wordproblems1Sols

wordproblems1Sols - Math 105 Word Problems 1 Answers...

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Answers Present Value 1. Z T 0 K ( t ) e - rt dt 2. Z 4 0 (5 , 000 - 100 t ) e - 0 . 16 t dt = 14 , 243 (use integration by parts) 3. Z 0 10000 e 0 . 04 t e - 0 . 12 t dt = 125 , 000 4. Z 2 0 (50 + 7 t ) e - 0 . 1 t dt = 102 . 9, so $102 , 900. 5. Z 4 0 50 e - 0 . 08 t e - 0 . 12 t dt = 137 . 668, so $137 , 668. Capital Value and Improper Integrals 6. Z 0 5000 e - 0 , 1 t dt = 50 , 000 7. Z 0 Ke - rt dt = K r 8. (a) n X k =1 M ( t k t Z 2 0 M ( t ) dt , with Δ t = 2 n , t k = 2 k n (b) n X k =1 M ( t k ) e - 0 . 1 t k Δ t Z 2 0 M ( t ) e - 0 . 1 t dt , with Δ t = 2 n , t k = 2 k n 9. 80000 + Z 0 50000 e - rt dt ± = 80000 + 50000 r ² Understanding Differential Equations 10. The growth rate is y 0 = ( . 0004)(500)(1000 - 500) = 100 fish per month. 11. (a) It is decreasing at $1600 per year. (b) The account is earning interest at 4% per year compounded continuously, and money is being withdrawn steadily at the rate of $2000 per year (or if there are both deposits and withdrawals, the rate of withdrawals is $2000/year more than the rate
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This note was uploaded on 01/14/2012 for the course MATH 105 taught by Professor Malabikapramanik during the Fall '10 term at UBC.

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wordproblems1Sols - Math 105 Word Problems 1 Answers...

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