Tony makes deposits of $320 monthly (at the end of each month) into an account that pays 1.8% interest compounded monthly.
How much interest will he have earned after 3 years?
A=p(1+r(12)(100))12×n=320(1+1.8(12)(100))12×3=320×1.0554=$337.728A=p(1+\frac{r}{(12)(100)})^{12\times n}=320(1+\frac{1.8}{(12)(100)})^{12\times 3}=320\times1.0554=\$337.728A=p(1+(12)(100)r)12×n=320(1+(12)(100)1.8)12×3=320×1.0554=$337.728
I=A−P=337.728−320=$17.728I=A-P=337.728-320=\$17.728I=A−P=337.728−320=$17.728
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