Solution manual for finite mathematics an applied approach 3rd edition by young

Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.1 The Cartesian Plane and Graphing Chapter Applications of Linear Functions Exercises Section 1.1 The given points are plotted on the graph below: The given points are plotted on the graph below: -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 (-2, -1) -2 (-1, -2)-3 -4 -5 -6 -7 -8 -9 (3,5) (-2,5) (6,1) x -9 The given points are plotted on the graph below: -8 -7 -6 -5 -4 (-2,0) -6 -5 -4 -3 -2 -1 -9 -8 -7 -6 -5 -4 -3 -2 -1 x -1 -2 (0,-2) -3 -4 -5 -6 -7 -8 -9 -9 -8 -7 -1 -2 -3 -4 -5 -6 -7 -8 -9 -6 -5 -4 -3 -2 -1 (-1,-4) (5,0) 1 (6,0) x 9 (0,5)5 x -9 -8 -7 -6 -5 -4 -3 -2 -1 (-3,-2) (4,-3) Full file at https://TestbankDirect.eu/ -1 -2 -3 -4 -5 -6 -7 -8 -9 (4.5,2.5) x The given points are plotted on the graph below: y -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 (-1/2,5) (0,0.5) (4,4) The given points are plotted on the graph below: (-7,1) -1 y (0,6) -3 -2 (4,2) The given points are plotted on the graph below: y -9 -8 -7 10 y (0, (0,6) 6) y -1 -2 -3 -4 -5 -6 -7 -8 -9 y (3/4,4) x (3,-5) Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Applications of Linear Functions 10 The coordinates are shown next to the appropriate point in the graph below: The given points are plotted on the graph below: (0,0) -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 y (4,1/2) (-1,3) x (4,0) 10 y (0,5) (-4,2) (4,-0.5) -9 -8 -7 -6 -5 -4 -3 -2 (-4,-3) -1 (5,1) x -1 -2 (0,-2) -3 -4 (2,-4) -5 -6 -7 -8 -9 -10 The given points are plotted on the graph below: 11 The data is shown below: y (0,4.5) -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 (-2,-3) -4 -5 -6 -7 -8 -9 1050 x (3,-4) Dollars (billions) (-3.5,0.5) U.S Exports 950 1996 The coordinates are shown next to the appropriate point in the graph below: (-3,3) -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 (-2,-3) -4 -5 -6 -7 -8 -9 -10 1997 1998 2000 Year 1999 12 The data is shown below: y Civilian Labor Forces 80 (1,2) (4,0) x (2,-2) U.S Males (millions) 10 1000 70 60 50 Females (millions) 40 Full file at https://TestbankDirect.eu/ 30 40 50 60 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.1 The Cartesian Plane and Graphing 16 The data is shown below: 13 The data is shown below: Gestation Periods vs Life Span 30 20 10 Gestation (days) 300 400 500 600 x % of U.S High School Graduates Life Span (years) 40 200 College Enrollment 70 50 68 66 64 62 60 58 Year 56 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 17 The Data is shown below: 14 The data is shown below: Median Household Income Daily English Newspapers 1400 Median income 37500 Evening 1200 1000 35000 32500 30000 800 1988 400 500 600 700 1990 1992 1996 1994 18 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use 15 The data is shown below: U.S Smokers % of Smokers (18 and older) 50 45 x y 2(0) − = −3 40 ( 32 ) − = ( x, y ) (0, −3) ( , 0) The graph is shown at the top of the next page 35 30 25 Year 20 1965 1970 1975 1980 1998 Morning 1985 1990 1995 Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Applications of Linear Functions Plotting the points and connecting them with a smooth curve we see the following graph: y -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 (0,-3) -5 -6 -7 -8 -9 x (1.5,0) -9 -8 -7 19 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 y x (0,0) ( x, y ) x y 0 − = −4 −1 ( −1) = −5 4−4 = ( 0, −4) ( 4, 0) ( x, y ) ( −1, −5) (1,5) -8 -7 -6 -5 -4 -3 -2 -1 (4,0) -1 -2 -3 -4 -5 (0,-4) -6 -7 -8 -9 -10 y ( x, y ) −2 ( ) = ( 0, 0) ( 2, −4) −2 ( 2) = −4 -9 -8 -7 -6 -5 -4 -3 -2 -1 (-1,-5) x Full file at https://TestbankDirect.eu/ Plotting the points and connecting them with a smooth curve we see the following graph: x 20 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use (1) = y 21 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use y 10 7 (2,-4) x Plotting the points and connecting them with a smooth curve we see the following graph: -9 Plotting the points and connecting them with a smooth curve we see the following graph: -1 -2 -3 -4 -5 -6 -7 -8 -9 y (1,5) x 22 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use x y ( x, y ) ( 0) − = −6 ( 0, −6) ( 3, 0) ( 3) − = Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.1 The Cartesian Plane and Graphing Plotting the points and connecting them with a smooth curve we see the following graph: -9 -8 -7 -6 -5 -4 -3 -2 -1 Plotting the points and connecting them with a smooth curve we see the following graph: y (3,0) -1 -2 -3 -4 -5 -6 -7 (0,-6) -8 -9 x -9 -8 -7 23 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 y x y ( x, y ) x y ( x, y ) −2 ( ) + = ( 0, 7) ( 3,1) −5 ( −5 ) + = ( 0) + = ( −5, 0) ( 0,5) −2 ( 3) + = Plotting the points and connecting them with a smooth curve we see the following graph: -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 24 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use y ( x, y ) ( 0) + = −1 ( 4) + = ( 0, 2) ( 4, 0) x −1 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 x 26 Because the exponent on the variable is one, this is a linear function Therefore, we only need to plot two points Notice the representative table below: x f ( x) = x − f ( 0) = ( 0) − = −1 f ( 2) = ( 2) − = The graph is shown at the top of the next page Full file at https://TestbankDirect.eu/ 10 y (0,5) (-5,0) x Plotting the points and connecting them with a smooth curve we see the following graph: y 25 Recognizing that the equation is a linear equation, we simply need to find two points on the line So we will create a representative table using any two values of x we wish to use x ( x, f ( x ) ) ( 0, −1) ( 2,5) Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Applications of Linear Functions Plotting the points in the plane and connecting them with a smooth curve we see the following graph: -9 -8 -7 -6 -5 -4 -3 -2 -1 f(x) f(t) x -1 -2 (0,-1) -3 -4 -5 -6 -7 -8 -9 x f ( x) = −2 x −1 f ( −1) = −2 ( −1) = -9 -8 -7 ( x, f ( x ) ) ( −1, 2) ( 2, −4) f ( ) = −2 ( ) = − Plotting the points in the plane and connecting them with a smooth curve we see the following graph: -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 (0,4) (2,5) 27 Because the exponent on the variable is one, this is a linear function Therefore, we only need to plot two points Notice the representative table below: Plotting the points in the plane and connecting them with a smooth curve we see the following graph: f(x) x -6 -5 -4 -3 -2 -1 f (t ) = −t + f ( 0) = − ( 0) + = 4 f ( 4) = − ( 4) + = ( t , f ( t )) ( 0, 4) ( 4, 0) Full file at https://TestbankDirect.eu/ (4,0) t ( t , P ( t )) t P (t ) = t − −2 P ( −2) = ( −2) − = −1 P ( −1) = ( −1) − = −2 P ( ) = ( ) − = −3 P (1) = (1) − = −2 P ( 2) = ( 2) − = ( −2,1) ( −1, −2) ( 0, −3) (1, −2) ( 2,1) 2 2 Plotting the points in the plane, and connecting them we a smooth curve, we graph the parabola (-2,1) t 29 The exponent on the independent variable is two, the graph will be a parabola We need to plot several points to get the basic shape To this we will simply choose more values of t However, we still create the same representative table below: -9 -8 -7 28 Because the exponent on the variable is one, this is a linear function Therefore, we only need to plot two points Notice the representative table below: -1 -2 -3 -4 -5 -6 -7 -8 -9 -6 -5 -4 -3 -2 -1 P(t) t (2,1) -1 -2 (-1,-2) -3 (1,-2) -4 (0,-3) -5 -6 -7 -8 -9 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.1 The Cartesian Plane and Graphing 30 Once again, looking at the exponent on the independent variable, we see that it is a one This is a linear equation, and we only need to plot two points The representative table is as follows: p Q ( p) = p − Q ( 0) = ( 0) − = −4 ( 0, −4) ( 2, 0) Plotting the appropriate points and connecting the plots with a smooth curve, we get the graph of the line below: -9 -8 -7 -6 -5 -4 -3 -2 -1 For Problems 32–37, I will be using the standard graphing window on the calculator ( p, Q ( p ) ) Q ( 2) = ( 2) − = Q(p) (2,0) 32 Entering y = following graph: x into the graphing calculator, shows the p -1 -2 -3 -4 -5 (0,-4) -6 -7 -8 -9 33 Entering y = x + into the graphing calculator, shows the following graph: 31 Once again, we have a linear function The representative table is shown below: p Z ( p) = p − −2 Z ( −2) = ( −2) − = −3 ( p, Z ( p ) ) ( −2, −3) ( 5, 4) Z ( 5) = ( ) − = Plotting the points and connecting them with a smooth curve, we see the graph of the line below: -9 -8 -7 -6 -5 -4 -3 -2 -1 (-2,-3) -1 -2 -3 -4 -5 -6 -7 -8 -9 Z(p) (5,4) p 7 Full file at https://TestbankDirect.eu/ 34 Entering y = x − into the graphing calculator, shows the following graph: Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Applications of Linear Functions 35 Entering y = x − into the graphing calculator, shows the following graph: t f (t ) 0 f ( 0) = 50 (1 − 120 ) = 50 ( t , f ( t )) ( 0,50) ( 60, 25) (120, 0) 60 f ( 60) = 50 (1 − 120 ) = 25 60 f (120) = 50 (1 − 120 120 ) = 120 Plotting these arbitrary points and connecting them with a smooth curve, we see the graph of the function: f(t) (pollutant in ppm) 36 Entering y = x into the graphing calculator, shows the following graph: 60 50 (0,50) 40 30 (60,25) 20 10 t (days) 20 40 60 80 100 120 (120,0) 37 Entering y = ( x − 1) into the graphing calculator, shows the following graph: 39 The independent variable in this function is thousands of miles x This application implies that x ≥ It also would not make sense to have a negative tread depth, so f ( x ) ≥ This implies that (1 − 40x ) ≥ ⇒ x ≤ 40 So an appropriate domain for this function would be ≤ x ≤ 40 Creating a representative table using appropriate values from the domain we see: 38 The independent variable in this case is time, t We would believe that time cannot be negative so the minimum value is 0, the time the removal process began It would also make sense that there could not be a negative amount of pollutant So f (t ) ≥ Which means (1 − ) ≥ ⇒ t ≤ 120 So an appropriate domain for t 120 this application would be ≤ t ≤ 120 Now create a representative table like the one shown at the top of the next column Full file at https://TestbankDirect.eu/ x f ( x) f ( 0) = (1 − 400 ) = 20 40 f ( 20) = (1 − 20 40 ) = f ( 40) = (1 − 40 40 ) = The graph is shown at the top of the next page ( x, f ( x ) ) ( 0, 2) ( 20,1) ( 40, 0) Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.1 The Cartesian Plane and Graphing Plotting the points from the previous page and connecting them with a smooth curve gives us the graph of the function: p in hundreds of dollars Since price must be positive, p ≥ It also makes 41 The independent variable is price sense to have a positive number of stereos sold This means ( that 16 − f(x) (tread depth in cm's) p2 ) ≥ ⇒ −8 ≤ p ≤ However since price is positive, and appropriate domain would be ≤ Create a representative table: p ≤8 (40,0) 10 S ( p ) = 30 16 − 20 30 40 Plotting the points and connecting them with a smooth curve we get the graph of the function: 500 representative table using appropriate points from the domain: 400 ( p, S ( p ) ) S ( 0) = 25 − ( 0) = 25 S (1) = 25 − (1) = 24 S ( 3) = 25 − ( 3) = 16 S ( 5) = 25 − ( 5) = ( 0, 25) (1, 24) ( 3,16) ( 5, 0) 2 2 Plotting the points and connecting them with a smooth curve we see the graph of the function: 28 S(p) (thousands of DVD's) 26 (0,25) S(p) (# of steroes sold each year) 22 300 (6,210) 200 100 (8,0) (3,16) p (100's of dollars) 10 11 C ( 0) = 500 + 120 ( 0) = 500 10 25 C ( x) 12 10 x 16 14 12 13 C (10) = 500 + 120 (10) = 1700 C ( 25) = 500 + 120 ( 25) = 3500 The graph is shown at the top of the next page p (dollars) (5,0) Full file at https://TestbankDirect.eu/ 14 42 The independent variable in this application is the number of inspections during one week x Since there are only five inspectors the maximum number of inspections in a given week is 25 So an appropriate domain for this function is ≤ x ≤ 25 Create a representative table: 20 18 (2,450) (0,480) (1,24) 24 (8, 0) 82 p 25 − p ≥ ⇒ −5 ≤ p ≤ However since p ≥ , we establish the appropriate domain of ≤ p ≤ Create a S ( p) ( 2, 450) ( 6, 210) 62 p ( 0, 480) 22 Thus it makes sense that p ≥ Likewise, the number of DVD’s sold must be positive This means that ( p, S ( p ) ) 50 40 The independent variable in this problem is price ) ( ) S ( 2) = 30 (16 − ) = 450 S ( 6) = 30 (16 − ) = 210 S (8) = 30 (16 − ) = x (1000's of miles) p2 S ( 0) = 30 16 − 04 = 480 (20,1) ( p (0,2) ( x, C ( x ) ) ( 0,500) (10,1700) ( 25,3500) Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 10 Chapter Applications of Linear Functions Plotting the points and connecting them with a smooth curve, we get the graph of the function: b The graph is: C(x) (cost in dollars) (25,3500) 3000 -9 -8 -7 -6 -5 -4 -3 -2 -1 2000 (10,1700) 1000 (0,500) -1 -2 -3 -4 -5 -6 -7 -8 -9 y (4,5) x (6,-1) x (# of inspections) 10 15 20 25 43 The independent variable in this application is the number of shifts needed each week x Since there are eight students available to work one shift a day, there is a maximum number of 40 shifts per week The appropriate domain for this function is ≤ x ≤ 40 a The slope is m = −1 − (−1) = =0 −2 − −5 b The graph is: Create a representative table using appropriate points from the domain x C ( x ) = 80 + 40 x C ( 0) = 80 + 40 ( 0) = 80 ( x, C ( x ) ) ( 0,80) C ( 20) = 80 + 40 ( 20) = 880 20 -6 -5 -4 ( 20,880) C ( 40) = 80 + 40 ( 40) = 1680 40 -9 -8 -7 ( 40,1680) Plotting the points and connecting them with a smooth curve, we get the graph of the function: -3 -2 -1 -1 (-2,-1) -2 -3 -4 -5 -6 -7 -8 -9 C(x) (cost in dollars) (40,1680) 1500 a The slope is m= y x 9 (3,-1) 5−5 = −4 = −1 − 3 b The graph is: 1000 (-1,5) (20,880) 500 x (# of shifts) (0,80) 10 15 20 25 30 35 40 45 Exercises Section 1.2 a The slope is m = −1 − −6 = = −3 6−4 Full file at https://TestbankDirect.eu/ -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 -8 -9 y (1/3,5) x Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 68 Chapter Applications of Linear Functions n 10 Sum x 102 110 102 98 100 101 96 100 104 94 1007 y 45 50 48 38 40 45 42 45 39 38 430 x*x 10404 12100 10404 9604 10000 10201 9216 10000 10816 8836 101581 x*y y*y 4590 2025 5500 2500 4896 2304 3724 1444 4000 1600 4545 2025 4032 1764 4500 2025 4056 1521 3572 1444 43415 18652 The least squares regression line is calculated to be y = 0.647 x − 22.891 Plot this line on the graph on the previous page c Using the formula from earlier or a spreadsheet, the correlation coefficient is r = 0.675 24 a Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line The regression line and the scatter plot are shown below: 82 85 6724 6970 7225 10 87 90 7569 7830 8100 11 72 80 5184 5760 6400 12 80 91 6400 7280 8281 13 79 82 6241 6478 6724 14 81 90 6561 7290 8100 15 82 91 6724 7462 8281 16 92 97 8464 8924 9409 17 91 95 8281 8645 9025 18 83 90 6889 7470 8100 19 89 86 7921 7654 7396 7225 6800 6400 20 85 80 Sum 1667 1696 140023 141945 144954 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 20b + 1667 m = 1696 1667b + 140023m = 141945 To solve this system multiply the first equation by −83.35 and add it to the second equation to get 1078.55m = 583.4 which implies that m = 0.541 Substitute this value back into one of the original equations ( ) to get 20b + 1667 0.541 = 1696 Solve this equation for b 115 y (performance rating) 110 20b + 901.7 = 1696 20b = 794.3 b = 39.715 105 100 95 90 85 The least squares regression line is calculated to be y = 0.541x + 39.715 Where y is the employees performance rating, and x is the employees test result Plot this line on the graph in the previous column 80 75 70 65 60 55 50 x (aptitude test score) 45 45 50 55 60 65 70 75 80 85 90 95 c Using the correlation coefficient formula: 100 105 110 115 b Create a chart like the following: (Note: chart continues on the next column.) n x y x*x x*y y*y 80 85 6400 6800 7225 86 71 7396 6106 5041 87 90 7569 7830 8100 84 75 7056 6300 5625 91 89 8281 8099 7921 60 70 3600 4200 4900 83 74 6889 6142 5476 93 85 8649 7905 7225 Full file at https://TestbankDirect.eu/ r= 20 (141, 945) − (1667 )(1696) 20 (140, 023) − (1667 ) 20 (144, 954 ) − (1696) The correlation coefficient is r = 0.5277 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.5 Regression and Correlation 25 a Create a table like the one shown below: n 10 Sum x 17 19 21 23 25 28 32 38 39 41 283 y 19 25 33 57 71 113 123 252 260 293 1246 x*x 289 361 441 529 625 784 1024 1444 1521 1681 8699 x*y 323 475 693 1311 1775 3164 3936 9576 10140 12013 43406 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 10b + 283m = 1246 283b + 8699m = 43406 ) to get 10b + 283 11.801 = 1246 divide by Therefore the least squares regression line for these data points is y = 11.801x − 209.382 where x is the diameter of the tree in inches and y is volume in hundreds of board feet that can be retrieved from each tree ( ∑ xy) − ( ∑ x)( ∑ y) n ( ∑ x ) − ( ∑ x) n ( ∑ y ) − ( ∑ y ) 2 r = 0.981 The correlation coefficient is r = 0.981 b The rate of change between board feet and the inches in diameter of the tree is the slope of the regression line For each additional inch in diameter, the regression model predicts an increase in 11.801 hundred feet of boards, or 1,180.1 additional board feet c Substitute x = 45 into the regression equation to get y = 11.801( 45) − 209.382 = 321.663 The regression model predicts that a tree with a diameter of 45 inches will produce 321.663 hundred feet of boards or 32,166 feet of boards d To find out how large a tree was in diameter that produced 250 hundred (25000) board feet, substitute y = 250 into the regression equation and solve for x 459.382 = 11.801( x ) x = 38.93 The regression model predicts that the tree that produces 25000 feet of boards is 38.93 inches in diameter y (Sales volume in hundreds of thousands) n 10 ( 255216) − (1246) To find the correlation coefficient use the formula: r= 26 a Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line The regression line and the scatter plot are shown below: Solve this equation for b 10b + 3339.818 = 1246 10b = −2093.818 b = −209.382 10 (8699) − ( 283) 250 = 11.801( x ) − 209.382 To solve this system multiply the first equation by −28.3 and add it to the second equation to get 690.1m = 8144.2 which implies that m = 11.801 Substitute this value back into one of the original equations ( 10 ( 43406) − ( 283)(1246) r= y*y 361 625 1089 3249 5041 12769 15129 63504 67600 85849 255216 69 2 x (wages in $1000's) And substitute the values from the table as shown at the top of the next column Full file at https://TestbankDirect.eu/ 10 15 20 25 30 35 40 45 50 55 60 65 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 70 Chapter Applications of Linear Functions b Create a chart like the following n x y x*x x*y 30.2 3.7 912.04 111.74 13.69 52.1 4.8 2714.41 250.08 23.04 23.6 2.7 556.96 63.72 7.29 38.2 3.5 1459.24 133.7 12.25 45.8 4.7 2097.64 215.26 22.09 59.2 5.1 3504.64 301.92 26.01 41.7 3.9 1738.89 162.63 15.21 62.1 5.4 3856.41 335.34 29.16 Sum 352.9 33.8 16840.23 1574.39 148.74 27 a Create a table like the ones shown below for each set of data y*y Current Dollars n Sum Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 8b + 352.9m = 33.8 352.9b + 16840.23m = 1574.39 x 10 15 19 49 y 4.88 5.94 6.75 7.69 9.08 34.34 x*x 25 100 225 361 711 x*y y*y 23.8144 29.7 35.2836 67.5 45.5625 115.35 59.1361 172.52 82.4464 385.07 246.243 Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation for current dollars to be y = 0.210 x + 4.81 , where y is average hourly earnings in current dollars and x is the year since 1980 To solve this system multiply the first equation by −44.1125 and add it to the second equation to get 1272.93m = 83.39 which implies that m = 0.0655 Substitute this value back into one of the original equations To find the correlation coefficient use the formula and substitute the values from the table: to get 8b + 352.9 0.0655 = 33.8 r= ( ) Solve this equation for b ( 246.243) − ( 34.34) The correlation coefficient between hourly average earnings in current dollars and the year since 1980 is r = 0.991 The least squares regression line is calculated to be y = 0.0655 x + 1.335 , where y is sales volume in hundreds of thousands of dollars, and x is the employees wage in thousands of dollars Plot this line on the graph above Using the correlation coefficient formula (1574.39) − ( 352.9)( 33.8) (16840.23) − ( 352.9) ( 711) − ( 49) r = 0.991 8b + 23.12 = 33.8 8b = 10.68 b = 1.335 r= ( 385.07 ) − ( 49)( 34.34) (148.74) − ( 33.8) The correlation coefficient is r = 0.959 c The value y = corresponds to $700,000 Substitute this value into the regression equation and solve for x = 0.0655 x + 1.335 5.665 = 0.0655x x = 86.49 According to the regression model, wages of $86,490 will be spent to support a sales volume of $700,000 Full file at https://TestbankDirect.eu/ Constant Dollars n Sum x 10 15 19 49 y 5.7 5.39 5.07 4.97 5.39 26.52 x*x 25 100 225 361 711 x*y y*y 32.49 26.95 29.0521 50.7 25.7049 74.55 24.7009 102.41 29.0521 254.61 141 Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation for constant dollars to be y = −0.0229 x + 5.53 , where y is average hourly earnings in constant dollars and x is the year since 1980 To find the correlation coefficient use the formula and substitute the values from the table as shown at the top of the next page Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.5 Regression and Correlation ( 254.61) − ( 49)( 26.52) r= ( 711) − ( 49) 71 y (# of seconds to count 50 chirps) (141) − ( 26.52) 80 r = −0.599 70 60 50 The correlation coefficient between hourly average earnings in constant dollars and the year since 1980 is r = 0.991 40 30 b Plot the year since 1980 on the x-axis and then draw the regressions lines on the same coordinate plane The graph is shown below: 20 10 x (Temp degree's F) 9.5 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5 0.5 y (Hourly earnings in $) -0.5 Constant Dollars x (year since 1980) 10 15 20 30 40 50 60 70 80 90 100 110 120 130 b Create a chart like the following Current Dollars 10 20 c Substitute x = 25 into the appropriate regression equation to get ( ) Current Dollars: y = 0.210 25 + 4.81 = 10.06 ( ) Constant Dollars: y = −0.0229 25 + 5.53 = 4.96 Subtract the two numbers to find the difference between current dollar and constant dollar average hourly earnings in 2005 We see that the difference is 10.84 − 4.96 = 5.88 or there is a difference of $5.88 between current dollar and constant dollar average hourly earnings 28 a Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line The regression line and the scatter plot are shown at the top of the next column n 10 11 12 Sum x 45 52 55 58 60 62 65 68 70 75 82 83 775 y 94 59 42 36 32 32 26 24 21 20 17 16 419 x*x 2025 2704 3025 3364 3600 3844 4225 4624 4900 5625 6724 6889 51549 x*y 4230 3068 2310 2088 1920 1984 1690 1632 1470 1500 1394 1328 24614 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 12b + 775m = 419 775b + 51549m = 24614 To solve this system multiply the first equation by −64.583 and add it to the second equation to get 1497.175m = −2446.3 which implies that m = −1.6339 Substitute this value back into one of the original equations ( ) to get 12b + 775 −1.6339 = 419 Solve this equation for b 12b − 1266.64 = 419 12b = 1685.64 b = 140.47 The solution is continued on the next page Full file at https://TestbankDirect.eu/ y*y 8836 3481 1764 1296 1024 1024 676 576 441 400 289 256 20063 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 72 Chapter Applications of Linear Functions The least squares regression line is calculated to be y = −1.63 x + 140.47 , where y is number of seconds it takes to count to 50 chirps, and x is the temperature in degrees Fahrenheit Plot this line on the graph on the previous page 12 ( 24614) − ( 775)( 419) 12 ( 51549) − ( 775) 12 ( 20063) − ( 419) The correlation coefficient is r = −0.858 c Substitute x = into the regression equation to get y = −1.63 ( 0) + 140.47 = 140.47 The model predicts it would take approximate 140.47 seconds to count to 50 chirps when the temperature is zero degrees Fahrenheit d Substitute y = into the regression equation to get = −1.63 x + 140.47 and then solve for x 1.63 x = 139.47 x = 85.6 The model predicts in order to count to 50 chirps in second, the temperature would have to be 85.6 degrees Fahrenheit 29 a Create a chart like the following n Sum x 10 12 36 y 79 135 135 163 279 791 x*x 25 81 100 144 350 x*y 675 1215 1630 3348 6868 y*y 6241 18225 18225 26569 77841 147101 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 5b + 36m = 791 36b + 350m = 6868 Using the correlation coefficient formula r= ( 6868) − ( 36)( 791) ( 350) − ( 36) (147101) − ( 791) The correlation coefficient is r = 0.83 b Substitute x = 30 into the regression equation to get y = 12.916 ( 30) + 65.2 = 452.68 According to the regression equation there will be $452.68 million spent on magazine advertising for drugs and remedies 30 a Create a chart like the following n 10 Sum x 10 55 y 5.47 4.98 4.17 3.81 3.92 4.05 3.93 3.9 3.95 3.97 42.15 x*x 16 25 36 49 64 81 100 385 x*y 5.47 9.96 12.51 15.24 19.6 24.3 27.51 31.2 35.55 39.7 221.04 y*y 29.9209 24.8004 17.3889 14.5161 15.3664 16.4025 15.4449 15.21 15.6025 15.7609 180.4135 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 10b + 55m = 42.15 55b + 385m = 221.04 To solve this system multiply the first equation by −7.2 and add it to the second equation to get 90.8m = 1172.8 which implies that m = 12.916 Substitute this value back into one of the original equations ( 5b + 464.976 = 791 5b = 326.024 b = 65.2 The least squares regression line is calculated to be y = 12.916 x + 65.2 , where y is number of millions of dollars spent on magazine advertising for drugs and remedies, and x is number of years since 1980 Using the correlation coefficient formula: r= Solve the equation in the previous column for b ) to get 5b + 36 12.916 = 791 Full file at https://TestbankDirect.eu/ To solve this system multiply the first equation by −5.5 and add it to the second equation to get 82.5m = −10.785 which implies that m = −.131 Substitute this value back into one of the original equations ( ) to get 10b + 55 −.131 = 42.15 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Section 1.5 Regression and Correlation Solve the equation on the previous page for b 10b − 7.19 = 42.15 10b = 49.34 b = 4.934 The least squares regression line is calculated to be y = −0.131x + 4.934 , where y lead emissions in thousands of metric tons, and x is number of years starting with 1988 r= Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation gross waste generated to be y = 3.57 x + 84.925 , where the variable y is the amount of waste generated in millions of tons and x is the year since 1960 To find the correlation coefficient use the formula, and substitute the values from the table r= Using the correlation coefficient formula 10 ( 221.04) − ( 55)( 42.15) 10 ( 385) − ( 55) 10 (180.4135) − ( 42.15) The correlation coefficient is r = −0.716 c Substitute y = into the regression equation to get = −0.131x + 4.934 Solve this equation for x 0.131x = 4.934 x = 37.66 The regression equation predicts that it will take 37.66 years after 1988, or by the year 2026 lead emissions will be reduced to zero 31 a Create a table like the ones shown below for each set of data Gross waste generated (in millions of tons) n Sum x 10 20 25 27 30 35 38 185 y 87.8 121.9 151.5 164.4 178.1 205.2 211.4 220.2 1340.5 x*x 100 400 625 729 900 1225 1444 5423 x*y y*y 7708.84 1219 14859.61 3030 22952.25 4110 27027.36 4808.7 31719.61 6156 42107.04 7399 44689.96 8367.6 48488.04 35090.3 239552.7 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 8b + 185m = 1340.5 185b + 5423m = 35090.3 Full file at https://TestbankDirect.eu/ 73 ( 35181.3) − (185)(1343.1) ( 5423) − (185) ( 240658.8) − (1343.1) 2 r = 0.989 The correlation coefficient between tons of waste generated and years since 1960 is r = 0.989 Materials recovered (in millions of tons) n Sum x 10 20 25 27 30 35 38 185 y 5.9 8.6 14.5 16.4 20.4 33.6 54.9 62.2 216.5 x*x x*y y*y 0 34.81 100 86 73.96 400 290 210.25 625 410 268.96 729 550.8 416.16 900 1008 1128.96 1225 1921.5 3014.01 1444 2363.6 3868.84 5423 6629.9 9015.95 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 8b + 185m = 216.5 185b + 5423m = 6629.9 Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation waste generated to be y = 1.418 x − 5.727 , where the variable y is amount of waste recovered in millions of tons, and x is the year since 1960 To find the correlation coefficient use the formula, and substitute the values from the table as shown at the top of the next page Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 74 Chapter Applications of Linear Functions r= ( 6629.9) − (185)( 216.5) ( 5423) − (185) ( 9015.95) − ( 216.5) 140 y (Max length in mm) 130 120 110 r = 0.854 100 90 80 The correlation coefficient between amount of waste recovered and the number of years since 1960 is r = 0.854 70 60 50 40 30 b The solution to this problem is found by setting the equation you found in part (a) for waste recovery equal to times the equation you found in part (b) for waste generation In other words solve 1.418 x − 5.727 = ( 3.574 x + 84.626) 1.418 x − 5.727 = 1.787 x + 42.313 369 x = −48.04 x = −130.2 If you could use this model to predict into the past as well as the future, it predicts that the last time we recovered half as much waste as we generated was the year 1830 An intuitive solution to the previous problem would be to recognize that the slope for the equation that models the generation of waste is larger than the slope that models the recovery of waste This means that instead of recovering a greater percentage of generated waste each year, we are actually recovering a smaller percentage of waste each year c Substitute x = 50 into each regression equation The waste generation regression equation predicts that the amount waste generated in 2010 will be y = 3.574 ( 50) + 84.626 = 263.326 million tons The waste recovery regression equation predicts that the amount of waste recovered in 2010 will be y = 1.418 ( 50) − 5.727 = 65.173 million tons 32 a The scatter plot is plotted at the top of the next column with the regression equation: 20 10 x (Max Wingspan in mm) 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 It appears that the data is not actually linear The data appears to curve upwards as x increases b Create a chart like the following n Sum x 115 140 70 100 150 150 135 860 y 80 90 75 75 130 110 100 660 x*x 13225 19600 4900 10000 22500 22500 18225 110950 x*y 9200 12600 5250 7500 19500 16500 13500 84050 y*y 6400 8100 5625 5625 16900 12100 10000 64750 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 7b + 860m = 660 860b + 110950m = 84050 The solution to this system is m = 0.56 and b = 25.479 The least squares regression line is calculated to be y = 0.56 x + 25.479 , where y is the maximum caterpillar length, in mm, and x is the maximum wingspan, in mm c Using the correlation coefficient formula: r= (84050) − (860)( 660) (110950) − (860) ( 64750) − ( 660) The correlation coefficient is r = 0.811 This correlation coefficient describes how well the linear regression line fits the data A correlation coefficient of 0.81 implies that the line is a good fit to describe the relationship between the maximum wingspan and the maximum length of the caterpillar, but normally you would like to see a higher coefficient Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Summary Exercises d Substitute y = 150 into the regression equation to get 150 = 0.56 x + 25.479 Solving this equation for x yields: x = 222.36 The model predicts the maximum wingspan of a giant silkworm moth that had a maximum caterpillar length of 150 mm would be 222.36 mm Solve the equation for y to get the equation in slopeintercept form x + y = 13 subtract x y = 13 − x y = 132 − 52 x divide by The coefficient in front of the x variable is the slope The slope of this equation is m = e Substitute x = 70 into the regression equation to ( ) 75 −5 ( ) get y = 0.56 70 + 25.479 = 64.679 Substitute the expression x + into the equation for x The regression model would predict a caterpillar with a maximum wingspan of 70 mm would have a maximum caterpillar length of 64.679 mm This leads us to believe that the linear model is not a good predictor of the relationship between the maximum caterpillar length and maximum wingspan to find the y-coordinate Chapter Summary Exercises y = 3x + y = into the equation to find the x -intercept 3x − ( 0) = 30 Solving this equation for x by dividing by yields x = 10 The x -intercept is the point (10, 0) ) y = ( x + 2) + simplify the right hand side y = 3x + + y = 3x + 10 f ( −5) = + Substitute ( substitute x = x + f ( −5) = + f ( −5) = ( −5) −10 −7 Substitute x = into the equation to find the y -intercept x + The slope of the line that is parallel to y = x + 10 is m = Using point-slope ( 0, −6) formula, the equation of the line that passes through Marginal cost is the additional cost associated with producing an additional item For linear functions marginal cost is associated with the slope of the line The marginal cost for this function is $7 y − = 52 ( x − 0) y = 52 x Fixed cost is the cost that does not change regardless of output For linear functions it is associated with the constant term of the function The fixed cost for this function is $430 Solve ( 0) − y = 30 Solving this equation for y by dividing by -5 yields y = −6 The y -intercept is the point y = x + 10 ⇒ y = ( 0, 0) with slope m = The horizontal line that passes through the point ( −3,8) will have the equation y = ( ) The vertical line that passes through the point −3,8 will have the equation x = −3 Full file at https://TestbankDirect.eu/ 5 is x + y = for x by subtracting y x = − y Now substitute this into x + y = (7 − y) + y = 14 − y + y = 14 − y = − y = −9 y=9 The solution is continued at the top of the next page Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 76 Chapter Applications of Linear Functions Now substitute the value of y back into the original equation to get the value of x x + ( 9) = x = −2 4b + 14m = 28 14b + 54m = 99 ( ) The point of intersection is −2, To solve this system multiply the first equation by Add the two equations together to eliminate the y variable 3x + y = 13 +5 x − y = 11 −14 and add it to the second equation to get 5m = which implies that m = 0.2 Substitute this value back into one of the original equations ( ) to get 4b + 14 0.2 = 28 x + y = 24 Solve this equation for b 4b + 2.8 = 28 subtract 2.8 4b = 25.2 divide by Solve this equation for x x= Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is b = 6.3 24 = Now substitute the value of x back into one of the original equations to get the value for y ( 3) + y = 13 6y = y= Therefore the least squares regression line for these data points is y = 0.2 x + 6.3 Looking at a scatter plot of the data we see that the data is not in a nice line This leads us to believe that the correlation coefficient would be closer to zero then to one or negative one y ( ) The point of intersection is 3, 23 10 The slopes of the two equations are equal, but the y -intercept of x + y = is ( 0,3) while the y -intercept of 10 x + y = 10 is ( 0, 53 ) Therefore the lines are parallel and never intersect They are inconsistent and have no solution x 11 The slopes of the two equations are equal The y -intercept of x + y = is ( 0,3) while the y -intercept of 10 x + y = 18 is ( 0,3) as well Therefore the lines are equivalent, thus the common points are all the points that satisfy x + y = 12 Create the table as shown We have four points -1 -1 13 a Substitute x = 60 into the cost function C ( 60) = 7.9 ( 60) + 2520 = 2994 The total cost of producing 60 items is $2994 b Take the total cost of producing 60 items in found in part n Sum x 14 y 28 x*x 16 25 54 x*y 16 18 20 45 99 y*y 64 36 25 81 206 Full file at https://TestbankDirect.eu/ (a) and divide by 60 2994 = 49.9 The average cost of 60 producing the first 60 items is $49.90 c The marginal cost is the additional cost of producing the 61st item The marginal cost is $7.90 (the slope of the linear cost function) Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Summary Exercises 14 a Let x be the number of years since 1990, and let y bet the sales of the discount store in millions of dollars The 1.8 million dollars in sales in 1990 can be formulated as the ( ) point 0,1.8 and the 2.7 million dollars in sales in 1992 ( ) $ R(x) 9000 8000 (100,6500) 7000 C(x) 6000 5000 4000 can be formulated as the point 2, 2.7 Use these two 3000 points to calculate the slope of the linear function 1000 2.7 − 1.8 m= = 45 2−0 77 P(x) 2000 x (# of items) (100,0) 20 40 60 80 100 120 140 -1000 ( ) Now use the slope and the y -intercept, 0,1.8 , to find the equation of the line that will estimate sales in future years is y = 45 x + 1.8 Where x and y are defined as above y = 20 into the function found in part (a) to get 20 = 45 x + 1.8 and solve for x 20 = 45 x + 1.8 18.2 = 45x x = 40.44 -2000 -3000 -4000 16 a Create a table like the one shown below: b Substitute It will take 40.44 years to reach 20 million dollars Therefore the first year sales can be expected to reach 20 million dollars is the year 2031 (1990+41=2031) 15 a Profit is revenue minus cost P ( x) = R ( x) − C ( x) P ( x ) = 65 x − ( 25 x + 4000) P ( x ) = 40 x − 4000 ( ) b The break-even point is when P x = 40 x − 4000 = 40 x = 4000 x = 100 The company will break-even when they sell 100 items c The functions are graphed at the top of the next column n 10 11 12 Sum x 12 15 81 y 14 15 17 21 13 25 33 19 185 x*x 16 25 36 16 49 81 25 144 225 81 711 x*y 10 32 70 90 18 36 119 189 65 300 495 171 1595 y*y 25 64 196 225 36 81 289 441 169 625 1089 361 3601 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 12b + 81m = 185 81b + 711m = 1595 To solve this system multiply the first equation by −6.75 and add it to the second equation to get 164.25m = 346.25 which implies that m = 2.108 Substitute this value back into one of the original equations ( ) to get 12b + 81 2.108 = 185 Solve this equation for b 12b + 170.753 = 185 12b = 14.247 b = 1.187 Therefore the least squares regression line for these data points is y = 2.108 x + 1.187 Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 78 Chapter Applications of Linear Functions b To find the correlation coefficient use the formula: ( ∑ xy) − ( ∑ x)( ∑ y) n ( ∑ x ) − ( ∑ x) n ( ∑ y ) − ( ∑ y ) Sample Test Answers n r= 2 The points are plotted below: And substitute the values from the table: (-2,6) 12 (1595) − (81)(185) r= 12 ( 711) − (81) 12 ( 3601) − (185) (-5,0) -9 -8 -7 r = 0.987 -6 -5 -4 -3 -2 -1 (-4,-1) The correlation coefficient is r = 0.987 c The scatter plot and the regression line are shown below: 34 32 30 28 26 24 22 20 18 16 14 12 10 -2 -4 -1 -2 -3 -4 -5 -6 -7 -8 -9 y (2,7) x (3, -2) y The equations are graphed below: -9 -8 -7 -6 -5 -4 -3 -2 -1 x 10 12 14 ( ) 17 The first thing we notice is the y -intercept is 0, so -1 -2 -3 -4 -5 -6 -7 -8 -9 y y=3x-6 y=2x-4 y=x-2 x that eliminates choices (c), (d), and (e) Now using ( 0, 4) and ( 2,8) we see the slope of the relation is 8− = This eliminates answer (a), and 2−0 reconfirms the correct choice of answer (b) y = x + m= 18 The correlation coefficient must be a number between −1 ≤ r ≤ This eliminates answers (b) and (c) The line is increasing, which means there is a positive relationship, which eliminates answer (a) and (e) The stronger the positive relationship the closer the correlation is to one Therefore the answer must be (d) r = 0.89 Full file at https://TestbankDirect.eu/ Use the slope-intercept formula to find the equation The slope is the per unit change or m = 3.7 The equation of the line will be y = 3.7 x − 2.8 Let x be the number of items and let y be the total cost of production The two points associated with this relation ( ) ( ) are 50, 650 and 70, 714 The slope of the line is m= 714 − 650 = 3.2 Now use the point slope formula 70 − 50 to find the equation of the line y − 650 = 3.2 ( x − 50) y − 650 = 3.2 x − 160 y = 3.2 x + 490 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter Sample Test Answers Let x be the number of years from the current year and let y be the population of the town If the town is growing at a rate of 250 per year, the slope of the line is m = 250 The current population (population when x = ) is 2300, which is the y -intercept Using the slope-intercept formula, the linear function that will give the approximate population x years from now is y = 250 x + 2300 The equation of the vertical line that passes through the ( ) point −2, will be x = −2 x − 2y = −2 y = − x 9− x y= −2 y = x − 92 ( ) ( ) associated with this line are 0,5600 and 10, 2100 The slope of the line is m = 5600 − 2100 = −350 − 10 Use the slope and the y -intercept to find the equation of the line that models the value of the tractor as a function of the number of years from the date of purchase The function is y = −350 x + 5600 Solve x + y = for x x = 5− y ( ) m = 12 and the y -intercept is 0, −29 To find the x -intercept set y = in the original equation x − ( 0) = x =9 The x -intercept is ( 0,9) ( ) ( x − y = 12 ( − y ) − y = 12 10 − y − y = 12 −3 y = y = −32 Now substitute the value of y back into one of the original equations to get x x + −32 = Substitute x into ( ) ) x = 173 The two points on the line are −2, and 0, 6.4 − 6.4 = 3.2 Use the slope −2 − and the y -intercept in the slope-intercept formula to get the y = 3.2 x + 6.4 The solution is ( 17 , −32 ) Elimination method Add the two equations together to eliminate Let x be the price of the special, and y be the number of ( ) ( ) specials sold Use the points 4.75, 78 and 5.50, 63 The slope of the line is m = 10 Let x be the number of years after the purchase of the tractor, and let y be the value of the tractor Two points 12 Substitution method The equation is now in slope-intercept form The slope is equation The equation that will model the number of lunch specials y sold for a given price x is y = −20 x + 173 11 The slopes of the two equations are different; therefore there is only one intersection point Solve the equation for y The slope of the line is m = 79 63 − 78 = −20 5.50 − 4.75 Use the point-slope formula to find the equation of the line The calculations are shown below: y − 78 = −20 ( x − 4.75) y − 78 = −20 x + 95 y = −20 x + 173 Full file at https://TestbankDirect.eu/ x+ y =5 +2 x − y = 12 3x + y = 17 Solve for x x = 17 x = 173 The solution is continued on the next page y Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 80 Chapter Applications of Linear Functions Substitute x back into one of the original equations to find y 17 + y=5 y= −2 The solution is ( 17 , −32 ) S ( p) = D ( p) 0.03 p − = 17.5 − 0.015 p Solve for p 0.045 p − = 17.5 0.045 p = 26.5 p = 588.89 Substitute the price back into supply or demand to get the equilibrium quantity 13 Solve both equations for y S ( 588.89) = 0.03 ( 588.89) − = 8.67 Equation 0.2 x + 0.8 y = 0.8 y = −0.2 x + y = −41 x + 154 Equilibrium is reached when the price is $588.89 At this price the annual supply is 8.67 quilts, and the annual demand for quilts is 8.67 16 Let x start with year 1, be the year the mutual fund was invested, and let y be the return (in percentage terms) Equation ( x + y = 15 y = − x + 15 y = −41 x + 154 ) achieved by the mutual fund y = ⇒ 4% return a Using the above variables four data points will be (1,8) ; ( 2,9) ; ( 3,12) ; ( 4,14) over the first four years of The slope of both equations is m = ( ) −1 The y-intercept of both equations is 0, 154 Therefore there are infinitely many solutions All points given by the equation y = −41 x + 154 will solve the system 14 Revenue is price times quantity The revenue function is R ( x) = 98 x The cost function is direct cost plus fixed cost The Cost ( ) function is C x = 73 x + 40, 000 Profit is revenue minus cost The profit function will be P ( x) = R ( x) − C ( x) P ( x ) = 98 x − ( 73 x + 40, 000) the fund These points are plotted on the axis below: 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 y (%return on investment) -1 x (# of years) b Create a chart like the following using the data points P ( x ) = 25 x − 40, 000 ( ) To find the break-even point set P x = and solve for x 25 x − 40, 000 = 25 x = 40, 000 x = 1600 The company will break-even if they sell 1600 calculators 15 To find the equilibrium price and quantity set supply equal to demand and solve for p as shown at the top of the next column Full file at https://TestbankDirect.eu/ n Sum x 10 y 12 14 43 x*x 16 30 x*y 18 36 56 118 y*y 64 81 144 196 485 Let m and b represent the slope and y-intercept of the least squares regression line The system of equations that give us these values is 4b + 10m = 43 10b + 30m = 118 Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ Chapter In–Depth Application To solve the system on the previous page multiply the first equation by −10 and add it to the second equation to get 5m = 10.5 which implies that m = 2.1 14 81 y (Unemployed per 100) 13 12 11 10 Substitute this value back into one of the original equations ( ) to get 4b + 10 2.1 = 43 Solve this equation for b 4b + 21 = 43 subtract 21 4b = 22 divide by b = 5.5 -1 Therefore the least squares regression line for these data points is y = 2.1x + 5.5 c To find the correlation coefficient use the formula from the text and substitute the appropriate values from the table r= (118) − (10)( 43) ( 30) − (10) ( 485) − ( 43) r = 0.984 The correlation coefficient is r = 0.984 d Substitute x = into the regression equation to get y = 2.1( 5) + 5.5 = 16 The regression model predicts a 16% return on investments for the fifth year From the plot we can see unemployment rose from 1980 to 1982 ( < x < ) and in the years 1989 to 1992 ( < x < 12 ) Unemployment was falling in the years 1982 to 1989 ( < x < ) and again in the years 1992 to 1999 ( 12 < x < 19 ) Let x be the number of years since 1980 and let y be the number of homicides per 1000,000 people The scatter plot for the data is shown below: 14 12 11 10 2.1x + 5.5 = ( 20) 2.1x = 14.5 x = 6.9 -1 In–Depth Application: Unemployment and Homicides in the United States Let x be the number of years since 1980 and let y be the number of unemployed per 100 people The scatter plot for the data is shown at the top of the next column Full file at https://TestbankDirect.eu/ y (Homicides per 100000) 13 e Substitute y = 20 into the regression equation and solve for x The regression model predicts that in the 7th year the fund will return 20% on the investment x (years since 1980) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 x (years since 1980) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 From the plot we can see that the homicide rate generally increased in the years 1984 to 1991( < x < 11 ) In 1987 there was a one year decline, but the rates continued to increase after that Homicide rates were falling in the years 1980 to 1984 ( < x < ) and in the years 1991 to 1999 ( 11 < x < 19 ) In 1993 we see a one year increase, but the rates continued to decrease after that Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at https://TestbankDirect.eu/ 82 Chapter Applications of Linear Functions c It appears that unemployment rates and homicide rates have the same pattern When unemployment rates are falling, so are the homicide rates This is not an exact match, but it looks close d First let x be the unemployment rate, and y be the number of homicides per 100,000 Create a chart like before and use the correlation coefficient formula to find n 10 11 12 13 14 15 16 17 18 19 20 Sum x 7.1 7.6 9.7 9.6 7.5 7.2 6.2 5.5 5.3 5.6 6.8 7.5 6.9 6.1 5.6 5.4 4.9 4.5 4.2 130.2 y x*x x*y y*y 10.22 50.41 72.562 104.4484 9.83 57.76 74.708 96.6289 9.07 94.09 87.979 82.2649 8.25 92.16 79.2 68.0625 7.91 56.25 59.325 62.5681 7.95 51.84 57.24 63.2025 8.55 49 59.85 73.1025 8.26 38.44 51.212 68.2276 8.41 30.25 46.255 70.7281 8.66 28.09 45.898 74.9956 9.42 31.36 52.752 88.7364 9.79 46.24 66.572 95.8441 9.31 56.25 69.825 86.6761 9.51 47.61 65.619 90.4401 8.96 37.21 54.656 80.2816 8.22 31.36 46.032 67.5684 7.41 29.16 40.014 54.9081 6.8 24.01 33.32 46.24 6.28 20.25 28.26 39.4384 5.7 17.64 23.94 32.49 168.51 889.38 1115.219 1446.852 The correlation coefficient calculated from the formula is r = 0.542 This is a positive correlation, but it is not very high There does not appear to be a significant relationship between unemployment rates and homicide rates in the US over this time period For the years 1980 to 1988, create a chart using that data n x y x*x x*y y*y 7.1 10.22 50.41 72.562 104.4484 7.6 9.83 57.76 74.708 96.6289 9.7 9.07 94.09 87.979 82.2649 9.6 8.25 92.16 79.2 68.0625 7.5 7.91 56.25 59.325 62.5681 7.2 7.95 51.84 57.24 63.2025 8.55 49 59.85 73.1025 6.2 8.26 38.44 51.212 68.2276 5.5 8.41 30.25 46.255 70.7281 Sum 67.4 78.45 520.2 588.331 689.2335 Full file at https://TestbankDirect.eu/ The correlation coefficient is r = 0.09 which implies that there is almost no linear relationship between the unemployment rate and the homicide rate For the years 1989 to 1999 create a chart like the following n 10 11 Sum x 5.3 5.6 6.8 7.5 6.9 6.1 5.6 5.4 4.9 4.5 4.2 62.8 y 8.66 9.42 9.79 9.31 9.51 8.96 8.22 7.41 6.8 6.28 5.7 90.06 x*x x*y y*y 28.09 45.898 74.9956 31.36 52.752 88.7364 46.24 66.572 95.8441 56.25 69.825 86.6761 47.61 65.619 90.4401 37.21 54.656 80.2816 31.36 46.032 67.5684 29.16 40.014 54.9081 24.01 33.32 46.24 20.25 28.26 39.4384 17.64 23.94 32.49 369.18 526.888 757.6188 The correlation coefficient calculated from the formula is r = 0.87 this is a very strong positive correlation The data suggest that there is a strong relation between the unemployment and the homicide rate over the years 1989 to 1999 There are an unlimited number of factors that could influence the trends in the data Economic reasons could vary from the recession in the mid 80’s to the technology bubble of the 90’s Political events have seen Republicans occupy the White House from 1980–1992, and Democrats occupy the White House from 1992–1999 There have been several conflicts including the Gulf War All of these factors and many more will influence the data that we have