To calculate the ending balance for a deposit compounded daily, we use the formula for compound interest:

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

Where:

• $A$ is the ending balance (the amount you're solving for),
• $P$ is the initial deposit (principal) = 10,000,
• $r$ is the annual interest rate (as a decimal) = 1.85% = 0.0185,
• $n$ is the number of times the interest is compounded per year (daily compounding means $n = 365$),
• $t$ is the number of years = 3.

Let's calculate:

$A = 10,000 \left(1 + \frac{0.0185}{365}\right)^{365 \times 3}$

I'll calculate the exact value for you.The ending balance after 3 years, with a $10,000 deposit at 1.85% interest compounded daily, would be approximately $10,570.68.

Would you like more details or have any questions? Here are 5 related questions for deeper exploration:

1. How would the balance change with different compounding frequencies (e.g., monthly, quarterly)?
2. What would the ending balance be with a different interest rate, say 2.5%?
3. How does daily compounding compare to annual compounding in terms of growth?
4. What would the total interest earned over the 3 years be?
5. How would the outcome differ if the investment period were 5 years instead of 3?

Tip: The more frequently interest is compounded, the greater the final amount due to more opportunities for the interest to grow.