Question
Spencer owns a small manufacturing business. In 5 years, he wants to buy a machine that costs $7,000.00. If Spencer opens a savings account that earns 5% interest compounded continuously, how much will he have to deposit as principal to have enough money in 5 years to buy the machine?
Round your answer to the nearest cent.
Round your answer to the nearest cent.
Answers
GPT-4o mini
To determine how much Spencer needs to deposit as principal to buy the machine in 5 years, we'll use the formula for continuous compounding interest:
\[
A = Pe^{rt}
\]
Where:
- \(A\) is the amount of money accumulated after time \(t\) (the future value),
- \(P\) is the principal amount (the initial amount of money),
- \(r\) is the annual interest rate (as a decimal),
- \(t\) is the time the money is invested or borrowed (in years),
- \(e\) is the base of the natural logarithm (approximately equal to 2.71828).
Here, we know:
- \(A = 7,000\) (the future value, or the cost of the machine),
- \(r = 0.05\) (the annual interest rate of 5%),
- \(t = 5\) (the time in years).
We need to solve for \(P\):
Rearranging the formula for \(P\):
\[
P = \frac{A}{e^{rt}}
\]
Plugging in the known values:
\[
P = \frac{7000}{e^{0.05 \times 5}}
\]
Calculating \(e^{0.05 \times 5}\):
\[
e^{0.25} \approx 1.284025
\]
Now substituting back into the equation for \(P\):
\[
P = \frac{7000}{1.284025}
\]
Calculating \(P\):
\[
P \approx \frac{7000}{1.284025} \approx 5454.95
\]
Thus, rounding to the nearest cent, Spencer needs to deposit approximately:
\[
\boxed{5454.95}
\]
So, Spencer should deposit $5,454.95 to have enough money to buy the machine in 5 years.
\[
A = Pe^{rt}
\]
Where:
- \(A\) is the amount of money accumulated after time \(t\) (the future value),
- \(P\) is the principal amount (the initial amount of money),
- \(r\) is the annual interest rate (as a decimal),
- \(t\) is the time the money is invested or borrowed (in years),
- \(e\) is the base of the natural logarithm (approximately equal to 2.71828).
Here, we know:
- \(A = 7,000\) (the future value, or the cost of the machine),
- \(r = 0.05\) (the annual interest rate of 5%),
- \(t = 5\) (the time in years).
We need to solve for \(P\):
Rearranging the formula for \(P\):
\[
P = \frac{A}{e^{rt}}
\]
Plugging in the known values:
\[
P = \frac{7000}{e^{0.05 \times 5}}
\]
Calculating \(e^{0.05 \times 5}\):
\[
e^{0.25} \approx 1.284025
\]
Now substituting back into the equation for \(P\):
\[
P = \frac{7000}{1.284025}
\]
Calculating \(P\):
\[
P \approx \frac{7000}{1.284025} \approx 5454.95
\]
Thus, rounding to the nearest cent, Spencer needs to deposit approximately:
\[
\boxed{5454.95}
\]
So, Spencer should deposit $5,454.95 to have enough money to buy the machine in 5 years.