To determine the total interest Mary would pay in each scenario, we can calculate as follows:
(a) Total interest with 5.5% simple interest for 3 years
The formula for simple interest is: \[ I = P \times r \times t \] where:
- \( I \) is the interest,
- \( P \) is the principal amount (the initial amount of money),
- \( r \) is the rate of interest (in decimal),
- \( t \) is the time in years.
For Mary:
- \( P = 1000 \)
- \( r = 5.5% = 0.055 \)
- \( t = 3 \) years
Plugging in the values: \[ I = 1000 \times 0.055 \times 3 = 1000 \times 0.165 = 165 \]
So, the total interest paid at 5.5% simple interest would be $165.
(b) Total interest with 3% interest compounded continuously for 3 years
The formula for continuous compounding is: \[ A = P \times e^{rt} \] 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 rate of interest (in decimal),
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm (approximately 2.71828).
First, let’s find \( A \):
- \( P = 1000 \)
- \( r = 3% = 0.03 \)
- \( t = 3 \) years
Calculating \( A \): \[ A = 1000 \times e^{0.03 \times 3} \] \[ A = 1000 \times e^{0.09} \approx 1000 \times 1.09417 \approx 1094.17 \]
Now, to find the total interest \( I \), we subtract the principal from the amount accumulated: \[ I = A - P = 1094.17 - 1000 \approx 94.17 \]
So, the total interest paid at 3% compounded continuously would be approximately $94.17.
(c) Comparison of total interests
- Total interest with 5.5% simple interest: $165
- Total interest with 3% compounded continuously: $94.17
Thus, the option with 3% compounded continuously results in less total interest, specifically $94.17, compared to $165 for the simple interest option.
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