Asked by Alexis
I cannot figure this out for the life of me. Maybe I am overthinking it.
Money is transferred into an account at the rate of R(t)=5000t(e^-0.6t)
If the account pays 2% interest compounded continuously, how much will accumulate in the account over a 9-year period?
Money is transferred into an account at the rate of R(t)=5000t(e^-0.6t)
If the account pays 2% interest compounded continuously, how much will accumulate in the account over a 9-year period?
Answers
Answered by
Steve
Money at 2% grows like e^0.02t
The amount deposited at time t gets to grow for (9-t) years. Seems to me like the account balance after 9 years is thus
∫[0,9] R(t) * e^(.02(9-t)) dt
The amount deposited at time t gets to grow for (9-t) years. Seems to me like the account balance after 9 years is thus
∫[0,9] R(t) * e^(.02(9-t)) dt
Answered by
Steve
I may have misspoken. Take a look here for a discussion of the topic.
faculty.atu.edu/mfinan/2243/business65.pdf
faculty.atu.edu/mfinan/2243/business65.pdf
Answered by
Steve
actually, it appears I was right after all...
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