Question #681d4

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Question #681d4

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Answer 1 · 2017-09-24T02:30:57

Well use a fomula that is used to calculate compound interest:

$A = D {(1 + \frac{r}{n})}^{n t}$ $A = D(1 + r/n)^(nt)$

A is the amount.
D is the deposit.
r is interest rate in decimal form.
n is the number of times that interest is compounded per year.
t is the number of years

Let's substitute now:

$A = 4500 {(1 + \frac{0.1}{2})}^{2 \cdot 10}$ $A = 4500(1+(0.1)/2)^(2*10)$

This gives us an answer of

$\approx $ 11, 939.80$ $~~ $11,939.80$

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Question #681d4

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