To solve this compound interest problem, we will start with part a, where the interest is compounded semiannually, and then proceed to part b, where the interest is compounded monthly.
### Part (a): Compounded Semiannually
1. Given Data:
- Principal ([tex]\( P \)[/tex]): \[tex]$10,000
- Annual Interest Rate (\( r \)): 0.065 (6.5%)
- Time (\( t \)): 3 years
- Compounding Periods per Year (\( n \)): 2 (since it's semiannually)
2. Formula for Compound Interest:
\[
A = P \left(1+\frac{r}{n}\right)^{ nt }
\]
3. Plugging in the Values:
\[
A = 10,000 \left( 1 + \frac{0.065}{2} \right)^{2 \cdot 3}
\]
4. Simplifying:
\[
A = 10,000 \left( 1 + 0.0325 \right)^6
\]
\[
A = 10,000 \left( 1.0325 \right)^6
\]
5. Calculating the Accumulated Value:
\[
A \approx 12,115.47
\]
Therefore, the accumulated value if compounded semiannually is \(\$[/tex] 12,115.47\).
### Part (b): Compounded Monthly
1. Given Data:
- Principal ( [tex]\( P \)[/tex]): \[tex]$10,000
- Annual Interest Rate (\( r \)): 0.065 (6.5%)
- Time (\( t \)): 3 years
- Compounding Periods per Year (\( n \)): 12 (since it's monthly)
2. Formula for Compound Interest:
\[
A = P \left( 1 + \frac{r}{n} \right)^{n t}
\]
3. Plugging in the Values:
\[
A = 10,000 \left( 1 + \frac{0.065}{12} \right)^{12 \cdot 3}
\]
4. Simplifying:
\[
A = 10,000 \left( 1 + 0.0054167 \right)^{36}
\]
\[
A = 10,000 \left( 1.0054167 \right)^{36}
\]
5. Calculating the Accumulated Value:
\[
A \approx 12,146.72
\]
Therefore, the accumulated value if compounded monthly is \(\$[/tex] 12,146.72\).
By following the steps above, we've found the accumulated value for both semiannual and monthly compounding. Now let's fill in the answer:
### Final Answer for Part (b):
[tex]\[
\boxed{12,146.72}
\][/tex]
