Questions: Find the interest earned. Assume 3 frac12 % interest compounded daily. Assume a non-leap year. Amount Date Deposited Date Withdrawn Interest Earned ------------ 6500 February 12 April 21

Solution Steps

To find the interest earned, we need to use the formula for compound interest. The formula for compound interest is:

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

where:

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

In this case:

• \( P = 6500 \)
• \( r = 3.5\% = 0.035 \)
• \( n = 365 \) (compounded daily)
• \( t \) is the number of days between February 12 and April 21, divided by 365.

First, we need to calculate the number of days between February 12 and April 21. Then, we can use the compound interest formula to find the interest earned.

The number of days between the deposit date (February 12, 2023) and the withdrawal date (April 21, 2023) is calculated as follows: \[ \text{days\_between} = 68 \]

To find the time in years, we divide the number of days by the number of days in a year: \[ t = \frac{68}{365} \approx 0.1863 \]

Using the compound interest formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where:

• \( P = 6500 \)
• \( r = 0.035 \)
• \( n = 365 \)
• \( t \approx 0.1863 \)

We calculate the total amount: \[ A \approx 6500 \left(1 + \frac{0.035}{365}\right)^{365 \times 0.1863} \approx 6542.52 \]

The interest earned is the total amount minus the principal: \[ \text{interest\_earned} = A - P \approx 6542.52 - 6500 \approx 42.52 \]

Final Answer