To solve this problem, we can use the compound interest formula:

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

Where:

• $A$ is the amount of money in the account after interest.
• $P$ is the principal (initial deposit).
• $r$ is the annual interest rate (in decimal form).
• $n$ is the number of times the interest is compounded per year.
• $t$ is the number of years the money is invested or borrowed for.

Given:

• $P = 11,000$
• $r = 2\% = 0.02$
• $n = 2$ (since the interest is compounded semiannually, i.e., twice a year)
• $t = 3$ years

Step 1: Calculate the amount after 3 years

Substitute these values into the compound interest formula:

$A = 11,000 \left(1 + \frac{0.02}{2}\right)^{2 \times 3}$ $A = 11,000 \left(1 + 0.01\right)^6$ $A = 11,000 \left(1.01\right)^6$

Now, calculate $(1.01)^6$:

$1.01^6 \approx 1.061520$

Now, multiply by the principal:

$A = 11,000 \times 1.061520$ $A \approx 11,676.72$

A. The amount of money in the account after 3 years is $11,676.72.

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Step 2: Find the interest earned

The interest earned is the total amount $A$ minus the initial principal $P$:

$\text{Interest} = A - P$ $\text{Interest} = 11,676.72 - 11,000$ $\text{Interest} = 676.72$

B. The interest earned is $676.72.

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Would you like any further clarification on how I arrived at the solution or any other specific details?

Here are 5 related questions:

1. How do you calculate compound interest when the interest is compounded annually, quarterly, or continuously?
2. What is the difference between compound interest and simple interest?
3. How does the frequency of compounding affect the final amount of money in the account?
4. What happens to the interest earned if the rate is increased from 2% to 5%?
5. How would the formula change if the interest were compounded continuously?

Tip: When solving compound interest problems, always make sure the rate and time are in the correct units corresponding to the frequency of compounding.