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The question in the image asks for the balance of an investment after 9 years, using compound interest. The investment amount is $15,000 with an annual interest rate of 7%, compounded annually.

The formula for compound interest is:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $A$ is the amount of money accumulated after $n$ years, including interest.
- $P$ is the principal amount (the initial investment), which is $15,000.
- $r$ is the annual interest rate (7% or 0.07).
- $t$ is the number of years (9 years in this case).
- $n$ is the number of times the interest is compounded per year (since it's compounded annually, $n = 1$).

Let’s calculate the balance:

\[
A = 15000 \times \left(1 + \frac{0.07}{1}\right)^{1 \times 9}
= 15000 \times (1.07)^9
\]

Now, calculating this:

\[
A = 15000 \times 1.838459
\]
\[
A \approx 27,576.89
\]

So, the account balance after 9 years will be approximately **$27,576.89**.

Would you like more details, or do you have any other questions? Here are some related questions to explore further:

1. What happens if the interest is compounded semi-annually instead of annually?
2. How much interest would be earned after 9 years?
3. What is the difference between simple and compound interest?
4. How does the rate of return change with different compounding intervals (monthly vs. yearly)?
5. How can one calculate the number of years needed to double an investment using compound interest?

**Tip:** Compounding more frequently (like quarterly or monthly) will yield a higher balance compared to compounding annually.

Suppose your friend's parents invest $15,000 in an account paying 7% compounded annually. What will the balance be after 9 years?

Learn how to calculate compound interest for a $15,000 investment over 9 years with a 7% annual interest rate. Understand the formula used and see the step-by-step calculation, resulting in a balance of approximately $27,576.89.

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