We can calculate the balance using the formula for compound interest:

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

Where:

• $A$ is the balance (final amount).
• $P$ is the principal (starting amount).
• $r$ is the annual interest rate (as a decimal).
• $n$ is the number of times per year that the interest is compounded.
• $t$ is the time in years.

From the problem:

• $P = 3000$
• $r = 15\% = 0.15$
• $n = 12$ (since the interest is compounded monthly).
• $t = 9$ years.

Now, substituting the values into the formula:

$A = 3000 \left(1 + \frac{0.15}{12}\right)^{12 \times 9}$

Let's solve for $A$:

$A = 3000 \left(1 + 0.0125\right)^{108}$ $A = 3000 \left(1.0125\right)^{108}$ $A = 3000 \times 3.474004$ $A \approx 10422.01$

So, the balance after 9 years is approximately $10,422.01.

Would you like further clarification on any steps?

Here are 5 related questions:

1. How would the balance change if the interest rate were 10% instead of 15%?
2. How does the number of compounding periods affect the final balance?
3. What is the impact of increasing the time (t) on compound interest?
4. If the interest were compounded quarterly instead of monthly, how would that change the result?
5. What would the balance be after 5 years under the same conditions?

Tip: The more frequently interest is compounded (the larger $n$ is), the greater the final balance will be.