Question

1. suppose you invest $2000 at an annual interest rate of 5.1% compounded continuously. how much will you have in the account after 3 years? round the solution to the nearest dollar. $2331 $2453 $2443 $2324

Explanation:

Step1: Recall the formula for continuous compounding

The formula for continuous compounding is \( A = Pe^{rt} \), where \( P \) is the principal amount, \( r \) is the annual interest rate (in decimal), \( t \) is the time in years, and \( A \) is the amount after time \( t \).

Step2: Identify the values

Here, \( P = 2000 \), \( r = 5.1\% = 0.051 \), and \( t = 3 \) years.

Step3: Calculate the exponent

Calculate \( rt = 0.051 \times 3 = 0.153 \).

Step4: Calculate \( e^{rt} \)

Using a calculator, \( e^{0.153} \approx 1.1653 \) (more accurately, using a calculator for \( e^{0.153} \)).

Step5: Calculate the amount \( A \)

Multiply \( P \) by \( e^{rt} \): \( A = 2000 \times 1.1653 \approx 2330.6 \), which rounds to \$2331. So the correct option is \$2331.

Answer:

\$2324 (assuming the correct option is the last one, but let's calculate properly)
Wait, let's do the calculation. The formula for continuous compounding is \( A = Pe^{rt} \), where \( P = 2000 \), \( r = 0.051 \) (5.1% as a decimal), \( t = 3 \).

So \( A = 2000 \times e^{0.051 \times 3} \). First, calculate the exponent: \( 0.051 \times 3 = 0.153 \). Then \( e^{0.153} \approx 1.1652 \). Then \( 2000 \times 1.1652 = 2330.4 \), which is close to \$2331. Wait, maybe my approximation of LXI4 is off. Let's use a calculator: LXI5. Then LXI6, which is approximately \$2331. So the correct option is \$2331.