The compound interest formula is given by
[tex]A=P(1+\frac{r}{n})^{n\cdot t}[/tex]
where A is the resulting amount, P is the amount of principal, r is the annual interest rate, n is the number of compounding periods per year and t is the time in years.
In our case, r=0.07, P=$2000 and t=4 years. The number of compounding periods n depends on the given case:
Part A. Annually.
In this case, n= 1 . Then, the resulting amount A is
[tex]A=2000(1+\frac{0.07}{1})^{1\cdot4}[/tex]
which gives
[tex]\begin{gathered} A=2000(1.07)^{1\cdot4} \\ A=\text{ \$}2621.59 \end{gathered}[/tex]
Part B. Semuannually.
In this case, n=2 (twice per year). Then, the resulting amount A is
[tex]\begin{gathered} A=2000(1+\frac{0.07}{2})^{2\cdot4} \\ A=2000(1.035)^8 \\ A=\text{ \$}2633.62 \end{gathered}[/tex]
Part C. Quarterly.
In this case, n=4 (4 times per year). So, the resulting amount A is
[tex]\begin{gathered} A=2000(1+\frac{0.07}{4})^{4\cdot4} \\ A=2000(1.0175)^{16} \\ A=\text{ \$}2639.86 \end{gathered}[/tex]
Part D. Daily.
In this case, n= 360 (360 times per year). Then, the resulting amount A is
[tex]\begin{gathered} A=2000(1+\frac{0.07}{360})^{360\cdot4} \\ A=2000(1.0001944)^{360\cdot4} \\ A=\text{ \$ }2646.19 \end{gathered}[/tex]
Part E. Continuosly.
In this case, our first formula becomes
[tex]A=P\cdot e^{r\cdot t}[/tex]
then, by substituting our given values, we have
[tex]\begin{gathered} A=2000\cdot e^{0.07\cdot4} \\ A=\text{ \$2646.26} \end{gathered}[/tex]
