Exam 1 - S09-1

# Exam 1 - S09-1 - Math 1260 Exam 1 Solutions – Spring 2009...

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Unformatted text preview: Math 1260 Exam 1 Solutions – Spring 2009 1.) Taking the integral of both sides of the equation Q ( x ) = 200 x- 1 / 2 we find that Q ( x ) = R 200 x- 1 / 2 dx = 400 x 1 / 2 + c . We can find c by pluging Q = 500 and x = 4 into the above equation: 500 = 800 + c or c =- 300. Therefore Q ( x ) = 400 x 1 / 2- 300, and Q (25) = 400 · 5- 300 = 1700. 2.) The average gas price is equal to [2 . 5 + 2 . 98 + 3 . 05 + 3 . 2 + 4 . 1 + 2 . 25] · 1 6 ≈ 3 . 01 dollars per gallon. 3.) We have P (100)- P (50) = R 100 50 (- . 2 x +30) dx = (- . 1 x 2 +30 x ) 100 50 = [- . 1(100) 2 +30 · 100]- [- . 1(50) 2 + 30 · 50] = 750. 4.) Applying the future value formula FV = PV e rt in our situation gives FV = 825 e (0 . 02625)10 ≈ 1 , 072 . 65 million dollars. This is the amount to be paid by the Taxpayers. 5.) Using midpoints, which is the only choice available here, we have R 2 f ( x ) dx ≈ [ f (0 . 2)+ f (0 . 6)+ f (1) + f (1 . 4) + f (1 . 8)] · Δ x = [7 . 9 + 8 . 2 + 8 . 1 + 7 . 5 + 6 . 3](0 . 4) = 15 . 2 . 6.) From the shape of the graph of the population p ( t ) we conclude that p ( t ) is modeled by a logistic differential equation of the form dp dt = rp (1- p K ). Since p ( t ) goes to 30 as t gets large we conclude that K must be 30. Therefore, we must havemust be 30....
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## This note was uploaded on 03/02/2012 for the course MATH 10260 taught by Professor Himonas during the Fall '09 term at Notre Dame.

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Exam 1 - S09-1 - Math 1260 Exam 1 Solutions – Spring 2009...

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