#### Answer

(a) After 10 years, there will be \$9479.19 in the account.
(b) After 10 years, there will be \$9527.79 in the account.
(c) After 10 years, there will be \$9560.92 in the account.
(d) After 10 years, there will be \$9577.70 in the account.

#### Work Step by Step

We can use this formula:
$A = P~(1+\frac{r}{n})^{nt}$
$A$ is the final amount in the account
$P$ is the principal (the amount of money invested)
$r$ is the interest rate
$n$ is the number of times per year the interest is compounded
$t$ is the number of years
(a) $A = P~(1+\frac{r}{n})^{nt}$
$A = (\$5,000)~(1+\frac{0.065}{2})^{(2)(10)}$
$A = \$9479.19$
After 10 years, there will be \$9479.19 in the account.
(b) $A = P~(1+\frac{r}{n})^{nt}$
$A = (\$5,000)~(1+\frac{0.065}{4})^{(4)(10)}$
$A = \$9527.79$
After 10 years, there will be \$9527.79 in the account.
(c) $A = P~(1+\frac{r}{n})^{nt}$
$A = (\$5,000)~(1+\frac{0.065}{12})^{(12)(10)}$
$A = \$9560.92$
After 10 years, there will be \$9560.92 in the account.
(d) If the money is compounded continuously, we can use this formula:
$A = P~e^{rt}$
$A = (\$5,000)~e^{(0.065)(10)}$
$A = \$9577.70$
After 10 years, there will be \$9577.70 in the account.