To determine the accumulated value of an investment of \[tex]$10,000 over a period of 3 years at an annual interest rate of 6.5%, we will use the compound interest formulas provided. We will calculate the accumulated value for three different compounding methods: semiannually, monthly, and continuously.
#### Given Values:
- Principal amount (\(P\)): \$[/tex]10,000
- Time ([tex]\(t\)[/tex]): 3 years
- Annual interest rate ([tex]\(r\)[/tex]): 6.5% or 0.065 in decimal
### Part (a): Compounded Semiannually
For semiannual compounding, the interest is compounded twice a year ([tex]\(n = 2\)[/tex]).
The formula used is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt}\][/tex]
Plugging in the given values:
[tex]\[ n = 2 \][/tex]
[tex]\[ A = 10000 \left(1 + \frac{0.065}{2}\right)^{2 \times 3} \][/tex]
Calculate step-by-step:
[tex]\[ A = 10000 \left(1 + 0.0325\right)^{6} \][/tex]
[tex]\[ A = 10000 (1.0325)^6 \][/tex]
[tex]\[ A \approx 12115.47 \][/tex]
So, the accumulated value if the money is compounded semiannually is [tex]\(\$12,115.47\)[/tex].
### Part (b): Compounded Monthly
For monthly compounding, the interest is compounded twelve times a year ([tex]\(n = 12\)[/tex]).
The formula used is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Plugging in the given values:
[tex]\[ n = 12 \][/tex]
[tex]\[ A = 10000 \left(1 + \frac{0.065}{12}\right)^{12 \times 3} \][/tex]
Calculate step-by-step:
[tex]\[ A = 10000 \left(1 + 0.0054167\right)^{36} \][/tex]
[tex]\[ A = 10000 (1.0054167)^{36} \][/tex]
[tex]\[ A \approx 12146.72 \][/tex]
So, the accumulated value if the money is compounded monthly is [tex]\(\$12,146.72\)[/tex].
### Part (c): Compounded Continuously
For continuous compounding, the formula used is:
[tex]\[ A = P e^{rt} \][/tex]
Plugging in the given values:
[tex]\[ A = 10000 e^{0.065 \times 3} \][/tex]
Calculate step-by-step using the exponential function:
[tex]\[ A = 10000 e^{0.195} \][/tex]
[tex]\[ A \approx 10000 \times 1.2153 \][/tex]
[tex]\[ A \approx 12153.11 \][/tex]
So, the accumulated value if the money is compounded continuously is [tex]\(\$12,153.11\)[/tex].
