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demand function: 3000 f(n) = 4 + (100 ≤ n ≤1500) n + 100 where n is the number of tee shirts that can be sold per month at a price of f(n) dollars per shirt. 41. At a price level of $19 per shirt, compare the predictions for monthly sales obtained using each model. Hint: You need to solve each of the equations p(n) = 19 and f(n) = 19 and interpret the results. 42. Show that at a price level of $10 per shirt, the model p predicts more than twice as many sales per month as does the model t Hint: Solve each of the equations n(n) = 10 1 2 $ @ 4 3 % olo5 66 & 7 8 9 0 O P { delete

demand function: 3000 f(n) = 4 + (100 ≤ n ≤1500) n + 100 where n is the number of tee shirts that can be sold per month at a price of f(n) dollars per shirt. 41. At a price level of $19 per shirt, compare the predictions for monthly sales obtained using each model. Hint: You need to solve each of the equations p(n) = 19 and f(n) = 19 and interpret the results. 42. Show that at a price level of $10 per shirt, the model p predicts more than twice as many sales per month as does the model t Hint: Solve each of the equations n(n) = 10 1 2 $ @ 4 3 % olo5 66 & 7 8 9 0 O P { delete

Algebra for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter8: Functions
Section8.CR: Review Problem Set
Problem 50CR
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demand function: 3000 f(n) = 4 + (100 ≤ n ≤1500) n + 100 where n is the number of tee shirts that can be sold per month at a price of f(n) dollars per shirt. 41. At a price level of $19 per shirt, compare the predictions for monthly sales obtained using each model. Hint: You need to solve each of the equations p(n) = 19 and f(n) = 19 and interpret the results. 42. Show that at a price level of $10 per shirt, the model p predicts more than twice as many sales per month as does the model t Hint: Solve each of the equations n(n) = 10 1 2 $ @ 4 3 % olo5 66 & 7 8 9 0 O P { delete
Transcribed Image Text:demand function: 3000 f(n) = 4 + (100 ≤ n ≤1500) n + 100 where n is the number of tee shirts that can be sold per month at a price of f(n) dollars per shirt. 41. At a price level of $19 per shirt, compare the predictions for monthly sales obtained using each model. Hint: You need to solve each of the equations p(n) = 19 and f(n) = 19 and interpret the results. 42. Show that at a price level of $10 per shirt, the model p predicts more than twice as many sales per month as does the model t Hint: Solve each of the equations n(n) = 10 1 2 $ @ 4 3 % olo5 66 & 7 8 9 0 O P { delete
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