Samantha deposits $1,500 into the Park Street Bank. The account pays 1.12% annual interest, compounded daily. To the nearest cent, how much is in the account at the end of three non-leap years?

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

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

Where:
- $ A $ is the amount of money accumulated after n years, including interest.
- $ P $ is the principal amount (initial investment).
- $ r $ is the annual interest rate (decimal).
- $ n $ is the number of times interest is compounded per year.
- $ t $ is the time the money is invested for in years.

Given values:
- $ P = 1500 $
- $ r = 1.12\% = 0.0112 $ (converted to decimal)
- $ n = 365 $ (since it's compounded daily)
- $ t = 3 $ years

Let's substitute these values into the formula:

\[
A = 1500 \left(1 + \frac{0.0112}{365}\right)^{365 \times 3}
\]

Now we will calculate this.The amount in the account at the end of three non-leap years, compounded daily, is approximately **$1,551.26**.

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How would the result change if interest were compounded monthly instead of daily?
2. What would be the total interest earned over the three years?
3. How much would the amount be after 5 years instead of 3 years?
4. How does the compound interest formula differ from simple interest?
5. What if the annual interest rate was 2% instead of 1.12%?

**Tip:** The more frequently interest is compounded, the greater the final amount will be.

Learn how to calculate the compound interest for a deposit of $1,500 at an annual interest rate of 1.12%, compounded daily. This problem involves using the compound interest formula and is suitable for students in Grades 9-12. Find out the final amount in the account after three years.

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