## Thinking Mathematically (6th Edition)

(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.
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.