Answer :
Sure, let's solve this problem step-by-step:
Problem Statement:
Suppose that [tex]$1200 is borrowed for six years at an interest rate of 5% per year, compounded continuously. Find the amount owed, assuming no payments are made until the end. Do not round any intermediate computations, and round your answer to the nearest cent. Solution: 1. Identify the variables: - Principal (P): $[/tex]1200
- Annual interest rate (r): 5% or 0.05
- Time (t): 6 years
2. Use the formula for continuously compounded interest:
The formula to calculate the amount [tex]\( A \)[/tex] using continuous compound interest is:
[tex]\[ A = P \times e^{(r \times t)} \][/tex]
Where:
- [tex]\( P \)[/tex] is the principal amount
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828)
- [tex]\( r \)[/tex] is the annual interest rate
- [tex]\( t \)[/tex] is the time the money is invested or borrowed for
3. Substitute the given values into the formula:
[tex]\[ A = 1200 \times e^{(0.05 \times 6)} \][/tex]
4. Calculate the exponent:
[tex]\[ r \times t = 0.05 \times 6 = 0.30 \][/tex]
5. Find [tex]\( e^{0.30} \)[/tex]:
[tex]\[ e^{0.30} \approx 1.34986 \][/tex]
6. Calculate the final amount:
[tex]\[ A = 1200 \times 1.34986 \approx 1619.8305690912039 \][/tex]
7. Round the final amount to the nearest cent:
[tex]\[ A \approx 1619.83 \][/tex]
Therefore, after six years, the amount owed is approximately [tex]\( \mathbf{1619.83} \)[/tex].
Problem Statement:
Suppose that [tex]$1200 is borrowed for six years at an interest rate of 5% per year, compounded continuously. Find the amount owed, assuming no payments are made until the end. Do not round any intermediate computations, and round your answer to the nearest cent. Solution: 1. Identify the variables: - Principal (P): $[/tex]1200
- Annual interest rate (r): 5% or 0.05
- Time (t): 6 years
2. Use the formula for continuously compounded interest:
The formula to calculate the amount [tex]\( A \)[/tex] using continuous compound interest is:
[tex]\[ A = P \times e^{(r \times t)} \][/tex]
Where:
- [tex]\( P \)[/tex] is the principal amount
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828)
- [tex]\( r \)[/tex] is the annual interest rate
- [tex]\( t \)[/tex] is the time the money is invested or borrowed for
3. Substitute the given values into the formula:
[tex]\[ A = 1200 \times e^{(0.05 \times 6)} \][/tex]
4. Calculate the exponent:
[tex]\[ r \times t = 0.05 \times 6 = 0.30 \][/tex]
5. Find [tex]\( e^{0.30} \)[/tex]:
[tex]\[ e^{0.30} \approx 1.34986 \][/tex]
6. Calculate the final amount:
[tex]\[ A = 1200 \times 1.34986 \approx 1619.8305690912039 \][/tex]
7. Round the final amount to the nearest cent:
[tex]\[ A \approx 1619.83 \][/tex]
Therefore, after six years, the amount owed is approximately [tex]\( \mathbf{1619.83} \)[/tex].