 6 3 73 113xxx= +== 31. To find x when( ) 4.9 f x = , substitute 4.9 for ( ) f xand solve for x. 4.9 3 73 11.94.0xxx = += 33. To find x when( )23f x = , substitute 23for ( ) f xand solve for x. SSM: Intermediate AlgebraHomework 2.3 37 22 531323136xxx = == 35. To find x when( ) f x a = , substitute a for ( ) f xand solve for x. 2 55 21 52 2a xa xx a= + == + 37.Since the line contains the point (6, 4), f(6) = 4. 39.Since the line contains the point (3, 1), f(3) = 1. 41.Since the line contains the point (2.5, 1.2),f(2.5) = 1.2. 43.Since the line contains the point (6, 0), x = 6when f(x) or y = 0. 45.Since the line contains the point (3, 3), x = 3 when f(x) or y = 3. 47. Since the line contains the point 9 1,2 2 , x = 92 or 4.5 when f(x) or y = 12. 49. ( ) 2 4 f = 51. When( ) 2 f x = , x is 1. 53. ( ) or 5 8 f x y x = To find the y-intercept, set x = 0 and solve for y. ( ) 5 0 80 88yyy= = = To find the x-intercept, set y = 0 and solve for x. 0 5 88 585xxx= == 55. ( ) or 3 f x y x =To find the y-intercept, set x = 0 and solve for y. ( ) 3 0 0 y = =To find the x-intercept, set y = 0 and solve for x. 0 30xx== 57. ( ) or 5 f x y =is the equation of a horizontal line.The y-intercept of the line will be y = 5.There is no x-intercept. 59. ( ) or 2.1 4.2 f x y x = To find the y-intercept, set x = 0 and solve for y. ( ) 2.1 0 4.2 4.2 y = = To find the x-intercept, set y = 0 and solve for x. 0 2.1 4.24.2 2.12xxx= = = 61. ( )1 or 32f x y x = To find the y-intercept, set x = 0 and solve for y. ( )10 3 32y = = To find the x-intercept, set y = 0 and solve for x. 10 321326xxx= == 63.1 2 3 9 173 1 6 179 1 179 182x x x xx xxxx + = + = + === 65. ( ) ( ) 3 6 2 43 12 6 49 6 2 49 4 44 554x x xx x xx xxxx = + = + + = + = == Homework 2.3 SSM: Intermediate Algebra 38 67. ( ) 0.5 3.9 2 2.12 70.5 1.95 2 2.12 70.5 0.05 2.12 71.62 7.054.35x xx xx xxx + = + = + = = 69. ( ) ( ) ( ) 4 2 3 1 2 64 8 3 3 2 125 2 121717x x xx x xx xxx = + + = + = + == 71. 1 1 23 4 31 1 212 123 4 34 3 84 11114xxxxx = = === 73. 3 5 14 8 23 5 18 84 8 26 5 46 99 36 2xxxxx = = = == = 75.10 1x x = += This is a contradiction.There is no value for x for which the equation is true. 77.Both solutions are correct since x = 2 in the equation2 4 0 x + = .However, Student 1 used unnecessary steps to find the solution. 79.a. ( ) 0.81 45.86 f t t = b. ( ) ( ) 108 0.81 108 45.8687.48 45.86 41.62f = = = When t = 108, p = 41.62.This means that in the year 1900 + 108 = 2008, the percentage of births out of wedlock will be 41.62%. c.43 0.81 45.8688.86 0.81110ttt= = When p = 43,110 t .This means that the percentage of out of wedlock births will hit 40% in the year 1900 + 110 = 2010. d.100 0.81 45.86145.86 0.81180ttt= = In the year 1900 + 180 = 2080, all births will be out of wedlock. e.Find p, when t = 1997 1900 = 97. ( ) ( ) 97 0.81 97 45.8678.57 45.86 32.71f = = = This means that in the year 1997, the percentage of births out of wedlock will be 32.71%.Since the actual percentage was 32.4%, the error was 32.71% 32.4% = 0.31%. 81.a. ( ) 5.85 81.18 f n n = + b.To find the T-intercept, set n = 0 and solve for T. ( ) 5.85 0 81.1881.18TT= += The T-intercept is (0, 81.18).This means that it took the student approximately 81 seconds to complete the task when sober. c.This model is an increasing function.This means that the more drinks the student consumes, the more time it takes to complete the task. d.From the graphing calculator table, we have: f(0) = 81.18,f(1) = 87.03, f(2) = 92.88,f(3) = 98.73, f(4) = 104.58,f(5) = 110.43, f(6) = 116.27,f(7) = 112.13, f(8) = 127.98,f(9) = 133.83, f(10) = 139.68. According to the model, f(4) = 104.58.This shows an error of 0.42 and this is the least error in the tables.The estimate for f(10) is 139.68, an error of 44.32, which is the greatest error in the table. This large error is due to the fact that (10, 184) does not fit the pattern of the rest of the data.SSM: Intermediate AlgebraHomework 2.3 39 83.a. b.Use the points (1, 452) and (10, 666) to find the equation of the line. 666 452 21423.810 1 9m= = So, y = 23.8x + b.Substitute 1 for x and 452 for y since the line contains (1, 452) and then solve for b.( ) 452 23.8 1452 23.8428.2y mx bbbb= += += += The equation of the line is y = 23.8x + 428.2.(Your equation may be slightly different ifyou chose different points.) Graphing the line with the scatterplot, we see that the model fits the data very well. c.Since 2001 1990 = 11, find f(11). ( ) ( ) 11 23.8 11 428.2261.8 428.2 690f = += + = According to the model, 690 million enplanements were made in 2001.The actual number of enplanements was 622 million.This is an error of 68 million enplanements. d.68 million enplanements is equivalent to 68174= million round trips.If on average each round trip is $500, the airlines lost 17 $500 $8500 = million or $8.5 billion. 85.a. The data points lie close to the line that passes through (0, 32) and (100, 212).To find the equation for f, start by finding the slope using these points. 212 32 1801.8100 0 100m= = = So, f(c) = 1.8c + b.To solve for b, substitute 0 for C and 32 for f(c) since the line contains (0, 32). 32 = 1.8(0) + b 32 = b The equation for f is f(c) = 1.8c + 32. This is also the linear regression equation. This model fits the extremely well. b. ( ) ( ) 25 1.8 25 32 45 32 77 f = + = + =If it is25 Co, it is77Fo. c.40 1.8 328 1.881.84.4cccc= +== If it is40Fo, it is4.4 Co. 87.a. The data points lie close to the line that passes through (94, 4.5) and (98, 5.7).To find the equation for f, start by finding the slope using these points. 5.7 4.5 1.20.398 94 4m= = = So, f(t) = 0.3t + b.To solve for b, substitute 94 for t and 4.5 for f(t) since the line contains (94, 4.5). ( ) 4.5 0.3 944.5 28.223.7bbb= += + = The equation for f is f(t)= 0.3c 23.7. This model fits the very well. Homework 2.3 SSM: Intermediate Algebra 40 b. ( ) ( ) 99 0.3 99 23.7 29.7 23.7 6 f = = =This means that in the year 1999, there were 6 thousand or 6,000 male-female pairs of bald eagles.

c. ( ) ( ) 15 0.3 15 23.7 5 23.7 18.7 f = = = This means that in the year 1915, there were a negative number of bald eagle pairs.This cannot be the case.Model breakdown has occurred. d.10 0.3 23.733.7 0.333.70.3112.3tttt= === This means that there will be 10,000 male-female pairs of bald eagles in the year 1900 + 112 = 2012. e.From year to year the changes are0.3,0.4,0.2,0.2,and 0.4 .On average it is 0.3 thousand pairs per year.This is the same as the slope.Answers may vary. 89.a. ( ) 2.48 23.64 f x x = b.100 2.48 23.64123.64 2.4850xxx= = This means that the cutoff score would have to be 50 (out of 50) to ensure that all students succeed in the intermediate algebra course.

c.0 2.48 23.6423.64 2.4810xxx= = This means that for scores 10 and under, no students would succeed in the intermediate algebra course. d.For the 16 20 range, we use a score of 18 to represent the group.Find the percentage for x = 18. ( ) 2.48 18 23.6444.64 23.64 21p = = = This means that 21% of the students who score in the 16 20 range pass the intermediate algebra course.If 145 students scored in this range, we could expect 21% or 0.21 145 31 students to pass. e.First calculate the percentages for each of the score groups. ( ) 2.48 23 23.64 33.4 p = =( ) 2.48 28 23.64 45.8 p = =( ) 2.48 33 23.64 58.2 p = =( ) 2.48 38 23.64 70.6 p = =( ) 2.48 43 23.64 83 p = =( ) 2.48 48 23.64 95.4 p = =Next, multiply each of the percentages by the number of students in that category. 0.334 94 31.4 =students 0.458 44 20.2 =students 0.582 19 11.1 =students 0.706 12 8.5 =students 0.83 9 7.5 =students 0.954 4 3.8 =students Add these results to obtain the total number of students, 82.5.This means that of students who scored at least 21 points on the placement test, approximately 83 students will pass the class. 91.a.From the information given, we have the data points (0, 640) and (4, 0).Use these points to find the slope of the equation for f. 640 0 6401600 4 4m= = = So, f(t) = 160t + b.To solve for b, substitute 0 for t and 640 for f(t) since the line contains (0, 640). ( ) 640 160 0640bb= += The equation for f is f(t) = 160t + 640. SSM: Intermediate AlgebraHomework 2.4 41 b.

c.Since it takes 4 hours to pump out the water, the domain is0 4 t . Since the water level is at 640 cubic feet before starting to pump, the range is( ) 0 640 f t . 93.When f(3) = 5, the input for f is 3 and the output is 5.Possible equations for f will vary.Sample equations are as follows. ( )( )( )2 114353f x xf x xf x x= = += Homework 2.4 1.a.Since the distance the student drives increases by 70 miles per hour, the slope is 70. b.An equation for f is in the form f(t) = mt + b.Since the distance is 0 when t = 0, b = 0.So, f(t) = 70t. 3.a.Number Paying Bills Online tn 0 ( ) 1.7 0 2.1 +1 ( ) 1.7 1 2.1 +2 ( ) 1.7 2 2.1 +3 ( ) 1.7 3 2.1 +4 ( ) 1.7 4 2.1 +5 ( ) 1.7 5 2.1 +t ( ) 1.7 2.1 t + b. ( ) 1.7 2.1 n t = +or2.1 1.7 n t = + c. d.The slope of the model is 2.1.This means that 2.1 million more U.S. households each year are paying bills online. 5.a.Since the cost per hour is $2.50, the slope is 2.5. b.An equation for f is in the form f(t) = mt + b.Since the cost is $4.95 when the customer is online for 0 hours, b = 4.95.So, f(t) = 2.5t + 4.95. c. d.Find n when the cost of the regular plan will be $23.90. 23.90 2.5 4.9518.95 2.57.58ttt= +== When a customer is online more than 7.58 hours, the flat rate plan is a better deal. 7.a.Since the rental fee is $15.75 per textbook, the slope is 15.75. b.An equation for f is in the form f(n) = mn + b.Since the cost of tuition is $1224.60 even when no textbooks are rented, b = 1224.60.So, f(n) = 15.75t + 1224.60. c. Find when ( ) f nis $1303.35. 1303.35 15.75 1224.6078.75 15.755nnn= +== The total cost, ( ) f n , is $1303.35 when 5 textbooks are rented. Homework 2.4 SSM: Intermediate Algebra 42 9.a.Since 0.05 gallons of gas are being used each mile, the slope is 0.05.This means that for every mile that is driven, the amount of gas in the tank decreases by 0.05 gallons. b.An equation for g is in the form g(x) = mx + b.Since at the beginning of the trip, there are 15.3 gallons of gas in the tank, when x = 0, b = 15.3.So, f(t) = 0.05t + 15.3. c.To find the x-intercept, let g(x) = 0 and solve for x. 0 0.05 15.315.3 0.05306xxx= + = = The x-intercept is (306, 0).This means that when 306 miles have been driven, the gas tank will be empty. d.Since the car can be driven between 0 and 306 miles on the 15.3 gallon tank of gas, the domain is0 306 x . e.Find x when g(x) = 1. 1 0.05 15.314.3 0.05286xxx= + = = The car can be driven 286 miles before refueling. 11.The slope of the model is 1.70.This means that the average salary for professors at four-year public colleges is increasing by $1.7 thousand or $1700 per year. 13.The slope of the model is 0.27.This means that the record time for the womens 400 -meter run is decreasing by 0.27 seconds each year. 15.Yes, t and c are linearly related.The slope is 0.99.This means that for every minute the person talks, there is an additional charge of $0.99. 17.a.Create a scattergram using the data in the table. Use the first and last data points, (0, 70) and (50, 52) to find the slope. 70 52 180.360 50 50m= = = So, the equation is of the form ( ) 0.36 f t t = b + .Substitute 0 for t and 70 for( ) f tand solve for b. ( ) 70 0.36 070 070bbb= += += The equation is( ) 0.36 70 f t t = + . Graphing the equation on the same graph as the scattergram, we see that the equation fits the data fairly well. b.The slope of f is 0.36.This means that the percentage of the worlds population that lives in rural areas decreases by 0.36 of one percent each year. c.When no one lives in a rural area, f ( t) = 0. 0 0.36 7070 0.36194ttt= + = This means that in the year 1950 + 194 = 2144, no one in the world will live in a rural area.It is highly improbable that there will ever be a time that no one lives in a rural area.Also, it is highly unlikely that a mathematical model will be valid for a span of nearly 200 years. 19.a.Create a scattergram using the data in the table. Use the first and last data points, (5, 2977) and (10, 3774) to find the slope. 3774 2977 797159.410 5 5m= = = SSM: Intermediate AlgebraHomework 2.4 43 So, the equation is of the form ( ) 159.4 f t t = b + .Substitute 5 for t and 2977 for( ) f tand solve for b. ( ) 2977 159.4 52977 7972180bbb= += += The equation is( ) 159.4 2180 f t t = + . Graphing the equation on the same graph as the scattergram, we see that the equation fits the data fairly well. b.Create a scattergram using the data in the table. Use the first and last data points, (5, 14,537) and (10, 19,312) to find the slope. 19, 312 14, 537 477595510 5 5m= = = So, the equation is of the form ( ) 955 g t t =b + .Substitute 5 for t and 2977 for( ) g tand solve for b. ( ) 14, 537 955 514, 537 47759762bbb= += += The equation is( ) 955 9762 g t t = + .Graphing the equation on the same graph as the scattergram, we see that the equation fits the data very well. c.The slope of f is $159.40 and the slope of g is $955.This tells us that the tuition at private schools is increasing more rapidly that the tuition at public schools. d.Calculate the tuition for each of the 4 years.Since t is in years since 1990, calculate f(17), f(18), f(19) and f(20), as well as, g(17), g(18), g(19) and g(20). We can use the table feature of the graphing calculator to determine these values. So, for public colleges, the cost of a four-year program will be $4890 + $5049 + $5209 + $5368 + $5527 = $20,994.Similarly, for private colleges, the cost of a four-year program will be $25,977 + $26,952 + $27,907 + $28,862 + $29,817 = $139,515. 21.a.tf(t) 00 1500 21000 31500 41900 52300 b. c.It will take some amount of time to decelerate from 500 mph to 400 mph.Answers may vary. d.The constant rate of change property does not apply in this exercise.Since the rate is 500 miles per hour for the first few hours and 400 miles per hour for the next 2 hours, the rate of change is not constant.Since we do not have a constant rate of change, we cannot say that the relationship between the variables is linear. Chapter 2 Review Exercises SSM: Intermediate Algebra 44 23.Answers may vary.One example is as follows. 25.Answers may vary.One example is as follows. 27.Answers may vary. Chapter 2 Review Exercises 1.1. 3x 5 = x +83x 5 x = x +8 x2x 5 = 82x 5 + 5 = 8 +52x =13x =132 2. 3(x 2) +1 = x 7(x +1)3x 6 +1 = x 7x 73x 5 = 6x 73x 5+ 6x = 6x 7 +6x9x 5 = 79x 5 + 5 = 7 +59x = 2x =29 3. 1.2(x 3.1) = 4.8x + 7.41.2x 3.72 = 4.8x + 7.41.2x 3.72 4.8x = 4.8x + 7.4 4.8x3.6x 3.72 = 7.43.6x 3.72 +3.72 = 7.4 + 3.723.6x =11.12x =11.123.6x 3.09 4. 23 x 1 =3423 x 1+1 =34+123x =743223x =7432x =218 5. f (3) = 5(3) + 9= 15+ 9= 6 6.f (0) = 5(0) +9= 0 +9= 9 7. f (2) = 5(2) +9=10 +9=19 8.f (6.89) = 5(6.89) + 9= 34.45 +9= 25.45 9. f25 = 525 + 9= 2 + 9=11 10. f(a +1)= 5(a +1)+9= 5a 5+9= 5a +4 11. 9 = 2x + 39 3 = 2x + 3 36 = 2x3 = xx = 3 12. 0 = 2x +30 2x = 2x +3 2x2x = 3x =32 13.6 2 36 3 2 3 39 29 9 or 2 2xxxx x = + = + = = = SSM: Intermediate AlgebraChapter 2 Review Exercises 45 14. 1.81 = 2x + 31.81 3 = 2x + 3 31.19 = 2x1.192= x0.595 = xx = 0.595 15. 22 3323 2 3 337231 7 122 3 27 7or6 6xxxxx x= + = + = = = = 16. a = 2x +3a 3 = 2x +3 3a 3 = 2x12(a 3) =12 2xa 32= xx = a 32 17.Since the point (3, 2) lies on the graph, f(3) = 2. 18. Since the point 10,2 lies on the graph, ( )102f = . 19.Since the point (1, 0) lies on the graph, f(1) = 0. 20.Since the point (3, 1) lies on the graph, f(3) = 1. 21.Since the point (7, 3) lies on the graph, f(7) = 3. 22.Since the point (1, 0) lies on the graph, f(1) = 0. 23.Since the point (5, 2) lies on the graph, f(5) = 2. 24.Since the point (7, 4) lies on the graph, f(7) = 4. 25.Since x is 0 when f(x) = 1, f(0) = 1. 26.Since x is 1 when f(x) = 2, f(2) = 1. 27.When f(x) = 0, x = 4. 28.When f(x) = 2, x = 1. 29. ( ) or 7 3 f x y x = +To find the y-intercept, set x = 0 and solve for y. ( ) 7 0 30 33yyy= += += To find the x-intercept, set y = 0 and solve for x. 0 7 33 73 37 7xxx= + = = = 30. ( ) or 3.1 5.7 f x y x = To find the y-intercept, set x = 0 and solve for y. ( ) 3.1 0 5.70 5.75.7yyy= = = To find the x-intercept, set y = 0 and solve for x. 0 3.1 5.75.7 3.15.73.11.84xxxx= = = 31. ( ) or 4 f x y =is the equation of a horizontal line.The y-intercept of the line will be y = 4.There is no x-intercept. 32. ( )4 or 27f x y x = +To find the y-intercept, set x = 0 and solve for y. ( )40 270 22yyy= += += To find the x-intercept, set y = 0 and solve for x. Chapter 2 Review Exercises SSM: Intermediate Algebra 46 4 0 2742772472472xxxxx= + = = = = 33. The "New line" has an increased slope since it is steeper than the "Model." The y-intercept is lower in the "New line" equation since it intersects the y-axis at a lower point than the "Model." 34.a.The students car has 13 gallons of gas in the tank when no hours have been driven, so when t= 0, b = 13.Since the car uses 1.8 gallons of gas per hour, the slope is 1.8.An equation for f is( ) 1.8 13 f t t = + . b.The slope of f is1.8.This means that the amount of gasoline decreases at a constant rate of 1.8 gallons per hour of driving. c.To find the A-intercept of f, lett = 0.( ) ( ) 1.8 0 13 0 13 13 A f t = = + = + =The A-intercept is (0, 13). This represents the fact that before the student leaves the gas station and before any time has been spent driving (t = 0), the car has 13 gallons of gas in the tank. d.To find the t-intercept of f, letf (t) = 0. 0 1.8 13 t = +0 1.8t 130 1.8t 1.8t 13 1.8t1.8t =13t =131.8t 7.22 The t-intercept is (7.22, 0). This means that the student can drive for 7.22 hours before running out of gas. e.Since the car can be driven for 7.22 hours before running out of gas, the domain is 0 7.22 t .Since the gas tank has between 0 and 13 gallons of gas, the range is 0 13 f . 35.a.Since the average household income increases by $1309 per year, the slope of g is 1309. b.In 1995, t = 0 and the average household income was $50,500.Therefore, our intercept is 50,500.Since the slope is 1309, the equation for g is( ) 1309 50, 500 gt t = + . c.To find g(12), substitute 12 for t. ( ) ( ) 12 1309 12 50, 50015, 708 50, 500 66, 208g = += + = The average household income in 1995 + 12 = 2007 will be $66,208. d.Substitute 70,000 for g(t) and solve for t. 70, 000 1309 50, 50019, 500 130914.90ttt= +== The average household income will be $70,000 in the year 1995 + 15 = 2010. 36.a.Use the first three steps of the modeling process to find the equation for f. Begin by creating a scattergram of the data. A line that comes close to the data points passes through (59, 8.0) and (65, 11.0). Use these points to find the slope of the line. m =11.0 8.065 59=36 = .5Sof (t) = 0.5t + b.To solve for b, substitute 59 for t and 8.0 for f(t) since the line contains (59, 8.0). 8.0 =.5(59) + b8.0 = 29.5 + b21.5 = bb = 21.5 The equation of the line isf (t) = 0.5t 21.5.(Your equation may be slightly different if you chose 2 other points. The linear regression equation is This equation will be used for f in parts bd.) The graph of the model fits the data fairly well. SSM: Intermediate AlgebraChapter 2 Review Exercises 47 regression equation isf (t) = 0.55t 25.13.This equation will be used for f in parts bd.) The graph of the model fits the data fairly well. b.In the year 2008, t = 108.Find f(108). ( ) ( ) 108 0.55 108 25.1359.4 25.13 34.27f = = = This means that 34.27% of Californians will approve of the split in 2008. c.To find the t-intercept of f, letf (t) = 0. 0 0.55 25.13 t = 0 0.55t 25.130 25.13 0.55t 25.13 25.1325.13 = 0.55t25.130.55= tt 45.69 The t-intercept is (45.69, 0). This means that in 1900 + 45 = 1945, no Californians approved of splitting California into two states. d.A majority of Californians would be reached when the percent is greater than 50.To find when a majority will be reached, let f (t) = 50 and solve for t. 50 = 0.55t 25.1350 + 25.13 = 0.55t 25.13 + 25.1375.13 = 0.55t75.130.55= tt =136.6 According to this model, after the year 2037 (t =137),a majority of Californians will be in favor of splitting California into two states. Answers may vary. 37.a.Use the first three steps of the modeling process to find the equation for g. Begin by creating a scattergram of the data. A line that is close to the data points passes through (0, 27) and (87, 94). Use these points to find the slope of the line. So Since the g-intercept for this line is (0, 27), b = 27. The equation for g is (Your equation may be b. g(100) = 0.80(100) + 24.44=104.44 This means that in the year 2000(t =100),104 percent of CEOs and presidents of large companies will have at least a bachelor' s degree. Model breakdown has occurred since this is an impossible prediction.A percent cannot be over 100. c. 100 = 0.80t + 24.4475.56 = 0.80t75.560.80= tt = 94.45 According to this model, 100 percent of CEOs and presidents of large companies had at least a bachelor' s degree in 1900 + 94 = 1994. d.Since percents range from 0 to 100, model breakdown occurs when( ) 0 g t .We found in part c of this exercise that g(t) = 100 in 1994. Therefore, model breakdown occurs for years after 1994. To find wheng(t) < 0,solve for t when g(t)=0. 0 = 0.80t + 24.4424.44 = 0.80t24.440.80= tt = 30.55 Chapter 2 Test SSM: Intermediate Algebra 48 Since t is years since 1900, whent = 31,the year is 1869. As a result, model breakdown occurs in all years before 1869.It can be argued that since the model is for years since 1900, it is not valid for any year prior to 1900.So, model breakdown occurs for years prior to 1900 and subsequent to 1994. e. Chapter 2 Test 1.1.Since the line contains the point (6, 1), ( ) 6 1 f = . 2.Since the line contains the point (3, 0),f (3) = 0. 3.Since the line contains the point (0, 1), f (0) = 1. 4. Since the line contains the point 85,3 , 8(5)3f = or approximately 2.7. 5.Since the line contains the point (6, 3),x = 6 whenf (x) = 3. 6.Since the line contains the point (3, 2),x = 3 whenf (x) = 2. 7.Since the line contains the point (3, 0), x = 3 whenf (x) = 0. 8.Since the line contains the point (4.5, 0.5),x = 4.5 whenf (x) = 0.5. 9. To find( ) 3 f , substitute 3 for x. ( ) ( ) 3 4 3 7 12 7 19 f = + = + = 10. Find x when( ) 2 f x = , substitute 2 for( ) f x . 2 4 75 45 5or4 4xxx x= + = = = 11.To find the x-intercept, letf (x)=0and solve for x. 0 = 3x 77 = 3x73= xx =73 The x-intercept is 73, 0 . To find the y-intercept, let x = 0 and solve for y orf (x) since y =f (x) . y = 3(0) 7y = 7 The y-intercept is (0, 7). 12.To find the x-intercept, letg(x) = 0and solve for x. 0 = 2x0 = xx = 0 The x-intercept is (0, 0). To find the y-intercept, let x = 0 and solve for y org(x)sincey = g(x)y = 2(0)y = 0 The y-intercept is (0, 0). 13.To find the x-intercept, leth(x) = 0and solve for x. 0 = 2.5x +1010 = 2.5x102.5= x4 = xx = 4 The x-intercept is (4, 0). To find the y-intercept, let x = 0 and solve for y or h(x) sincey = h(x).y = 2.5(0) +10y =10 The y-intercept is (0, 10). 14.To find the x-intercept, letk(x) = 0 and solve for x. 013x 8813x24 xx 24 The x-intercept is (24, 0). To find the y-intercept, let x = 0 and solve for y or k(x) since y = k(x). The y-intercept is (0, 8). SSM: Intermediate AlgebraChapter 2 Test 49 0 =13x 88 =13x24 = xx = 24 The x-intercept is (24, 0). To find the y-intercept, let x = 0 and solve for y or k(x) since y = k(x). y =13(0) 8y = 8 The y-intercept is (0, 8). 15. 2(x 5) = 4x +32x 10 = 4x +32x 10 4x = 4x +3 4x2x 10 = 32x 10 +10 = 3+102x =13x = 132 16. 3 1 18 4 23 1 18 8 88 4 23 2 43 62xxxxx = = === 17.Answers may vary. 18.a.The rate of change of tuition each year is $750. When t = 0, tuition is $8320.Therefore, the equation f is f (t) = 750t +8320. b.The slope of f is 750. This means that tuition increases by $750 per year. c.To find the D-intercept, let t = 0 and solve for D.Since we know D = 8320 when t = 0, the D-intercept is (0, 8320). This means that the college' s tuition was $8320 in the year 1990. d.Let D(t) = 20,000 and solve for t. 20, 000 = 750t +832011680 = 750tt 15.6 Therefore, tuition is predicted to be $20,000 in the year 2000 + 16 = 2016. 19.a.Use the first three steps of the modeling process. Begin by creating a scattergram using the data from the table. A line close to the data points passes through (20, 150) and (70, 113). Use these points to find the slope of the line. m =11315070 20= 3750= .74Sof (A) = 0.74A + b.To solve for b, substitute 20 for A and 150 for f (A) since the line contains (20, 150). 150 = .74(20) + b150 = 14.8 + b164.8 = b The equation for f isf (A) = 0.74A +164.8.(Your equation may be slightly different if you chose 2 other points. The linear regression equation is f (A) = 0.74A +164.47.This equation will be used for f in parts be.) The graph of the model fits the extremely data well. b.Find f (18). ( ) (18) 0.74 18 164.4713.32 164.47 151.15f = += + = Therefore, the target pulse rate for someone who is 18 years old is approximately 151 beats per minute. Chapter 2 Test SSM: Intermediate Algebra 50 c.Find A when f (A) = 118. 118 = 0.74A +164.4746.47 = 0.74A46.470.74= AA 62.8 Therefore, according to the model, a person approximately 63 year of age has a target pulse rate of 118 beats per minute. d.To find the A-intercept, let T or f (A) = 0 and solve for A. 0 = 0.74A +164.47164.47 = 0.74A164.470.74= AA 222.26 The A-intercept is (222.26, 0). Model breakdown has occurred since a person cannot live to be 222 years old. e.To find the T-intercept, let A = 0 and solve for T or f (A). T = 0.74(0)+164.47T =164.47 The T-intercept is (0, 164.47). This means that a newborn baby (A = 0) has a target pulse rate of 164 beats per minute.It is possible that model breakdown has occurred.It is highly unlikely that one model will be valid for a persons entire lifespan. 20.a.Use the first three steps of the modeling process to find the equation. Begin by creating a scattergram using the data in the table. A line close to the data points passes through (65, 1.2) and (90, 11.1). Use those points to find the slope of the line. m =11.11.290 65=9.925 = .396 So f (t) = 0.396t + b. To solve for b, substitute 65 for t and 1.2 for f (t) since the line contains (65, 1.2). 1.2 = 0.396(65) +b1.2 = 25.74 +b24.54 = bb = 24.54 The equation for f is f (t) = 0.396t 24.54. (Your equation may be slightly different if you chose 2 other points. The linear regression equation is This equation will be used for f in parts be.)The graph of the model fits the data well. (Your equation may be slightly different if you chose 2 other points. The linear regression equation is( ) 0.40 24.82. f t t = This equation will be used for f in parts be.)The graph of the model fits the data well. b.The slope of the model is 0.40.This means that the percentage of the military who are women is increasing by 0.40% each year. c. f (100) = 0.40(100) 24.82=15.18 This means that 15.2 percent of the military will be women in the year 2000 (t = 100). d.Let f (t) = 100 and solve for t. 100 = 0.40t 24.82124.82 = 0.40t124.820.40= tt =312.05 This means that 100 percent of the military will be women in the year 1900 + 312 = 2212. e.Model breakdown occurs for certain when the percent is below 0 or above 100. Find t when f (t) = 0. 0 = 0.40t 24.8224.82 = 0.40t24.820.40= tt = 62.05 Therefore, model breakdown occurs in years before 1900 + 62 = 1962, since it is not possible for a percent to be negative in this context.We know that, according to the model, 100 percent of the military will be women in 2212.Therefore, model breakdown is occurring in years after 2212.Therefore, model breakdown occurs when t < 62.05 and t > 312.05.