Solution:
Given that an investment accout is opend with an initial deposit of $6500, and the account chosen compounds interest annually.
A) Equation to calculate the compunded interest.
The interest is expressed as
[tex]\begin{gathered} \text{Interest = Final amount - principal/initial deposit} \\ \text{where} \\ A=P(1+\frac{r}{n})^{nt} \\ A\Rightarrow\text{final amount} \\ P\Rightarrow pr\text{incipal or initial deposit} \\ r\Rightarrow\text{interest rate} \\ n\Rightarrow number\text{ of times interest applied per }time\text{ period} \\ t\Rightarrow period \\ \text{thus, we have} \\ I=P(1+\frac{r}{n})^{nt}-P \\ \text{where} \\ I\text{ is the compounded interest} \end{gathered}[/tex]
B) Value of the account after 10 years, if interest is paid at 3.7%, if no other withdrawals or deposits are made.
This implies that
[tex]\begin{gathered} t=10,\text{ n=1} \\ r=\frac{3.7}{100}=0.037 \\ \text{recall that} \\ P=6500 \end{gathered}[/tex]
thus,
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ =6500(1+\frac{0.037}{1})^{10(1)} \\ \Rightarrow A=9347.61723 \end{gathered}[/tex]
Thus, the value of the account after 10 years is
[tex]\$9347.61723[/tex]
C) Interest earned if the account is left for 20 years.
In this case,
[tex]t=20[/tex]
Thus, the interest is evaluated as
[tex]\begin{gathered} I=P(1+\frac{r}{n})^{nt}-P \\ =6500(1+\frac{0.037}{1})^{20(1)}-6500 \\ =13442.76121-6500 \\ I=\$6942.76121 \end{gathered}[/tex]
The interest when the account is left alone for 20 years is
[tex]\$6942.76121[/tex]
