Question 972863
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ P\left(1\ +\ \frac{r}{n}\right)^{nt}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ = \frac{\ln\left(\frac{A}{P}\right)}{n\ln\left(1\ +\ \frac{r}{n}\right)}]


Where *[tex \Large A] is the future value, *[tex \Large P] is the present value, *[tex \Large r] is the interest rate expressed as a decimal, *[tex \Large n] is the number of compounding periods per year, and *[tex \Large t] is the number of years]


For this problem *[tex \Large A\ =\ 3900], *[tex \Large P\ =\ 1100], *[tex \Large r\ =\ 0.04], and (assuming, because you don't specify) *[tex \Large n\ =\ 1].  Plug in the numbers. You can do your own arithmetic.


Of course, at this rate George will never get his car.  By the time that he accumulates $3900 in this account, the cost of the car he wants will have increased to over $12,000 (give or take, depending on make and model)


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \