To find the future value of the investments using the formula \( A = P(1 + \frac{r}{n})^{nt} \), we'll apply it to both cases, 4 and 5.
Exercise 4:
Investing $2000 at an annual interest rate of 6% compounded quarterly for 5 years.
- P (initial investment) = $2000
- r (annual interest rate) = 6% = 0.06
- n (number of times compounded per year) = 4 (quarterly)
- t (number of years) = 5
Using the formula: \[ A = 2000 \left(1 + \frac{0.06}{4}\right)^{4 \times 5} \]
Now, calculate step by step:
- Calculate \( \frac{0.06}{4} = 0.015 \)
- Calculate \( 4 \times 5 = 20 \)
- Calculate \( 1 + 0.015 = 1.015 \)
- Raise to the power of 20: \[ 1.015^{20} \approx 1.346855007 \]
- Now, multiply by $2000: \[ A \approx 2000 \times 1.346855007 \approx 2693.71 \]
So, the investment will be worth approximately $2693.71 after 5 years.
Exercise 5:
Investing $3000 at an annual interest rate of 5% compounded monthly for 8 years.
- P (initial investment) = $3000
- r (annual interest rate) = 5% = 0.05
- n (number of times compounded per year) = 12 (monthly)
- t (number of years) = 8
Using the formula: \[ A = 3000 \left(1 + \frac{0.05}{12}\right)^{12 \times 8} \]
Now, calculate step by step:
- Calculate \( \frac{0.05}{12} \approx 0.00416667 \)
- Calculate \( 12 \times 8 = 96 \)
- Calculate \( 1 + 0.00416667 \approx 1.00416667 \)
- Raise to the power of 96: \[ 1.00416667^{96} \approx 1.432364654 \]
- Now, multiply by $3000: \[ A \approx 3000 \times 1.432364654 \approx 4297.09 \]
So, the investment will be worth approximately $4297.09 after 8 years.
Summary:
- After 5 years, the investment in exercise 4 will be worth approximately $2693.71.
- After 8 years, the investment in exercise 5 will be worth approximately $4297.09.
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