Find the present value​ (the amount that should be invested now to accumulate the following​ amount) if the money is compounded as indicated.

​$10 comma 707.7510,707.75

at

8.98.9​%

compounded

annuallyannually

for

44

yearsyears

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Part 1

The present value is

​$enter your response here.

​(Do not round until the final answer. Then round to the nearest cent as​ needed.)

To find the present value (PV), we use the present value formula for compound interest:

\[
PV = \dfrac{A}{(1 + r)^t}
\]

Where:
- $ A $ is the accumulated amount ($10,707.75),
- $ r $ is the annual interest rate (8.9% or 0.089),
- $ t $ is the number of years (44 years).

Let's calculate:

\[
PV = \dfrac{10,707.75}{(1 + 0.089)^{44}}
= \dfrac{10,707.75}{(1.089)^{44}}
\]

First, calculate $ (1.089)^{44} $:

\[
1.089^{44} \approx 34.278
\]

Now, divide the accumulated amount by this value:

\[
PV = \dfrac{10,707.75}{34.278} \approx 312.38
\]

Thus, the present value is approximately **$312.38**.

Would you like further clarification or have any questions?

Here are 5 related questions:
1. How does changing the interest rate affect the present value?
2. What happens if the compounding is done quarterly instead of annually?
3. How would the present value change if the investment period was shorter?
4. What is the future value if the initial investment is known?
5. How does continuous compounding affect the present value?

**Tip:** Always ensure to convert percentage rates into decimal form when using them in formulas.

Find the present value (the amount that should be invested now to accumulate the following amount) if the money is compounded as indicated. $10,707.75 at 8.9% compounded annually for 44 years. The present value is $_____. (Do not round until the final answer. Then round to the nearest cent as needed.)

Learn how to calculate the present value of $10,707.75 with an annual interest rate of 8.9% compounded annually over 44 years. This guide walks through the compound interest formula step by step, providing clear explanations for students in Grades 10-12.

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