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Solution manual for applied CALC 2nd edition by wilson

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Solution Manual for Applied CALC 2nd Edition by Wilson Exercises 1-1 At any given instant in time, you can have only one weight For example, at age 19 years, months, days, 11 hours, minutes, 27 seconds you have only one weight Thus your weight is a function of your age Temperature is a function of the time of day since at any time of the day when the temperature is measured, the measured temperature will be a single value The input value 182 has two different output values (32 and 47) Therefore, the number of salmon in a catch is not a function of the number of fish caught It appears that near x  , the graph goes vertical A vertical line drawn at that point would touch the graph in multiple locations However, if the graph doesn’t actually go vertical near x  , then it is the graph of a function One drawback of reading a graph is that it is sometimes difficult to tell whether the graph goes vertical or not The graph represents a function since any vertical line drawn will touch the graph exactly once 10 The graph represents a function since any vertical line drawn will touch the graph exactly once 11 y   x2  4;   x  3,  10  y  C    39.95    159.80 The total cost of four pairs of shoes is $159.80 H    16    120  56 Two seconds after he jumped, the cliff diver is 56 feet above the water 12 y  2 x2  1;   x  2,   y  p 12   3.55 On June 16, 2012 (12 days after June 4, 2012), the average retail gas price was $3.55/gallon I (4)  11 During Q4 2011 (4 quarters after Q4 2010) 11 million iPads were sold 13 y  x  x ; x  From the graph, it appears that y  at x  We calculate the exact value of y algebraically ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part CHAPTER Functions and Models y   1 The domain of the function is all real numbers except t  That is, t | t  1  1 4 14 y    x2  4 ; x  2 x  x  3x  From the graph, it appears that y  at x  2 We calculate the exact value of y algebraically y    2    2    2  4  4 2    2   8    0 But division by zero is not a legal operation Therefore, the function is not defined when x  2 Graphically speaking, there is a “hole” in the graph at x  2  15 The function f ( p)  p  p  has the domain of all real numbers r 1 has a r2 1 domain of all real numbers since no value of r will make the denominator equal to zero 16 The function h  r   3t  is 4t  undefined when the denominator equals zero 4t   4t  t 1 17 The function g  t   18 The function f (a)  a3  a  is undefined when the radicand is negative So a 1  a  1 The domain of the function is all real numbers greater than or equal to –1 That is, a | a  1 2x  is x2  undefined when the radicand is negative or the denominator is equal to zero 2x   x  6 x  3 The denominator is always positive The domain of the function is the set of all real numbers greater than or equal to –3 That is,  x | x  3 19 The function f ( x)  20 Since it doesn’t make sense to sell a negative number of bags of candy, n is nonnegative The domain of the candy profit function is the set of whole numbers That is, n | n is a whole number ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1-2 Linear Functions Exercises 1-2 The constant rate of change (slope) of the line passing through (2, 5) and (4, 3) is 35 m 42 2   1    y  y  y  4 vertical (y-)intercept:  0, 4  3x     3x  x horizontal (x-)intercept: The constant rate of change (slope) of the line passing through (1.2, 3.4) and (2.7, 3.1) is 3.1  3.4 m 2.7  1.2 0.3  1.5  0.2 The constant rate of change (slope) of the line passing through (2, 2) and (5, 2) is 22 m 52  0 vertical (y-)intercept:  0,10   x  10 10  x x  2 horizontal (x-)intercept:  2,  vertical (y-)intercept:  0,11  x  11 11  x x  5.5 horizontal (x-)intercept:  5.5,   43 ,  The slope of the line passing through (2, 5) and (4, 3) is 35 m 42 2   1 y  mx  b y  1x  b  1   b substituting  2,5  b7 The slope-intercept form of the line is y   x  The standard form of the line is x  y  A point-slope form of the line is y   1 x   The slope of the line passing through (1.2, 3.4) and (2.7, 3.1) is 3.1  3.4 m 2.7  1.2 0.3  1.5  0.2 y  mx  b y  0.2 x  b 3.4  0.2 1.2   b substituting 1.2,3.4  3.4  0.24  b b  3.64 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part CHAPTER Functions and Models The slope-intercept form of the line is y  0.2 x  3.64 The standard form of the line is typically written with integer coefficients Therefore, y  0.2 x  3.64 0.2 x  y  3.64 20 x  100 y  364 x  25 y  91 (dividing by 4) is the standard form of the line 11 y   0.5  x   A point-slope form of the line is y  3.4  0.2  x  1.2  The slope of the line passing through  2,  and (5, 2) is m 12 x  y  22   2  0 Since the slope is equal to zero, this line is a horizontal line The slopeintercept form of the line is y  x  and is commonly written as y  The standard form of the line is x  y  and is also often written as y  A point-slope form  13 y   23 x  43 of the line is y    x   10 y  x  14  0,3 and  3,5  are two points on the line The slope of the line is 53 m 30  ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1-2 Linear Functions The slope-intercept form of the line is y  x  15 This is a vertical line with x-intercept  3,  The equation of the line is x  16 If the package weight increases from pound to pounds, the cost increases $(7.95 – 5.51) = $2.44 But if the package weight increases from pounds to pounds, the cost increases $(10.20 – 7.95) = $2.25 Since the rate of change (rate of price increase, or slope) is not constant, the table of data may not be represented by a linear function 17 As letter weight increases from to ounces, the cost increases $(1.70 – 1.50) = $0.20 Similarly, as letter weight increases from to ounces, to ounces, and to ounces, the cost increases $0.20 each time The rate of change (rate of price increase, or slope) is constant, so the table of data may be represented by a linear function The constant rate, $0.20/ounce, indicates that for each additional ounce of weight in a standard-sized letter shipped from Queen Creek, Arizona to Ellensburg, Washington, the cost will increase $0.20 18 If the table of data represents a linear function then a linear function passing through two of the points will also pass through all other points in the table Cost to Dispose Clean of Clean Wood Wood at Enumclaw (Pounds) Transfer Station 500 $18.75 700 $26.25 900 $33.75 1000 $37.50 Source: www.dnr.metrokc.gov The slope of the line passing through  500,18.75 and  700, 26.25 is 26.25  18.75 dollars 700  500 pound 7.5  200  $0.0375 per pound m The equation of the line passing through these points is given by y  0.0375 x  b 18.75  0.0375  500   b 18.75  18.75  b b0 y  0.0375 x We evaluate the linear equation at x  900 and x  1000 y  0.0375  900   33.75 y  0.0375 1000   37.50 These results match the table data The data table may be represented by a linear function It costs an average of $0.0375 per pound to dispose of clean wood 19 Let x be the number of servings of WheatiesTM and y be the grams of fiber consumed We have y  2.1x  3.3 since each serving of cereal contains 2.1 grams of fiber and the banana ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part CHAPTER Functions and Models contains 3.3 grams of fiber We must solve  2.1x  3.3 4.7  2.1x x  2.238 3 In order to consume 8-grams of fiber, you would need to eat servings ( 14 cups) of Wheaties along with the large banana 20 The slope of the line is   2  m 44   undefined Therefore, the line is a vertical line Although the y-values change, every point on a vertical line has the same x-value The x-value of each of the points is x  The equation of the vertical line is x  ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1-3 Quadratic Functions Exercises 1-3 concave up, y-intercept  0,1 , vertex 1,  concave down, y-intercept  0,  , vertex  0,  concave up, y-intercept  0,  , vertex  0.5, 0.75 concave down, y-intercept  0, 4.5 , vertex 1,5.7  concave up, y-intercept  0, 2.1 , vertex  0.25,1.925 The graph appears to pass through  0,  ,  2,10  , and  7,  The quadratic equation will satisfy y  a  x   x   since  0,  and  7,  are x-intercepts of the function Substituting  2,10  into this equation yields 10  a      10  a   5  a  1 So the quadratic equation is y  1 x  x     x2  x The graph appears to pass through  0,  , 1,8 , and  3,  Each of these points will satisfy y  ax  bx  c That is,  a 0  b 0  c 6c  a 1  b 1  c 8 abc  a  3  b  3  c  9a  3b  c Since c  6, the other two equations may be simplified and written as the system of equations ab  9a  3b  6 Rewriting the first equation as b   a and substituting  a for b in the second equation yields 9a    a   6 6a   6 6a  12 a  2 Substituting 2 for a in either of the equations containing just a and b yields b  Thus a  2 , b  , and c  The graph’s equation is y  2 x  x  The graph appears to pass through  0, 3 and have vertex  2,1 Each of these points will satisfy y  ax  bx  c Since the yintercept is  0, 3 , c  3 Since the x-coordinate of the vertex is b x , 2a b 2   2a 4a  b Thus y  ax  bx  c is equivalent to y  ax   4a  x  Substituting the coordinates  2,1 into the equation yields  a  2    4a  2    a  8a   4a a  1 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part CHAPTER Functions and Models 60  4a  2b Since b  4a, b  4 The graph’s equation is y   x2  x  60  4a    a  40  60  4a  2a  80 20  2a a  10 Back substituting this result into b  a  40 yields b  10  40  50 The graph’s equation is y  10 x  50 x The graph appears to pass through  0,5 and have vertex  5, 20  Each of these points will satisfy y  ax  bx  c Since the yintercept is  0,5 , c  Since the xcoordinate of the vertex is x   b , 2a b 2a 10a  b Thus y  ax  bx  c is equivalent to y  ax   10a  x  5 11 Substituting the coordinates  5, 20  into the equation yields 20  a     10a    20  25a  50a  25  25a a 1 Since b  10a, b  10 The graph’s equation is y  x  10 x  y y (y) 19 33 10 14 4 The second differences are constant, so the data represent a quadratic function 12 10 The graph appears to pass through  0,  , 1, 40  , and  2, 60  Each of these points will satisfy y  ax  bx  c Since the yintercept is  0,  , c  Substituting the coordinates of the other two points into the equation y  ax  bx  c yields the following system of equations: 40  a  b 60  4a  2b The first equation may be written as b  a  40 Substituting a  40 for b in the second equation yields: x x y y (y) 10 -4 -16 -36 -64 -100 -12 -20 -28 -36 -8 -8 -8 The second differences are constant, so the data represent a quadratic function 13 x y y (y) 10 15 20 25 10 2 2 0 The first differences are constant, so the data represent a linear function ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1-3 Quadratic Functions 14 x y y (y) 12 16 20 10 20 40 80 10 20 40 10 20 Neither the first nor the second differences are constant, so the data represent neither a linear nor a quadratic function 15 x y y (y) -3 -1 -10 -1 26 71 -9 27 45 18 18 18 The second differences are constant, so the data represent a quadratic function 16 The parameter a is 0.05 and represents that the rate at which the change in the number of members of the USAA per year is increasing is 0.10 million people per year The parameter b is 0.15 and represents the increase in the number of members of the USAA per year in the initial year of 2003 The parameter c is 5.0 million and represents the number of members of the USAA in the initial year of 2003 17 The parameter a is -19.56 and represents that the rate at which the population of the United States per year is changing is decreasing 39.12 thousand people per year The parameter b is 3407 and represents the increase in the number of thousand people in the United States per year in the initial year of 1990 The parameter c is 250,100 and represents the number of thousand people in the United States in 1990 18 The parameter a is 141.25 and represents that the rate at which the change in the number of students enrolled in the Arizona Virtual Academy is increasing is 282.5 students per year each year The parameter b is 358.75 and represents the increase in the number of students enrolled in the Arizona Virtual Academy in the initial year of 2003 The parameter c is 318.75 and represents the number of students enrolled in the Arizona Virtual Academy in the year 2003 19 The parameter a is -31.15 and represents that the rate at which the change in the amount of money spent by consumers on books classified as adult trade is decreasing is 62.30 million dollars per year each year The parameter b is 556.1 and represents the increase in the amount of money in millions spent by consumers on books classified as adult trade in the initial year of 2004 The parameter c is 14970 and represents the amount of money in millions spent by consumers on adult trade books in the year 2004 20 The parameter a is 14.99 and represents that the rate at which the change in the amount of yogurt produced in millions of pounds is increasing is 29.98 million pounds per year each year The parameter b is 62.14 and represents the increase in yogurt production in millions of pounds per year in the initial year of 1997 The parameter c is 1555 and ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 10 CHAPTER Functions and Models represents the millions of pounds of yogurt produced in 1997 21 For the vertex, 4 x 1 2(2) y  2(1)  4(1)   6 The vertex is 1, 6  The vertical intercept is found by substituting for x The vertical intercept is  0, 8 22 For the vertex, b x 2a 0.4 x  2 2(0.1) y  0.1(2)  0.4(2)  1.2 y  1.6 The vertex is  2, 1.6  The vertical intercept is found by substituting for x The vertical intercept is  0, 1.2  23 For the vertex, b x 2a 4 x  2 2(1) y  (2)2  4(2)  y 1 The vertex is  2,1 The vertical intercept is found by substituting for x The vertical intercept is  0,5 24 For the vertex, b 2a 0 x 0 2(9) x y  9(0)  y  4 The vertex is  0, 4  The vertical intercept is found by substituting for x The vertical intercept is  0, 4  (The vertical intercept is the vertex in this case.) 25 For the vertex, b x 2a 4 x   2(3)  2  2 y  3           3  3  1 The vertex is   ,    3 The vertical intercept is found by substituting for x The vertical intercept is  0,1 26 We substitute 150 for j  t  in the given function rule and solve for t: 150  0.0990t  1.28t  154  0.0990t  128t  t 1.28  1.282   0.0990    0.0990  1.28  1.795104 0.198 t  2.60 or t  15.53 Since 2.60 would represent a time before 1998, we use 15.53 and estimate that the number of jobs in the ship- and boat-building industry will be 150 thousand in mid-2014 t ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-4 Polynomial and Rational Functions 15 21 3( x  x  2) x  x  16 3( x  2)( x  1)  ( x  4)  no hole exists vertical asymptote: x  4 horizontal asymptote: y  vertical intercept: f (0)   83 or (0,  83 ) horizontal intercepts: (set numerator = 0) x  and x  1 or (2,0) and (1,0) 23 a d 0.1 0.5 10 20 30 I (w/m2) 0.000398 0.000016 0.00000398 0.00000996 0.000000159 0.0000000398 0.00000000996 0.00000000443 b 22 x2  2x  x  x  10 ( x  3)( x  1)  (2 x  5)( x  2) no hole exists vertical asymptote: x and x  2 f ( x)  c As you move closer to the person speaking, the sound intensity increases at an increasing rate vertical intercept: f (0)  0.3 or (0, 0.3) horizontal intercepts: (set numerator = 0) x  and x  1 or (3,0) and (1,0) horizontal asymptote: y  ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 16 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models 24 a B( w, h)  705w , w  190 h2 pounds, B(190, h)  705(190) 133,950  h2 h2 h 60 62 64 66 68 70 72 74 76 B 37.21 34.85 32.70 30.75 28.97 27.34 25.84 24.46 23.19 b 600n 110(169  0.6n) 8.8(169  0.6n)  600n 1487.2  594.72n 2.5  n A 110-pound person can have about 2.5 beers 0.08  b Yes, the lighter a person is the fewer beers can be consumed 26.a c b Domain: t  d The function decreases at a decreasing rate Each increase in height yields a decrease in body mass index, but the decreases are less and less 25 a 600n 200(169  0.6n) 16(169  0.6n)  600n 2704  590.4n 4.56  n 0.08  c Time cannot be negative, nor would it be reasonable to consider very large values of time or very small values of time d As t   , r (t )  This means that if the patient takes “forever” to complete the workout, then the pace of the workout is really slow A 200-pound person can have about 4.5 beers ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-5 Exponential Functions and Logarithms 17 d Exercises 1-5 y  4(0.25) x a Since 0.25  and  , the graph is decreasing b Since  , the graph is concave up c The y-intercept is  0,  d -1 y  0.5(2) a Since  and 0.5  , the graph is increasing b Since 0.5  , the graph is concave up c The y-intercept is  0, 0.5 d 180 160 140 120 100 80 60 40 20 -1 4 y  1.2(2.3) x a Since 2.3  and 1.2  , the graph is decreasing b Since 1.2  , the graph is concave down c The y-intercept is  0, 1.2  d x 16 14 12 10 -1 y  0.4(5) x a Since  and 0.4  , the graph is increasing b Since 0.4  , the graph is concave up c The y-intercept is  0, 0.4  -1 -5 -10 -15 -20 -25 -30 y  3(0.9) x a Since 0.9  and  , the graph is decreasing b Since  , the graph is concave up c The y-intercept is  0,3 d -5 10 15 20 25 30 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 18 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models The y-intercept is  0,  so a   a  12  We have b  62  The exponential function is y   3x  We will check our result by evaluating the function at  3,54  54   33   a  161  a  64 x Therefore, y  64  12  We will check our solution with the point  6,1 54  2(27) 54  54 y  a 2  10  a   x 10  2a a5 Therefore, y   x  We will check our solution with the point  3, 40  y   2x  40   23  y  64  12  x  64  12   64  641  20 We have b  10  So y  a  x  We will determine the value of a by plugging in the point 1,10  11 We have b  14  So y  a  x  We will determine the value of a by plugging in the point 1,1  a  41   4a a  0.25 Therefore, y  0.25  x  We will check our solution with the point  4, 64  y  0.25  x  40    64  0.25  44  40  40 64  0.25  256  We have  4 b   16  1   4  10 The y-intercept is  0, 256  so a  256 We have y  256  b x  We know 1  x 64  64 42 Thus y  a   We will determine the value of a by plugging in the point  4,    50 b   256   5    32   ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-5 Exponential Functions and Logarithms 19 Therefore, y  256  12  We will check our solution with the point  7,  x y  256  12  x  256  2  256  128 p  3  389.99 1  0.16   389.99  0.84    22 11 The exponential function is t S  t   45 1.035 where t is years from now and S is an instructor’s salary in thousands of dollars S  5  45 1.035   53.446 Five years from now, the instructor’s salary is projected to be $53,446 12 The exponential function is t B  t   235 1.0232  where B is the balance (in dollars) and t is the number of years from now 250  235 1.0232  1.0638  1.0232  t t ln 1.0638   ln 1.0232  t ln 1.0638  ln 1.0232  t  2.7 In just over year and months, the balance is projected to reach $250 3  231.15 A television that cost $389.99 in 2010 is projected to cost $231.15 in 2013 14 The exponential function is t p  t   94.70 1.28 where p is the Black Eyed Peas concert ticket price and t is the number of years since August, 2010 p 1  94.70 1.28  121.22 A concert ticket that costs $94.70 in August 2010 is expected to cost $121.22 in August 2011 15 The exponential function is t T  t   2024 1.12  where T is the annual tuition (in dollars) and t is the number of years from now t 3567  2024 1.12  1.762  1.12  ln 1.0638   t ln 1.0232  t where p is the price of the television and t is the number of years from the end of 2010 t ln 1.762   ln 1.12  t ln 1.762   t ln 1.12  t ln 1.762  ln 1.12  t  5.0 Annual tuition is projected to be $3567 five years from now 13 The exponential function is t p  t   389.99 1  0.16   389.99  0.84  t ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 20 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models 16 x  8.827 23 325.0  313.8 1.00899  t 325.0  1.00899t 313.8  325.0  t  log1.00899    313.8  log  325.0 / 313.8  t log1.00899 t  3.918 17 x  14.21 According to the model, the population of the United States will reach 325.0 million in late 2015 (about 3.918 years after 2011) 24 8.0  7.022 1.01096  t 8.0  1.01096t 7.022  8.0  t  log1.01096    7.022  18 x  log  8.0 / 7.022  log1.01096 t  11.962 t According to the model, the population of the world will reach 8.0 billion in late 2023 (about 11.962 years after 2011) 19 4x  15  x  log 15 log15 x  1.953 log 25 20 1.3  3.4  t  log1.3 3.4 log 3.4 t  4.664 log1.3 t 21 2.1y  5.0  y  log 2.1 5.0 log 5.0 y  2.169 log 2.1 22 4.3x  1.2  x  log 4.3 1.2 log1.2 x  0.125 log 4.3 300  200 1.02599  t 300  1.02599t 200  300  t  log1.02599    200  log  300 / 200  log1.02599 t  15.803 t According to the model, the item cost $300 in late 2006 (about 15.803 years after 1990) ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-5 Exponential Functions and Logarithms 21 26 25000  50000  0.9747  t 25000  0.9747 t 50000  25000  t  log 0.9747    50000  1 t  log 0.9747   2 log 1 /  t log 0.9747 t  27.049 Based on an inflation rate of 2.599% per year, a $50,000 salary will have buyer power of $25,000 in early 2018 (about 27.049 years after 1990) This means that what would have bought $50,000 worth of goods in 1990 will only buy $25,000 worth of goods in early 2018 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 22 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models Exercises 1-6 Using ExpReg, we find that t c  t   3.9433  0.9753 with a coefficient of determination of r  0.9952, where t is the number of years before 2012 and c is the cost of a gallon of milk (in dollars) To predict the price of a gallon of milk in 2016, we let t  4 Finding c  4   4.36 implies that a gallon of milk will cost $4.36 in 2016 Using QuadReg, we find that p  s   0.3s  0.95s  1.74 with a coefficient of determination of r  1, where s is the number of staples (in thousands) and p is the price (in dollars) Since p  7.5  25.74, the predicted price of 7500 staples is $25.74 Using QuadReg, we find that w  h   0.03314h2  0.3223h  11.07 with a coefficient of determination of r  0.9999, where h is the height (in inches) and w is the weight (in pounds) Since w  71  179, a person 71 inches tall with a BMI of 25 will have a weight of 179 pounds Using LinReg, we find that c  g   4.99 g  3.79 with a coefficient of determination of r  1, where g is the number of games played and c is the cost (in dollars) Since c  5  28.74, it costs $28.74 to bowl games expenditures E (in dollars) for apparel, footwear, and related products and services for girls to 15, and we measure time t as years since 1985: t 10 15 20 25 E 81 87 101 118 121 101 Using CubicReg, we find that E  t   0.01526t  0.4479t  0.8677t  81.22 Since E  24   107, the predicted expenditure in 2009 was $107 We restrict the data considered to the average annual household expenditures E (in dollars) for apparel, footwear, and related products and services for children under 2, and we measure time t as years since 1985: t 10 15 20 25 E 56 70 81 82 82 91 Using CubicReg, we find that E  t   0.006444t  0.2910t  4.651t  55.33 Since E  27   96, the predicted expenditure in 2012 was $96 We restrict the data considered to the average annual household ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-6 Function Modeling and Combining Functions 23 We construct a scatter plot for the data in the window 2  x  23,  y  950 : s 2.08 2.68 3.12 3.36 3.67 4.76 U 15.37 16.63 17.28 18.01 18.65 20.42 Using LinReg, we find that U  s   1.899s  11.50 with Using CubicReg, we find that J  t   0.2056t  5.933t  1.573t  906.9 with r  0.9984, where t is years since 1990 and J is the number of apparel manufacturing jobs (in thousands) Since J  25  372, the predicted number of jobs in 2015 is 372 thousand This doesn’t seem reasonable because it conflicts with the overall downward trend observed between 1990 and 2011 Using ExpReg, we find that t L  t   4.222  0.8864  with r  0.9939, where s is the compensation cost in Slovakia (in U.S dollars) and U is the compensation cost in the United States (in U.S dollars) The slope is 1.899 dollars of U.S compensation costs per dollar of Slovakian compensation costs This means that a $1.00 increase in Slovakian compensation costs corresponds with a $1.899 increase in U.S compensation costs 10 We construct a scatter plot of the data in the window 1  x  13, 4800  y  8200 : r  0.9957, where t is years since 1990 and L is the percentage of all mass layoff events that occurred in fashion-related industries Since L  22   0.3, we predict that of all the mass layoff events in 2012, 0.3% were in fashion-related industries We enter the following data: We estimate that the line y  4800 is a reasonable horizontal asymptote for the data set We align the data set by subtracting 4800 from the output values as shown below ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 24 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models t 10 12 E 4902 5160 5529 5923 6508 7380 8044 E - 4980 102 360 729 1123 1708 2580 3244 Using ExpReg on the first and third columns of the table, we find that t a  t   174.5 1.311 is a model for the aligned data set Since we subtracted 4800 to create the aligned data set, we must add 4800 to a  t  to get the model for the school expenditures: t E  t   174.5 1.311  4800 with r  0.9300, where t is the number of school years after the 1990 – 91 school year and E is the projected expenditure per pupil The projected school expenditures per pupil in 2010 – 11 would be found by computing E  20   44115, meaning that projected school expenditures per pupil in 2010 – 11 would be $44115 This does not seem reasonable Using ExpReg on the first and second columns of the table, we find t that E  t   4747.11.04  with r  0.9822 Using this model, the projected school expenditures per pupil in 2010 – 11 would be found by computing E  20  10401, meaning that projected school expenditures per pupil in 2010 – 11 would be $10401 This seems reasonable 11 We construct a scatter plot of the data in the window 1  x  5, 44  y  50 : We estimate that the line y  44 is a reasonable horizontal asymptote for the data set We align the data set by subtracting 44 from the output values as shown below t p 49.9 48.0 46.3 45.6 45.1 p - 44 5.9 4.0 2.3 1.6 1.1 Using ExpReg on the first and third columns of the table, we find that t a  t   5.853  0.6521 is a model for the aligned data set Since we subtracted 44 to create the aligned data set, we must add 44 to a  t  to get the model for the accidents resulting in injuries: t p  t   5.853  0.6521  44 with r  0.9945, where t is the number of years after 2000 and p is the percentage of highway accidents resulting in injuries The change factor indicates that the percentage of accidents resulting in injuries is decreasing at a rate of about 35% (  100  1  0.65 % ) of the prior year’s decrease ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-6 Function Modeling and Combining Functions 25 12 When 29951 homes are available, the median price for a home is predicted to be $92,586 With this number of homes there is sufficient supply to meet the demand 14 a f  g  x    3x  5   4 x    x  To find the equilibrium point, we set D  x   S  x  We have: 0.41073x  130780  1.94 x  8541 2.35073x  122239 x  52000 D  52000  S  52000  109420 The equilibrium point is approximately  52000,109420  When 52,000 homes are available, the median price for a home is predicted to be $109,420 With this number of homes, there is sufficient supply to meet the demand b f  g  x    3x  5   4 x    x  12 15 a f  g  x   3x    5x  10  8 x  16 f  3x    for x not b    x   x  10 g equal to 16 a  f  g  x    x  1 x  3  x2  x  b 13 17 a g x3  f   x  x 1   f  g  x    x  x  1   2 x  5   x2  x  b  f  g  x    x  x  1 2 x  5  x  2 x   2 x  2 x      x   To find the equilibrium point, we set D  x   S  x  We have: 6.2841x  280799  2.41x  20405 8.6941x  260394 x  29951 D  29951  S  29951  92586 The equilibrium point is approximately  29951,92586  2 x  x  x  10 x  x   2 x  x  3x  10 x  18 a  g  f  x    x     x  8  x  x  12 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 26 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models f  x2  b    x   2x  g 19 a b  f  g  x    f  g  x   x 2 5 x2 20 a  V  3.383t  154.8   t   2.821t  225.5 P represents the percentage of the U.S population that has a registered vehicle b Solving 3.383t  154.8  1: 2.821t  225.5 3.383t  154.8  2.821t  225.5 0.562t  70.7 t  126 In 126 years from 1980 (which would be the year 2106), every person in the United States would own a registered vehicle 21 a  S  P  t    0.22t  53.18   2870.9t  282426 and represents the total number of thousands of gallons of soda pop consumed in the United States in year t c  S  P 10  15861662 According to the model, 15,861,662 thousand gallons of soda were consumed in the United States in the year 2010 22 f ( g (t ))  3(5  4t )2  (5  4t )  f ( g (2))  34 23 f ( g (t ))  f (2t )  3e2 t f ( g (2))  3e2(2)  3e8  4(2 0.4t ) f ( g (2))  1.52 24 f ( g (t ))  25 f ( g (t ))  8( 3t  t )  8( 3t  t )  f ( g (2))  4047 26 f ( g (t ))  f (t  3t  6)  5(t  3t  6)   5t  15t  26 f ( g (2))  5(2)  15(2)  26  16 27 f ( g (t ))  f (t  5)  t  f ( g (2))    3 b 28 f ( g (t ))  6(t  2t  6)  f ( g (2))  33 29 a K (C ( F ))  ( F  32)  273 The function is increasing on the interval 0,70 , approximately b Input: Fahrenheit Output: kelvin c A temperature of 81 degrees Fahrenheit is equal to ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-6 Function Modeling and Combining Functions 27 Summarizing: Income Level $2400 $29,700 $59,300 $129,000 $345,000 81  32   273  300 degrees kelvin d  F  32   273 299   F  32  538.2  F  32 570.2  F 572  32 a 30 a w(a (h ))  0.034(38.05  24.39h  2137) 1.245(38.05  24.39h  2137) 5.077(38.05  24.39h  2137)  34.346 b The independent variable is h d (height) and the dependent variable is w (weight) The composition w(a(h)) is the average weight (in pounds) as a function of the height (in inches) 31 a b 0.10i if  i  7550 755  0.15(i  7550) if 7550  i  30650  4220  0.25(i  30650) if 30650  i  74200 T (i )   15107.50  0.28(i  74200) if 74200  i  154800 37675.50  0.33(i  154800) if 154800  i  336550  97653  0.35(i  336550) if i  336550 0.10(2400)  240 755  0.15(29700  7550)  4077.50 4220  0.25(59300  30650)  11382.50 15107.50  0.28(129000  74200)  30451.50 97653  0.35(345000  336550)  100610.50 $240 $4077.50 $11,382.50 $30,451.50 $100,610.50 c Answers will vary A kelvin temperature of 572 degrees is equal to 570.2 degrees Fahrenheit c w(a(60))  94.42; For a healthy child that is 60 inches tall, the weight would be 94.92 pounds Tax b Size 10 11 12 13 14 15 16 17 18 19 20 Cost 19.95 39.90 59.85 79.80 99.75 119.70 139.65 159.60 179.55 199.50 219.45 231.40 259.35 279.30 217.50 232.00 246.50 261.00 275.50 290.00 19.95n if n  15 C (n)   14.50n if n  15 c C 14  19.95 14  279.30, so it will cost $279.30 to admit a group of 14 people C 15  14.50 15  217.50, so it will cost $217.50 to admit a group of 15 people d A group of 19 will cost $275.50, the largest calculated amount below the 14-person cost of $279.30 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 28 at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson CHAPTER Functions and Models 33 a 0 if a  3.99 if  a   W (a )  4.99 if  a  12 6.99 if 13  a  65  5.99 if a  65 b W (3)  3.99 W (6)  3.99 W (9)  4.99 W (24)  6.99 All outputs are in dollars c Evaluating W   is impossible since no category in the table includes itself We assume that they meant to say that children “2 and under” eat for free d $[2 6.99   3.99  4.99 ]       35 a Length L 0.2 0.4 0.6 0.8 1.0 Cost C 2.5 2.9 3.3 3.7 4.1 b The cost of a 10-mile taxi ride would be $2.50  $2(9.8)  $22.10 c A person can ride for 0.25 + 0.2 = 0.45 miles for $3.00 2.5  2( m)  3.00 2m  0.50 m  0.25 If billing is in (1/5)-mile increments, however, the person can ride only 0.4 miles  $31.94 34 a 129 if  a  10 H (a )   159 if a  10 d We graph the function C  L   2.5  L b The domain is a  3; the range is the set 129,159 c  83 if d 122 if d  A( d )  159 if d 179 if d  189 if d 1 2 3 4 5 d The domain is the set 1,2,3,4,5 The range is the set 83,122,159,179,189 e Assuming one child free: $ 3 129   159    $864 36 a $ 15  6  10  9.50  $185 b No, the table cannot be used to determine the number of 16-yearolds who can go to the museum for $100 because the table only gives prices for groups of 20 or more We aren’t given the regular cost for a 16-year-old ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson 1-6 Function Modeling and Combining Functions 29 c d 0.00 if  a  6.00 if  a  11  P(a )   9.50 if 12  a  65 8.00 if a  65 ©2015 Cengage Learning All rights reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Applied-CALC-2nd-Edition-by-Wilson ... https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson. .. https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file 14 at https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson. .. https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson Solution Manual for Applied CALC 2nd Edition by Wilson Full file at https://TestbankDirect.eu /Solution- Manual- for- Applied- CALC- 2nd- Edition- by- Wilson

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