Given:
$12,300 at 2% compounded semiannually for 4 years
So, P = 12300
r = 2% = 0.02
compounded semiannually ⇒ n = 2
Time = t = 4 years
We will find the compound interest I, we will use the following formula:
![I=P(1+(r)/(n))^(nt)-P](https://img.qammunity.org/2023/formulas/mathematics/college/17kb62esssbjlp0e90rsoalq2psj5y9brr.png)
Substitute with the given values:
![\begin{gathered} I=12300\cdot(1+(0.02)/(2))^(2\cdot4)-12300 \\ I=12300\cdot1.01^8-12300 \\ I=12300\cdot(1.01^8-1)\approx1019.1375 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/o7xk1qpbh77wo49pr19ak48fmb8y4l1g02.png)
Rounding to the nearest cent
So, the answer will be the compound interest = $1,019.14