To find the total amount after 10 years for an investment of \($2000 at an annual interest rate of 5.4%, compounded monthly, we will use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (in decimal form).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the number of years the money is invested.
Given:
- \( P = $\)2000 \)
- \(\( r = 5.4 \% = 0.054 \)\) (as a decimal)
- \(\( n = 12 \)\) (monthly compounding)
- \(\( t = 10 \)\) years
Plug in these values into the compound interest formula:
\(\[ A = 2000 \left(1 + \frac{0.054}{12}\right)^{12 \cdot 10} \]\)
First calculate the value inside the parentheses:
\(\[ \frac{0.054}{12} = 0.0045 \]\)
Now add 1 to this value:
\(\[ 1 + 0.0045 = 1.0045 \]\)
Raise this result to the power of \(\( 12 \times 10 = 120 \)\):
\(\[ 1.0045^{120} \]\)
Using a calculator for this exponentiation, we find:
\(\[ 1.0045^{120} \approx 1.7139 \]\)
Now multiply this result by the principal amount:
\(\[ A = 2000 \times 1.7139 \]\)
\(\[ A \approx 3427.80 \]\)
So, the amount after 10 years, compounded monthly, will be approximately $3427.80. Make sure to round the result to the nearest cent as necessary.
