How much should be invested now at an interest rate of 6.5% per year, compounded continuously, to have $1500 in four years? Do not round any intermediate computations, and round your answer to the nearest cent.

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**How much should be invested now at an interest rate of 6.5% per year, compounded continuously, to have $1500 in four years?**

**Do not round any intermediate computations, and round your answer to the nearest cent.**

[Input Box: $___] [Options: X, ⟳, ?]

Explanation: This question involves calculating the present value of an investment with continuous compounding interest. You will need to apply the formula for continuous compounding, which is:

\[ A = Pe^{rt} \]

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 for, in years.
- \( e \) is the base of the natural logarithm, which is approximately equal to 2.71828.

Given:
- \( A = $1500 \)
- \( r = 6.5\% = 0.065 \)
- \( t = 4 \) years

You need to solve for \( P \):

\[ 1500 = Pe^{0.065 \times 4} \]
\[ P = \frac{1500}{e^{0.26}} \]

Calculate the present value \( P \) without rounding any intermediate computations, and then round the final answer to the nearest cent.
Transcribed Image Text:**How much should be invested now at an interest rate of 6.5% per year, compounded continuously, to have $1500 in four years?** **Do not round any intermediate computations, and round your answer to the nearest cent.** [Input Box: $___] [Options: X, ⟳, ?] Explanation: This question involves calculating the present value of an investment with continuous compounding interest. You will need to apply the formula for continuous compounding, which is: \[ A = Pe^{rt} \] 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 for, in years. - \( e \) is the base of the natural logarithm, which is approximately equal to 2.71828. Given: - \( A = $1500 \) - \( r = 6.5\% = 0.065 \) - \( t = 4 \) years You need to solve for \( P \): \[ 1500 = Pe^{0.065 \times 4} \] \[ P = \frac{1500}{e^{0.26}} \] Calculate the present value \( P \) without rounding any intermediate computations, and then round the final answer to the nearest cent.
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