Questions: Adriana Lima invests 8,175 into an account that earns 7.9% compounded continuously. If she checks her account 14 years later, what will the balance in her account be?

Adriana Lima invests 8,175 into an account that earns 7.9% compounded continuously. If she checks her account 14 years later, what will the balance in her account be?
Transcript text: Adriana Lima invests $\$ 8,175$ into an account that earns $7.9 \%$ compounded continuously. If she checks her account 14 years later, what will the balance in her account be?
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Solution

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Solution Steps

To solve this problem, we need to use the formula for continuous compounding, which is given by \( A = P \times e^{rt} \), where \( A \) is the amount of money accumulated after n years, including interest. \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (in decimal), \( t \) is the time the money is invested for in years, and \( e \) is the base of the natural logarithm.

  1. Identify the principal amount \( P = 8175 \).
  2. Convert the annual interest rate from a percentage to a decimal \( r = 7.9/100 = 0.079 \).
  3. Identify the time period \( t = 14 \) years.
  4. Use the continuous compounding formula to calculate the future value.
Step 1: Identify the Variables

We are given the following values:

  • Principal amount \( P = 8175 \)
  • Annual interest rate \( r = 0.079 \)
  • Time period \( t = 14 \)
Step 2: Apply the Continuous Compounding Formula

The formula for continuous compounding is given by:

\[ A = P \times e^{rt} \]

Substituting the known values into the formula:

\[ A = 8175 \times e^{0.079 \times 14} \]

Step 3: Calculate the Future Value

Calculating the exponent:

\[ 0.079 \times 14 = 1.106 \]

Now, substituting this back into the equation:

\[ A = 8175 \times e^{1.106} \]

Calculating \( e^{1.106} \):

\[ e^{1.106} \approx 3.017 \]

Now, substituting this value into the equation for \( A \):

\[ A \approx 8175 \times 3.017 \approx 24706.8545 \]

Final Answer

The balance in Adriana Lima's account after 14 years is approximately

\[ \boxed{A \approx 24706.85} \]

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