Question

1. 2.1/4 points $12,000 is invested at an interest rate of 13% per year, find the amount of the investment at the end of 2 years for the following compounding methods. (round your answers to the nearest cent.) (a) semiannual $ 15437.60 (b) quarterly (c) monthly (d) continuously $ 15583.17

Explanation:

Step1: Recall compound - interest formulas

For compound - interest, the general formula is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times interest is compounded per year, $t$ is the number of years. For continuous compounding, the formula is $A = Pe^{rt}$. Given $P=\$12000$, $r = 0.13$, and $t = 2$.

Step2: Calculate for semi - annual compounding ($n = 2$)

Substitute into the compound - interest formula: $A=P(1 +\frac{r}{n})^{nt}=12000(1+\frac{0.13}{2})^{2\times2}=12000(1 + 0.065)^{4}$. First, calculate $(1 + 0.065)^{4}=1.065^{4}\approx1.286466$. Then $A = 12000\times1.286466\approx\$15437.59$.

Step3: Calculate for quarterly compounding ($n = 4$)

$A=P(1+\frac{r}{n})^{nt}=12000(1+\frac{0.13}{4})^{4\times2}=12000(1 + 0.0325)^{8}$. Calculate $(1 + 0.0325)^{8}\approx1.29157$. Then $A=12000\times1.29157\approx\$15500.84$.

Step4: Calculate for monthly compounding ($n = 12$)

$A=P(1+\frac{r}{n})^{nt}=12000(1+\frac{0.13}{12})^{12\times2}=12000(1+\frac{0.13}{12})^{24}$. Let $x=\frac{0.13}{12}\approx0.010833$. Then $(1 + 0.010833)^{24}\approx1.29693$. So $A = 12000\times1.29693\approx\$15563.17$.

Step5: Calculate for continuous compounding

Use the formula $A = Pe^{rt}$. Substitute $P = 12000$, $r=0.13$, and $t = 2$. Then $A=12000e^{0.13\times2}=12000e^{0.26}$. Since $e^{0.26}\approx1.29693$, $A\approx12000\times1.29693\approx\$15563.17$.

Answer:

(a) $\$15437.59$
(b) $\$15500.84$
(c) $\$15563.17$
(d) $\$15563.17$