To calculate the future value of an investment with compound interest, we utilize the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money invested).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for, in years.
For this problem, we have the following values:
- Principal, [tex]\( P \)[/tex] = [tex]$300
- Annual interest rate, \( r \) = 6% or 0.06 (in decimal form)
- Number of times compounded per year, \( n \) = 12 (monthly compounding)
- Time, \( t \) = 10 years
Substitute these values into the formula:
\[ A = 300 \left(1 + \frac{0.06}{12}\right)^{12 \times 10} \]
First, calculate the monthly interest rate:
\[ \frac{r}{n} = \frac{0.06}{12} = 0.005 \]
Next, substitute this value back into the formula:
\[ A = 300 \left(1 + 0.005\right)^{120} \]
Then, calculate the exponent portion:
\[ 1 + 0.005 = 1.005 \]
Raise this to the power of 120 (since \( n \times t = 12 \times 10 \)):
\[ 1.005^{120} \]
Finally, multiply this result by the initial principal to get:
\[ A = 300 \times 1.819396227536954 \approx 545.82 \]
Thus, $[/tex]300 invested at an annual interest rate of 6%, compounded monthly, will be worth approximately $545.82 after 10 years.
