This is a compound interest problem. To calculate the initial deposit (also known as the present value) and the total interest earned, we use the compound interest formula:

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

Where:

• $A$ is the future value ($5,500.00),
• $P$ is the present value (the amount to deposit now),
• $r$ is the annual interest rate as a decimal ($4 \frac{3}{4}\% = 0.0475$),
• $n$ is the number of compounding periods per year ($n = 4$ for quarterly compounding),
• $t$ is the time in years ($t = 13$).

Step 1: Solve for $P$ (Present Value)

Rearrange the compound interest formula to solve for $P$:

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

Substitute the known values: $P = \frac{5500}{\left(1 + \frac{0.0475}{4}\right)^{4 \cdot 13}}$

$P = \frac{5500}{\left(1 + 0.011875\right)^{52}}$

$P = \frac{5500}{\left(1.011875\right)^{52}}$

Calculate the exponentiation and division:

$P = \frac{5500}{1.82482232055} \approx 3015.97$

Step 2: Calculate Total Interest Earned

The total interest earned is the difference between the future value ($A$) and the initial deposit ($P$):

$\text{Interest} = A - P$

$\text{Interest} = 5500 - 3015.97 = 2484.03$

Final Results:

• Initial deposit required (P): $3,015.97
• Total interest earned: $2,484.03

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Questions to Expand Understanding:

1. What happens if the interest rate changes to 5%? How would it affect the deposit and interest?
2. How does the number of compounding periods (e.g., monthly instead of quarterly) affect the outcome?
3. What if you wanted the same future amount ($5,500) in 10 years instead of 13? How much would you need to deposit?
4. How does simple interest differ from compound interest in this scenario?
5. Can you explain the impact of the interest rate being compounded quarterly rather than annually?

Tip: Compounding more frequently results in slightly higher final values for the same interest rate and time frame.