a1.1.4 How much will an ihifialinvestment Of 5,000 be worth in 35 years if it earns interest compounded continuousiY at an annualrate Of 8.5%.? O
a1.1.4 How much will an ihifialinvestment Of 5,000 be worth in 35 years if it earns interest compounded continuousiY at an annualrate Of 8.5%.? O
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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![**Problem 21.1.4: Continuous Compounding Investment**
How much will an initial investment of $5,000 be worth in 35 years if it earns interest compounded continuously at an annual rate of 8.5%?
*Concept Explanation:*
In financial mathematics, continuous compounding uses the formula:
\[ A = Pe^{rt} \]
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (initial investment).
- \( r \) is the annual interest rate (decimal).
- \( t \) is the time the money is invested for (years).
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2ac1715-6ccd-4f91-87eb-b09a0d0b46c6%2F3d8f4aa2-4a02-43a2-8212-aaf55ca0d02e%2Fzp6kynh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 21.1.4: Continuous Compounding Investment**
How much will an initial investment of $5,000 be worth in 35 years if it earns interest compounded continuously at an annual rate of 8.5%?
*Concept Explanation:*
In financial mathematics, continuous compounding uses the formula:
\[ A = Pe^{rt} \]
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (initial investment).
- \( r \) is the annual interest rate (decimal).
- \( t \) is the time the money is invested for (years).
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
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