To solve this problem we will use the formula for compound interest:
[tex]P_N=P_0\cdot\mleft(1+\frac{r}{k}\mright)^{N\cdot k}.[/tex]Where:
• P_N, is the balance in the account after N years,
,• P_0, is the starting balance of the account (also called an initial deposit, or principal),
,• r, is the annual interest rate in decimal form,
,• k, is the number of compounding periods in one year.
In this problem we have that:
• N = 13 (13 years),
,• P_N is the unknown,
,• P_0 = $9400,
,• r = 4.1/100 = 0.041,
,• k = 2 (because the interest compounded twice per year).
Replacing these values in the formula above, we find:
[tex]P_{13}=9400\cdot(1+\frac{0.041}{2})^{13\cdot2}_{}\cong15931.85.[/tex]Answer
[tex]P_{13}=9400\cdot(1+\frac{0.041}{2})^{13\cdot2}_{}\cong15931.85.[/tex]