To solve for the annual interest rate, we will use the compound interest formula:

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

Where:

• $A$ is the final amount ($43,200),
• $P$ is the initial principal ($3,800),
• $r$ is the annual interest rate (which we are solving for),
• $n$ is the number of times interest is compounded per year (monthly compounding means $n = 12$),
• $t$ is the time in years (11 years).

Rearranging the formula to solve for $r$:

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

Substituting the known values:

$\left(1 + \frac{r}{12}\right)^{12 \times 11} = \frac{43,200}{3,800}$

$\left(1 + \frac{r}{12}\right)^{132} = 11.3684$

Taking the 132nd root of both sides:

$1 + \frac{r}{12} = 11.3684^{\frac{1}{132}}$

Now, solving for $r$.

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In the meantime, feel free to upload another question or clarify anything about this problem. Here are five related questions you might find useful:

1. How is compound interest different from simple interest?
2. What effect does increasing the number of compounding periods have on the final amount?
3. How can we calculate continuous compounding interest?
4. How does the interest rate affect the time it takes to double an investment?
5. What other factors can affect the growth of an investment?

Tip: When working with compound interest, remember that the more frequently interest is compounded, the greater the final amount will be due to interest on the interest earned.