Math Problem Statement
π΄=π(1+ππ)ππ‘ and π΄=ππππ‘ to solve the exercises below. Round your answer to the nearest cent. Find the accumulated value of an investment of $14490 for 7 years at an annual interest rate of 7.1% if the money is (a) compounded semiannually: functionsequation editor dollars (b) compounded quarterly: functionsequation editor dollars (c) compounded monthly: functionsequation editor dollars (d) compounded continuously: functionsequation editor dollars
Solution
To solve these problems, weβll use both compound interest formulas:
-
For compounding a finite number of times per year:
where:- is the accumulated amount,
- is the principal (initial amount),
- is the annual interest rate,
- is the number of compounding periods per year,
- is the time in years.
-
For continuous compounding:
Given:
- years
Let's calculate each case.
(a) Compounded Semiannually ( )
(b) Compounded Quarterly ( )
(c) Compounded Monthly ( )
(d) Compounded Continuously
Summary of Results
- (a) Compounded Semiannually: $23,558.23
- (b) Compounded Quarterly: $23,617.59
- (c) Compounded Monthly: $23,657.46
- (d) Compounded Continuously: $23,800.56
Would you like further details on any specific part?
Related Questions
- How does changing the compounding frequency impact the accumulated amount?
- What would be the accumulated value if the interest rate were 8% instead of 7.1%?
- How much more interest is earned with continuous compounding versus semiannual compounding?
- How would the accumulated amount change for an investment period of 10 years instead of 7?
- What would be the difference in the accumulated amount if the initial investment was $20,000 instead?
Tip:
The more frequently interest is compounded, the more interest is accumulated, but the differences diminish with very frequent compounding (like monthly to continuous).
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Math Problem Analysis
Mathematical Concepts
Compound Interest
Continuous Compounding
Exponential Growth
Formulas
A = P(1 + r/n)^(nt)
A = Pe^(rt)
Theorems
Exponential Growth in Compound Interest
Suitable Grade Level
Grades 10-12
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