Pregunta
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Aubrey invested \( \$ 150 \) in an account paying an interest rate of \( 6 \frac{5}{8} \% \) compounded monthly. Miguel invested \( \$ 150 \) in an account paying an interest rate of \( 6 \frac{1}{4} \% \) compounded daily. After 7 years, how much more money would Aubrey have in her account than Miguel, to the nearest dollar?

Ask by Cox Moran. in the United States
Jan 23,2025

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Aubrey would have approximately \$6 more than Miguel after 7 years.

Solución

To determine how much more money Aubrey would have in her account compared to Miguel after 7 years, we'll calculate the future value of both investments using the compound interest formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) = the amount of money accumulated after \( t \) years, including interest. - \( P \) = the principal amount (\$150 for both investors). - \( r \) = annual interest rate (in decimal). - \( n \) = number of times that interest is compounded per year. - \( t \) = time the money is invested for in years. ### Aubrey's Investment - **Principal (\( P \))**: \$150 - **Annual Interest Rate (\( r \))**: \( 6 \frac{5}{8}\% = 6.625\% = 0.06625 \) - **Compounded Monthly (\( n \))**: 12 times a year - **Time (\( t \))**: 7 years Plugging these values into the formula: \[ A_{\text{Aubrey}} = 150 \left(1 + \frac{0.06625}{12}\right)^{12 \times 7} \] \[ A_{\text{Aubrey}} \approx 150 \times (1.0055208333)^{84} \] \[ A_{\text{Aubrey}} \approx 150 \times 1.588 \] \[ A_{\text{Aubrey}} \approx \$238.20 \] ### Miguel's Investment - **Principal (\( P \))**: \$150 - **Annual Interest Rate (\( r \))**: \( 6 \frac{1}{4}\% = 6.25\% = 0.0625 \) - **Compounded Daily (\( n \))**: 365 times a year - **Time (\( t \))**: 7 years Plugging these values into the formula: \[ A_{\text{Miguel}} = 150 \left(1 + \frac{0.0625}{365}\right)^{365 \times 7} \] \[ A_{\text{Miguel}} \approx 150 \times (1.0001712329)^{2555} \] \[ A_{\text{Miguel}} \approx 150 \times 1.548 \] \[ A_{\text{Miguel}} \approx \$232.26 \] ### Difference in Amounts \[ \text{Difference} = A_{\text{Aubrey}} - A_{\text{Miguel}} \] \[ \text{Difference} \approx 238.20 - 232.26 \] \[ \text{Difference} \approx \$5.94 \] Rounding to the nearest dollar: **Aubrey would have approximately \$6 more than Miguel after 7 years.** **Answer:** $6

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To find out how much more money Aubrey has than Miguel after 7 years, we'll calculate the future value of both investments. **Aubrey's Investment:** Aubrey's interest rate is \( 6 \frac{5}{8}\% \), which is \( 6.625\% \) or \( 0.06625\) as a decimal. Since it's compounded monthly, we use the formula for compounded 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 (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 number of years the money is invested. For Aubrey: - \( P = 150 \) - \( r = 0.06625 \) - \( n = 12 \) (monthly) - \( t = 7 \) \[ A = 150 \left(1 + \frac{0.06625}{12}\right)^{12 \times 7} \] Calculating that, we find: \[ A = 150 \left(1 + 0.00552083333\right)^{84} \] \[ A = 150 \left(1.00552083333\right)^{84} \approx 150 \times 1.747422 \approx 262.11 \] **Miguel's Investment:** Miguel's interest rate is \( 6 \frac{1}{4}\% \), which is \( 6.25\% \) or \( 0.0625 \) as a decimal. Compounded daily means \( n = 365 \): \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] For Miguel: - \( P = 150 \) - \( r = 0.0625 \) - \( n = 365 \) - \( t = 7 \) \[ A = 150 \left(1 + \frac{0.0625}{365}\right)^{365 \times 7} \] Calculating that, we have: \[ A = 150 \left(1 + 0.0001712328767\right)^{2555} \] \[ A = 150 \left(1.0001712328767\right)^{2555} \approx 150 \times 1.494200 \approx 224.13 \] **Difference**: Finally, we need to find how much more Aubrey has than Miguel: \[ \text{Difference} = 262.11 - 224.13 = 37.98 \] Rounding to the nearest dollar, Aubrey has **$38** more than Miguel.

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