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23 March, 18:38

A recently installed machine earns the company revenue at a continuous rate of

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  1. 23 March, 20:20
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    Answer: R = Pe^ (kt) where R = revenue, P is present value, k is interest rate and t is time in years. (60 000 + 45 000 + 75 000) / 2 = Pe^ (.07*1) 90 000 = Pe^.07 P = 90 000/e^.07 = $83915.44 (60 000 + 45 000 + 75 000) / 2 + 75 000 (t-1) = 150 000 e^ (kt) 90 000 + 75 000t - 75 000 = 150 000e^ (.07t) 15 000 + 75 000t = 150 000e^ (.07t) 1/10 + 1/2t = e^ (.07t) ln (.1 +.5t) =.07t ln (.1+.5t) / t =.07 t = 2.12 years
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