Question
Edgar accumulated $5,000 in credit card debt. If the interest rate is 20% per year, and he does not make any payments for 2 years, how much will he owe on this debt in 2 years by each method of compounding? Round your answer to the nearest cent with no commas or dollar signs.
Compound continuously
7459.12
7237.12
7387.28
7434.57
Compound continuously
7459.12
7237.12
7387.28
7434.57
Answers
Answered by
GPT-4o mini
To calculate the amount Edgar will owe after 2 years with continuous compounding, we can use the formula for continuous compounding:
\[ A = P e^{rt} \]
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 (decimal).
- \( t \) is the time the money is invested or borrowed for, in years.
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Given:
- \( P = 5000 \)
- \( r = 0.20 \) (20% as a decimal)
- \( t = 2 \)
Now, substituting the given values into the formula:
\[ A = 5000 \times e^{0.20 \times 2} \]
\[ A = 5000 \times e^{0.40} \]
Now, calculate \( e^{0.40} \):
Using a calculator or a mathematical software, we find that
\[ e^{0.40} \approx 1.49182 \]
Now, multiplying by the principal amount:
\[ A \approx 5000 \times 1.49182 \]
\[ A \approx 7459.10 \]
Rounding to the nearest cent, Edgar will owe approximately \( 7459.10 \) after 2 years.
Among the options you've provided, the closest answer appears to be:
**7459.12**.
\[ A = P e^{rt} \]
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 (decimal).
- \( t \) is the time the money is invested or borrowed for, in years.
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Given:
- \( P = 5000 \)
- \( r = 0.20 \) (20% as a decimal)
- \( t = 2 \)
Now, substituting the given values into the formula:
\[ A = 5000 \times e^{0.20 \times 2} \]
\[ A = 5000 \times e^{0.40} \]
Now, calculate \( e^{0.40} \):
Using a calculator or a mathematical software, we find that
\[ e^{0.40} \approx 1.49182 \]
Now, multiplying by the principal amount:
\[ A \approx 5000 \times 1.49182 \]
\[ A \approx 7459.10 \]
Rounding to the nearest cent, Edgar will owe approximately \( 7459.10 \) after 2 years.
Among the options you've provided, the closest answer appears to be:
**7459.12**.