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

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

Where:

• $A$ is the amount in the account after interest.
• $P$ is the principal (the initial deposit), which is $2500.
• $r$ is the annual interest rate, expressed as a decimal (6.5% = 0.065).
• $n$ is the number of times the interest is compounded per year (for this problem, it's compounded annually, so $n = 1$).
• $t$ is the number of years, which is 2.

Plugging the values into the formula:

$A = 2500 \left(1 + \frac{0.065}{1}\right)^{1 \times 2}$ $A = 2500 \left(1 + 0.065\right)^2$ $A = 2500 \times (1.065)^2$ Now, let's calculate the final result.The balance in Mr. Wilkins' account after 2 years, with 6.5% interest compounded annually, is approximately $2,835.56. Therefore, the correct answer is:

Option A: $2,835.56

Would you like further details or explanations? Here are five related questions to deepen your understanding:

1. How would the balance change if the interest was compounded monthly instead of annually?
2. What would be the balance after 5 years with the same interest rate?
3. How does the compound interest formula differ from simple interest?
4. What would be the interest earned in the first year alone?
5. How would the result change if Mr. Wilkins deposited $3000 instead?

Tip: In compound interest problems, the more frequent the compounding periods, the higher the final amount due to interest being calculated on previously earned interest.