Question 576357
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P\left(1\ +\ \frac{r}{n}\right)^{nt}]


where *[tex \Large A] is the future value, *[tex \Large P] is the present value or invested principal, *[tex \Large r] is the decimal representation of the annualized interest rate, *[tex \Large n] is the number of compounding periods per year, and *[tex \Large t] is the number of years.


Plug in your values. . .  and do the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ 2000\left(1\ +\ \frac{0.06}{12}\right)^{12\,\cdot\,15}]


 . . . and do the arithmetic.  I would get out the calculator if I were you, unless you are a <i><b>serious</b></i> glutton for punishment.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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