Answer
$ 9.25\%$
Work Step by Step
If a principal $P$ is invested in an account paying an annual interest rate $r$, compounded $n$ times a year, then after $t$ years the amount $A(t)$ in the account is
$A(t)=P(1+\displaystyle \frac{r}{n})^{nt}$
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Solve for r after inserting given values
$1435.77=1000(1+\displaystyle \frac{r}{2})^{2(4)} \qquad$ ... $/\div 1000$
$1.43577=(1+\displaystyle \frac{r}{2})^{8} \qquad$ ... $/ (...)^{1/8}$
$1+\displaystyle \frac{r}{2}=(1.43577)^{1/8} \qquad$ ... $/-1$
$\displaystyle \frac{r}{2}=(1.43577)^{1/8} -1\qquad$ ... $/\times 2$
$r=2((1.43577)^{1/8} -1)\approx 0.0925$.
The rate was about $ 9.25\%$.
