Chapter Modeling with Linear Functions Homework.1 1. a. b. c. In 199, t = 1. Notice on the scattergram, when t = 1, p is approximately 4.5. Therefore, 4.5% of deaths due to car accidents in 199 were due to drunk driving. 5. a. 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 7, which represents the year 007. Therefore, it is estimated that in 007, 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 (0.4, 0). This means that in the year 010 (t = 0), it is predicted that there will be no collisions. This is extremely unlikely and gives an obviously false prediction. Model breakdown has occurred.. a. 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. b. 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 0 million users per year. 1
Homework.1 SSM: Intermediate Algebra 7. a. e. When n = 87, t is approximately 1, which represents 1995 + 1 = 007. This means that in the year 007, 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. 9. a. b. b. According to the model, the value of a share on January 1, 001 was approximately $150. c. c. In 1999, t = 14. Notice on the scattergram, when t = 14, n is approximately $175. This is an overestimate of $5 by the model. d. In 001, t = 16. Notice on the scattergram, when t = 16, n is approximately $190. This is an overestimate of $0 by the model. e. Answers may vary. f. d. The difference between the estimated value and the actual value of the stock on January 1, 001 is $150 $4 = $107. The model did not take into account the investors sudden lack of confidence in dot-com stocks that occurred after January 000. 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.). This point represents a 5.% difference in the percentage of sales between pop and rap/hiphop. In the year 1990, pop sound recordings outsold rap/hip-hop sound recordings by approximately 5.%.
SSM: Intermediate Algebra Homework. 1. 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. 1. 5. Check the scatterplot obtained from the graphing calculator. 7. a. 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. In this model, m will be the same as the original model. However, the value of b will increase.. Use (, 5) and (7, 15) to write the equation of the line. First, find the slope. 15 5 10 5 m = = = 7 4 So, y =.5x + b. Substitute for x and 5 for y since the line contains (, 5) and then solve for b. y = mx + b 5 5 = ( ) + b 10 15 = + b 5 = b The equation of the line is y =.5x.5. (Your equation may be slightly different if you chose different points.) Use the graphing calculator to check your results. b. Use the points (75, 14.) and (100,.) to find the equation of the line.. 14. 18.9 m = = 0.77 100 75 5 So, y = 0.77x + b. Substitute 75 for x and 14. for y since the line contains (75, 14.) and then solve for b. y = mx + b c. 14. = 0.77 75 + b 14. = 57.75 + b 4.45 = b The equation of the line is y = 0.77x 4.45. (Your equation may be slightly different if you chose two other points.) 9. a.
Homework. SSM: Intermediate Algebra 11. a. 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, 15) to find the equation of the line. 15 80 55 m = = 6.1 9 0 9 So, y = 6.1x + b. Substitute 0 for x and 80 for y since the line contains (0, 80) and then solve for b. y = mx + b 80 = 6.1 0 + b 80 = b The equation of the line is y = 6.1x + 80. (Your equation may be slightly different if you chose different points.) 1. a. b. Use the points (69, 51.7) and (8, 47.99) to find the equation of the line. 47.99 51.7.71 m = = 0.7 8 69 14 So, y = 0.7x + b. Substitute 69 for x and 51.7 for y since the line contains (69, 51.7) and then solve for b. y = mx + b 51.7 = 0.7 69 + b 51.7 = 18.6 + b 70. = b The equation of the line is y = 0.7x +70.. (Your equation may be slightly different if you chose different points.) c. b. Use the points (6, 97) and (11, 78) to find the equation of the line. 97 78 19 m = = =.8 6 11 5 So, y =.8x + b. Substitute 6 for x and 97 for y since the line contains (6, 97) and then solve for b. y = mx + b c. 97 =.8 6 + b 97 =.8 + b 119.8 = b The equation of the line is y =.8x + 119.8. (Your equation may be slightly different if you chose different points.) 15. a. b. Yes, the model predicts that the women s record time will equal the men s record time. This will happen in the year 00 with a record time of approximately 4.6 seconds. c. Yes, the model predicts that the women s record time will be less than the men s record time in years 004 and after. 4
SSM: Intermediate Algebra Homework. 17. a. 19. a. b. Use the points (6, 1.7) and (10, 0.5) to find the equation of the line. 1.7 0.5 1. m = = = 0. 6 10 4 So, y = 0.x + b. Substitute 6 for x and 1.7 for y since the line contains (6, 1.7) and then solve for b. y = mx + b c. 1.7 = 0. 6 + b 1.7 = 1.8 + b.5 = b The equation of the line is y = 0.x +.5. (Your equation may be slightly different if you chose different points.) b. Use the points (, 4) and (4, 84) to find the equation of the line. 84 4 50 m = = =.5 4 0 So, y =.5x + b. Substitute for x and 4 for y since the line contains (, 4) and then solve for b. y = mx + b c. 4 =.5 + b 4 = 57.5 + b.5 = b The equation of the line is y =.5x.5. (Your equation may be slightly different if you chose different points.) d. Homework. e. The regression equation from the graphing calculator is y = 0.8x +.. The models are so similar that it is difficult to distinguish them on the graphing calculator screen. 1. y + x = 17 Substitute for x and solve for y. y + = 17 y = 15 y = 5. y x = 7 Substitute 4 for x and solve for y. y ( 4) = 7 y 8 = 7 y = 19 19 y = 5. y 6x = 49 Substitute 5 for x and solve for y. 5
Homework. SSM: Intermediate Algebra y 6 5 = 49 y + 0 = 49 y = 79 y = 79 7. ( y ) 5 + 5 = 11+ 4x Substitute 1 for x and solve for y. 5 y + 5 = 11+ 4 1 5y 5 = 11 4 5y 5 = 15 5y = 10 y = 9..7 y.8x = 5.9 Substitute 6.6 for x and solve for y..7 y.8 6.6 = 5.9.7 y 5.08 = 5.9.7 y = 0.98 y 11.47 11. f ( 1) = ( 1) + = + = 1 19. g 1 1 = 6 4 = 1 4 = 6 6 1. g = 6 4 = 4 4 = 8. f ( a) ( a) ( a) = 4 1 = 4 1 5. f a + 1 = 4 a + 1 1 = 4a 4 1 = 4a 5 7. f a + h = 4 a + h 1 = 4a 4h 1 1. f ( 9) = ( 9) + = 7 + = 5 15. f ( ) = ( ) + = 6 + = 8 17. g (.) = 6(.) 4 = 1.8 4 = 9.8 9. To find x when f ( x ) = 6, substitute 6 for f ( x ) and solve for x. 6 = x + 7 x = 1 1 x = 1. To find x when f ( x ) = 4.9, substitute 4.9 for f ( x ) and solve for x. 4.9 = x + 7 x = 11.9 x 4.0. To find x when f ( x ) =, substitute f ( x ) and solve for x. for 6
SSM: Intermediate Algebra Homework. = x 5 1 = x 1 x = 6 5. To find x when f ( x) = a, substitute a for f ( x ) and solve for x. a = x 5 a + 5 = x 1 5 x = a + 7. Since the line contains the point ( 6, 4), f( 6) = 4. 9. Since the line contains the point (, 1), f() = 1. 41. Since the line contains the point (.5, 1.), f(.5) = 1.. 4. Since the line contains the point (6, 0), x = 6 when f(x) or y = 0. 45. Since the line contains the point (, ), x = when f(x) or y =. 47. 9 1 Since the line contains the point,, x = 9 or 4.5 when f(x) or y = 1. 49. f ( ) = 4 51. When f ( x ) =, x is 1. 5. f ( x) or y = 5x 8 To find the y-intercept, set x = 0 and solve for y. y = 5 0 8 y = 0 8 y = 8 To find the x-intercept, set y = 0 and solve for x. 0 = 5x 8 8 = 5x 8 5 = x 55. f ( x) or y = x To find the y-intercept, set x = 0 and solve for y. y = 0 = 0 To find the x-intercept, set y = 0 and solve for x. 0 = x 0 = x 57. f ( x) or y = 5 is the equation of a horizontal line. The y-intercept of the line will be y = 5. There is no x-intercept. 59. f ( x) or y =.1x 4. 61. To find the y-intercept, set x = 0 and solve for y. y =.1 0 4. = 4. To find the x-intercept, set y = 0 and solve for x. 0 =.1x 4. 4. =.1x = x 1 f ( x) or y = x To find the y-intercept, set x = 0 and solve for y. 1 y = ( 0 ) = To find the x-intercept, set y = 0 and solve for x. 1 0 = x 1 = x 6 = x 6. x 1+ x = x 9x + 17 x 1 = 6x + 17 9x 1 = 17 9x = 18 x = 65. 6( x) = x ( 4 x) 1 + 6x = x 4 + x 9 + 6x = x 4 9 + 4x = 4 4x = 5 5 x = 4 7
Homework. SSM: Intermediate Algebra 67. ( x ) 0.5.9 + =.1x 7 0.5x 1.95 + =.1x 7 0.5x + 0.05 =.1x 7 1.6x = 7.05 x 4.5 69. 4( x ) ( x 1) = ( x + 6) 4x 8 x + = x + 1 x 5 = x + 1 x = 17 x = 17 71. 1 1 x = 4 1 1 1 x = 1 4 4x = 8 4x = 11 x = 11 4 7. 5 1 x = 4 8 5 1 8 x = 8 4 8 6x 5 = 4 75. x = x + 1 6x = 9 9 x = = 6 0 = 1 This is a contradiction. There is no value for x for which the equation is true. 77. Both solutions are correct since x = in the equation x + 4 = 0. However, Student 1 used unnecessary steps to find the solution. 79. a. f ( t) 0.81t 45.86 = b. f 108 = 0.81 108 45.86 = 87.48 45.86 = 41.6 When t = 108, p = 41.6. This means that in the year 1900 + 108 = 008, the percentage of births out of wedlock will be 41.6%. 8 c. 4 = 0.81t 45.86 88.86 = 0.81t t 110 When p = 4, t 110. This means that the percentage of out of wedlock births will hit 40% in the year 1900 + 110 = 010. d. 100 = 0.81t 45.86 145.86 = 0.81t t 180 In the year 1900 + 180 = 080, all births will be out of wedlock. e. Find p, when t = 1997 1900 = 97. f 97 = 0.81 97 45.86 = 78.57 45.86 =.71 This means that in the year 1997, the percentage of births out of wedlock will be.71%. Since the actual percentage was.4%, the error was.71%.4% = 0.1%. 81. a. f ( n) 5.85n 81.18 = + b. To find the T-intercept, set n = 0 and solve for T. T = 5.85 0 + 81.18 T = 81.18 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) = 87.0, f() = 9.88, f() = 98.7, f(4) = 104.58, f(5) = 110.4, f(6) = 116.7, f(7) = 11.1, f(8) = 17.98, f(9) = 1.8, f(10) = 19.68. According to the model, f(4) = 104.58. This shows an error of 0.4 and this is the least error in the tables. The estimate for f(10) is 19.68, an error of 44., 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 Algebra Homework. 8. a. b. Use the points (1, 45) and (10, 666) to find the equation of the line. 666 45 14 m = =.8 10 1 9 So, y =.8x + b. Substitute 1 for x and 45 for y since the line contains (1, 45) and then solve for b. y = mx + b 45 =.8 1 + b 45 =.8 + b 48. = b The equation of the line is y =.8x + 48.. (Your equation may be slightly different if you chose different points.) Graphing the line with the scatterplot, we see that the model fits the data very well. The data points lie close to the line that passes through (0, ) and (100, 1). To find the equation for f, start by finding the slope using these points. 1 180 m = = = 1.8 100 0 100 So, f(c) = 1.8c + b. To solve for b, substitute 0 for C and for f(c) since the line contains (0, ). = 1.8(0) + b = b The equation for f is f(c) = 1.8c +. This is also the linear regression equation. This model fits the extremely well. b. f ( 5) = 1.8( 5) + = 45 + = 77 If it is 5 o C, it is 77 o F. c. 40 = 1.8c + 8 = 1.8c 8 = c 1.8 c 4.4 If it is 40 o F, it is 4.4 o C. 85. a. c. Since 001 1990 = 11, find f(11). f ( 11) =.8( 11) + 48. = 61.8 + 48. = 690 According to the model, 690 million enplanements were made in 001. The actual number of enplanements was 6 million. This is an error of 68 million enplanements. d. 68 million enplanements is equivalent to 68 = 17 million round trips. If on average 4 each round trip is $500, the airlines lost 17 $500 = $8500 million or $8.5 billion. 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. m = = = 0. 98 94 4 So, f(t) = 0.t + 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. 94 + b 4.5 = 8. + b.7 = b The equation for f is f(t) = 0.c.7. This model fits the very well. 9
Homework. SSM: Intermediate Algebra b. f ( 99) = 0.( 99).7 = 9.7.7 = 6 This means that in the year 1999, there were 6 thousand or 6,000 male-female pairs of bald eagles. c. f ( 15) = 0.( 15).7 = 5.7 = 18.7 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.t.7.7 = 0.t.7 = t 0. t = 11. This means that there will be 10,000 malefemale pairs of bald eagles in the year 1900 + 11 = 01. e. From year to year the changes are 0., 0.4, 0., 0., and 0.4. On average it is 0. thousand pairs per year. This is the same as the slope. Answers may vary. 89. a. f ( x) =.48x.64 b. 100 =.48x.64 1.64 =.48x x 50 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 =.48x.64.64 =.48x x 10 This means that for scores 10 and under, no students would succeed in the intermediate algebra course. d. For the 16 0 range, we use a score of 18 to represent the group. Find the percentage for x = 18. p =.48 18.64 = 44.64.64 = 1 This means that 1% of the students who score in the 16 0 range pass the intermediate algebra course. If 145 students scored in this range, we could expect 1% or 0.1 145 1 students to pass. e. First calculate the percentages for each of the score groups. p =.48.64 =.4 p =.48( 8).64 = 45.8 p =.48( ).64 = 58. p =.48( 8).64 = 70.6 p =.48( 4).64 = 8 p =.48( 48).64 = 95.4 Next, multiply each of the percentages by the number of students in that category. 0.4 94 = 1.4 students 0.458 44 = 0. students 0.58 19 = 11.1 students 0.706 1 = 8.5 students 0.8 9 = 7.5 students 0.954 4 =.8 students Add these results to obtain the total number of students, 8.5. This means that of students who scored at least 1 points on the placement test, approximately 8 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 0 640 m = = = 160 0 4 4 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 0 + b 640 = b The equation for f is f(t) = 160t + 640. 40
SSM: Intermediate Algebra Homework.4 b. c. c. Since it takes 4 hours to pump out the water, the domain is 0 t 4. Since the water level is at 640 cubic feet before starting to 0 f t 640. pump, the range is 9. When f() = 5, the input for f is and the output is 5. Possible equations for f will vary. Sample equations are as follows. f x = x 1 1 f ( x) = x + 4 5 f ( x) = x Homework.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.. a. Number Paying Bills Online t n 1.7 + 0.1 b. n 1.7 t (.1) 0 1 1.7 + 1(.1) 1.7 + (.1) 1.7 + (.1) 4 1.7 + 4(.1) 5 1.7 + 5(.1) t 1.7 + t (.1) = + or n =.1t + 1.7 d. The slope of the model is.1. This means that.1 million more U.S. households each year are paying bills online. 5. a. Since the cost per hour is $.50, the slope is.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) =.5t + 4.95. c. d. Find n when the cost of the regular plan will be $.90..90 =.5t + 4.95 18.95 =.5t 7.58 = t 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 $14.60 even when no textbooks are rented, b = 14.60. So, f(n) = 15.75t + 14.60. c. Find when f ( n ) is $10.5. 10.5 = 15.75n + 14.60 78.75 = 15.75n n = 5 The total cost, textbooks are rented. f n, is $10.5 when 5 41
Homework.4 SSM: Intermediate Algebra 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. gallons of gas in the tank, when x = 0, b = 15.. So, f(t) = 0.05t + 15.. c. To find the x-intercept, let g(x) = 0 and solve for x. 0 = 0.05x + 15. 15. = 0.05x 06 = x The x-intercept is (06, 0). This means that when 06 miles have been driven, the gas tank will be empty. Use the first and last data points, (0, 70) and (50, 5) to find the slope. 70 5 18 m = = = 0.6 0 50 50 So, the equation is of the form f t = t + b. Substitute 0 for t and 0.6 70 for f ( t ) and solve for b. 70 = 0.6( 0) + b 70 = 0 + b 70 = b The equation is f ( t) 0.6t 70 = +. Graphing the equation on the same graph as the scattergram, we see that the equation fits the data fairly well. d. Since the car can be driven between 0 and 06 miles on the 15. gallon tank of gas, the domain is 0 x 06. e. Find x when g(x) = 1. 1 = 0.05x + 15. 14. = 0.05x 86 = x The car can be driven 86 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. 1. The slope of the model is 0.7. This means that the record time for the women s 400 -meter run is decreasing by 0.7 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. b. The slope of f is 0.6. This means that the percentage of the world s population that lives in rural areas decreases by 0.6 of one percent each year. c. When no one lives in a rural area, f ( t) = 0. 0 = 0.6t + 70 70 = 0.6t 194 t This means that in the year 1950 + 194 = 144, 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 00 years. 19. a. Create a scattergram using the data in the table. 17. a. Create a scattergram using the data in the table. Use the first and last data points, (5, 977) and (10, 774) to find the slope. 774 977 797 m = = = 159.4 10 5 5 4
SSM: Intermediate Algebra Homework.4 So, the equation is of the form f t = t + b. Substitute 5 for t and 159.4 977 for f ( t ) and solve for b. 977 = 159.4( 5) + b 977 = 797 + b 180 = b f t = t +. The equation is 159.4 180 Graphing the equation on the same graph as the scattergram, we see that the equation fits the data fairly 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(0), as well as, g(17), g(18), g(19) and g(0). We can use the table feature of the graphing calculator to determine these values. b. Create a scattergram using the data in the table. Use the first and last data points, (5, 14,57) and (10, 19,1) to find the slope. 19,1 14,57 4775 m = = = 955 10 5 5 So, the equation is of the form g t = t + b. Substitute 5 for t and 955 977 for g ( t ) and solve for b. 14,57 = 955( 5) + b 14, 57 = 4775 + b 976 = b g t = t +. The equation is 955 976 Graphing the equation on the same graph as the scattergram, we see that the equation fits the data very well. So, for public colleges, the cost of a fouryear program will be $4890 + $5049 + $509 + $568 + $557 = $0,994. Similarly, for private colleges, the cost of a four-year program will be $5,977 + $6,95 + $7,907 + $8,86 + $9,817 = $19,515. 1. a. t f(t) 0 0 1 500 1000 1500 4 1900 5 00 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 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. 4
Chapter Review Exercises SSM: Intermediate Algebra. Answers may vary. One example is as follows. 5. Answers may vary. One example is as follows. 4. x 1 = 4 x 1+ 1 = 4 + 1 x = 7 4 x = 7 4 x = 1 8 5. f() = 5() + 9 = 15+ 9 = 6 6. f(0) = 5(0) + 9 = 0 + 9 = 9 7. Answers may vary. Chapter Review Exercises 1. x 5 = x + 8 x 5 x = x + 8 x x 5 = 8 x 5 + 5 = 8 + 5 x = 1 x = 1. (x ) + 1 = x 7(x +1) x 6 + 1 = x 7x 7 x 5 = 6x 7 x 5+ 6x = 6x 7 + 6x 9x 5 = 7 9x 5 + 5 = 7 + 5 9x = x = 9. 1.(x.1) = 4.8x + 7.4 1.x.7 = 4.8x + 7.4 1.x.7 4.8x = 4.8x + 7.4 4.8x.6x.7 = 7.4.6x.7 +.7 = 7.4 +.7.6x = 11.1 x = 11.1.6 x.09 7. f( ) = 5( ) + 9 = 10 + 9 = 19 8. f(6.89) = 5(6.89) + 9 = 4.45 + 9 = 5.45 9. f = 5 + 9 5 5 = + 9 = 11 10. f(a + 1) = 5(a +1) + 9 = 5a 5 + 9 = 5a + 4 11. 9 = x + 9 = x + 6 = x = x x = 1. 0 = x + 0 x = x + x x = x = 1. 6 = x + 6 = x + 9 = x 9 9 = x or x = 44
SSM: Intermediate Algebra Chapter Review Exercises 14. 1.81 = x + 1.81 = x + 1.19 = x 1.19 = x 0.595 = x x = 0.595 15. = x + = x + 7 = x 1 7 1 = x 7 7 = x or x = 6 6 16. a = x + a = x + a = x 1 (a ) = 1 x a = x x = a 17. Since the point (, ) lies on the graph, f() =. 18. 1 Since the point 0, lies on the graph, 1 f ( 0) =. 19. Since the point ( 1, 0) lies on the graph, f( 1) = 0. 0. Since the point (, 1) lies on the graph, f( ) = 1. 1. Since the point ( 7, ) lies on the graph, f( 7) =.. Since the point ( 1, 0) lies on the graph, f( 1) = 0.. Since the point ( 5, ) lies on the graph, f( 5) =. 4. Since the point (7, 4) lies on the graph, f(7) = 4. 5. Since x is 0 when f(x) = 1, f(0) = 1. 6. Since x is 1 when f(x) =, f() = 1. 7. When f(x) = 0, x = 4. 8. When f(x) =, x = 1. 9. f ( x) or y = 7x + To find the y-intercept, set x = 0 and solve for y. y = 7 0 + y = 0 + y = To find the x-intercept, set y = 0 and solve for x. 0 = 7x + = 7x x = = 7 7 0. f ( x) or y =.1x 5.7 To find the y-intercept, set x = 0 and solve for y. y =.1 0 5.7 y = 0 5.7 y = 5.7 To find the x-intercept, set y = 0 and solve for x. 0 =.1x 5.7 5.7 =.1x 5.7 x =.1 x 1.84 1. f ( x) or y = 4 is the equation of a horizontal. line. The y-intercept of the line will be y = 4. There is no x-intercept. 4 f ( x) or y = x + 7 To find the y-intercept, set x = 0 and solve for y. 4 y = ( 0 ) + 7 y = 0 + y = To find the x-intercept, set y = 0 and solve for x. 45
Chapter Review Exercises SSM: Intermediate Algebra. 4 0 = x + 7 4 = x 7 7 = x 4 7 = x 4 7 x = e. Since the car can be driven for 7. hours before running out of gas, the domain is 0 t 7.. Since the gas tank has between 0 and 1 gallons of gas, the range is 0 f 1. 5. a. Since the average household income increases by $109 per year, the slope of g is 109. b. In 1995, t = 0 and the average household income was $50,500. Therefore, our intercept is 50,500. Since the slope is 109, the equation for g is g ( t) = 109t + 50,500. 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." 4. a. The student s car has 1 gallons of gas in the tank when no hours have been driven, so when t = 0, b = 1. Since the car uses 1.8 gallons of gas per hour, the slope is 1.8. An equation for f is f ( t) = 1.8t + 1. b. The slope of f is 1.8. 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, let t = 0. A = f t = 1.8 0 + 1 = 0 + 1 = 1 The A-intercept is (0, 1). 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 1 gallons of gas in the tank. d. To find the t-intercept of f, let f(t) = 0. 0 1.8t 0 = 1.8tt + 1 1.8t 1.8t = 1 t = 1 1.8 t 7. The t-intercept is (7., 0). This means that the student can drive for 7. hours before running out of gas. 46 c. To find g(1), substitute 1 for t. g ( 1) = 109( 1) + 50,500 = 15, 708 + 50, 500 = 66, 08 The average household income in 1995 + 1 = 007 will be $66,08. d. Substitute 70,000 for g(t) and solve for t. 70, 000 = 109t + 50,500 19, 500 = 109t 14.90 = t The average household income will be $70,000 in the year 1995 + 15 = 010. 6. 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 8.0 65 59 = 6 =.5 So f(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) + b 8.0 = 9.5 + b 1.5 = b b = 1.5 The equation of the line is f(t) = 0.5t 1.5. (Your equation may be slightly different if you chose other points. The linear regression equation is
SSM: Intermediate Algebra Chapter Review Exercises regression equation is f(t) = 0.55t 5.1. This equation will be used for f in parts b d.) The graph of the model fits the data fairly well. b. In the year 008, t = 108. Find f(108). f ( 108) = 0.55( 108) 5.1 = 59.4 5.1 = 4.7 This means that 4.7% of Californians will approve of the split in 008. c. To find the t-intercept of f, let f(t) = 0. 0 = 0.55t 5.1 5.1 = 0.55t 5.1 0.55 = t t 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 5.1 50 + 5.1 = 0.55t 5.1 + 5.1 75.1 = 0.55t 75.1 0.55 = t t = 16.6 According to this model, after the year 07 (t = 17), a majority of Californians will be in favor of splitting California into two states. Answers may vary. 7. a. Use the first three steps of the modeling process to find the equation for g. Begin by creating a scattergram of the data. b. g(100) = 0.80(100) + 4.44 = 104.44 This means that in the year 000 (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 + 4.44 75.56 = 0.80t 75.56 0.80 = t t = 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 g ( t ) < 0 or g ( t ) > 100. We found in part c of this exercise that g(t) = 100 in 1994. Therefore, model breakdown occurs for years after 1994. To find when g(t) < 0, solve for t when g(t)=0. 0 = 0.80t + 4.44 4.44 = 0.80t 4.44 0.80 = t t = 0.55 A line that is close to the data points passes through (0, 7) and (87, 94). Use these 47
Chapter Test SSM: Intermediate Algebra e. Since t is years since 1900, when t = 1, 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. = 4x + 7 5 = 4x 5 5 = x or x = 4 4 11. To find the x-intercept, let f(x) = 0 and solve for x. 0 = x 7 7 = x 7 = x Chapter Test 1. Since the line contains the point (6, 1), f 6 = 1.. Since the line contains the point (, 0), f() = 0.. Since the line contains the point (0, 1), f(0) = 1. 4. 8 Since the line contains the point 5,, 8 f ( 5) = or approximately.7. 5. Since the line contains the point ( 6, ), x = 6 when f(x) =. 6. Since the line contains the point (, ), x = when f(x) =. 7. Since the line contains the point (, 0), x = when f(x) = 0. 8. Since the line contains the point (4.5, 0.5), x = 4.5 when f(x) = 0.5. 9. To find f ( ), substitute for x. f ( ) = 4( ) + 7 = 1 + 7 = 19 10. Find x when f ( x ) =, substitute for f ( x ). x = 7 7 The x-intercept is, 0. To find the y- intercept, let x = 0 and solve for y or f(x) since y = f(x). y = (0) 7 y = 7 The y-intercept is (0, 7). 1. To find the x-intercept, let g(x) = 0 and solve for x. 0 = x 0 = x x = 0 The x-intercept is (0, 0). To find the y-intercept, let x = 0 and solve for y or g(x) since y = g(x) y = (0) y = 0 The y-intercept is (0, 0). 1. To find the x-intercept, let h(x) = 0 and solve for x. 0 =.5x + 10 10 =.5x 10.5 = x 4 = x x = 4 The x-intercept is ( 4, 0). To find the y-intercept, let x = 0 and solve for y or h(x) since y = h(x). y =.5(0) +10 y = 10 The y-intercept is (0, 10). 14. To find the x-intercept, let k(x) = 0 and solve for x. 1 48
SSM: Intermediate Algebra Chapter Test 0 = 1 x 8 8 = 1 x 4 = x x = 4 The x-intercept is (4, 0). To find the y-intercept, let x = 0 and solve for y or k(x) since y = k(x). y = 1 (0) 8 y = 8 The y-intercept is (0, 8). 15. (x 5) = 4x + x 10 = 4x + x 10 4x = 4x + 4x x 10 = x 10 + 10 = + 10 x = 1 x = 1 16. 1 1 x = 8 4 1 1 8 x 8 = 8 8 4 x = 4 x = 6 x = 17. Answers may vary. d. Let D(t) = 0,000 and solve for t. 0, 000 = 750t + 80 11680 = 750t t 15.6 Therefore, tuition is predicted to be $0,000 in the year 000 + 16 = 016. 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 (0, 150) and (70, 11). Use these points to find the slope of the line. 11 150 m = 70 0 = 7 50 =.74 So f(a) = 0.74A + b. To solve for b, substitute 0 for A and 150 for f (A) since the line contains (0, 150). 150 =.74(0) + b 150 = 14.8 + b 164.8 = b The equation for f is f(a) = 0.74A + 164.8. (Your equation may be slightly different if you chose other points. The linear regression equation is f(a) = 0.74A + 164.47. This equation will be used for f in parts b e.) The graph of the model fits the extremely data well. 18. a. The rate of change of tuition each year is $750. When t = 0, tuition is $80. Therefore, the equation f is f(t) = 750t + 80. 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 = 80 when t = 0, the D-intercept is (0, 80). This means that the college's tuition was $80 in the year 1990. b. Find f (18). f (18) = 0.74( 18) + 164.47 = 1. + 164.47 = 151.15 Therefore, the target pulse rate for someone who is 18 years old is approximately 151 beats per minute. 49
Chapter Test SSM: Intermediate Algebra c. Find A when f (A) = 118. 118 = 0.74A + 164.47 46.47 = 0.74A 46.47 0.74 = A A 6.8 Therefore, according to the model, a person approximately 6 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 + 164.47 164.47 = 0.74A 164.47 0.74 = A A.6 The A-intercept is (.6, 0). Model breakdown has occurred since a person cannot live to be years old. e. To find the T-intercept, let A = 0 and solve for T or f (A). T = 0.74(0) + 164.47 T = 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 person s entire lifespan. 0. 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.) and (90, 11.1). Use those points to find the slope of the line. 11.1 1. m = 90 65 = 9.9 5 =.96 So f (t) = 0.96t + b. To solve for b, substitute 65 for t and 1. for f (t) since the line contains (65, 1.). 1. = 0.96(65) + b 1. = 5.74 + b 4.54 = b b = 4.54 The equation for f is f (t) = 0.96t 4.54. (Your equation may be slightly different if 50 (Your equation may be slightly different if you chose other points. The linear regression equation is f ( t) = 0.40t 4.8. This equation will be used for f in parts b e.) 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) 4.8 = 15.18 This means that 15. percent of the military will be women in the year 000 (t = 100). d. Let f (t) = 100 and solve for t. 100 = 0.40t 4.8 14.8 = 0.40t 14.8 0.40 = t t = 1.05 This means that 100 percent of the military will be women in the year 1900 + 1 = 1. e. Model breakdown occurs for certain when the percent is below 0 or above 100. Find t when f (t) = 0. 0 = 0.40t 4.8 4.8 = 0.40t 4.8 0.40 = t t = 6.05 Therefore, model breakdown occurs in years before 1900 + 6 = 196, 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 1. Therefore, model breakdown is occurring in years after 1. Therefore, model breakdown occurs when t < 6.05 and t > 1.05.