Question

find the present value that will grow $3000 if the annual interest rate is 9.5% compounded quarterly for 9yr

Answer 1

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 3000\\ r=rate\to 9.5\%\to \frac{9.5}{100}\to &0.095\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\to &4\\ t=years\to &9 \end{cases} \\\\\\ A=3000\left(1+\frac{0.095}{4}\right)^{4\cdot 9}\implies A=3000(1.02375)^{36}\implies A\approx 6983.97$