To determine how much [tex]$300 invested at 9% interest compounded continuously would be worth after 3 years, we use the continuous compounding formula:
\[ A(t) = P \cdot e^{rt} \]
where:
- \( P \) is the principal amount (the initial amount of money),
- \( r \) is the annual interest rate (expressed as a decimal),
- \( t \) is the time the money is invested for (in years),
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828),
- \( A(t) \) is the amount of money accumulated after \( t \) years, including interest.
Given:
- \( P = 300 \) (the principal amount),
- \( r = 0.09 \) (the annual interest rate, 9% converted to a decimal),
- \( t = 3 \) (the time in years).
Substitute these values into the formula:
\[ A(3) = 300 \cdot e^{0.09 \times 3} \]
Calculate the exponent:
\[ 0.09 \times 3 = 0.27 \]
Now, evaluate \( e^{0.27} \). Using the approximate value for \( e \):
\[ e^{0.27} \approx 1.31025 \]
Then, multiply by the principal:
\[ A(3) = 300 \cdot 1.31025 \]
\[ A(3) \approx 393.075 \]
Rounding to the nearest cent, we get:
\[ A(3) \approx 392.99 \]
Therefore, the amount of money accumulated after 3 years is:
\[ \boxed{392.99} \]
So, the correct answer is:
A. \$[/tex] 392.99
