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Question 1 (3 points) For graduation, Emma receives \( \$ 875 \) from the money in the cardes she received from family and friends. If she invests it in an account that earns \( 6 \% \) interest compounded continuously, how much will this money be worth in 3 years? 2.) Write the equation (how you will set it up through substitution): ***Be sure to use parenthesis to show what all is included in the exponent. 3.) After 3 years, Emma will have will be \( \$ \) in her account. (round to the nearest whole dollar)

Ask by Harmon Reyes. in the United States
Feb 03,2025

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After 3 years, Emma will have $1048 in her account.

Solución

To solve this problem, we will use the formula for continuous compounding of interest: \[ A = P \times e^{r \times t} \] where: - \( A \) is the amount of money after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (in decimal form), - \( t \) is the time in years. Given: - \( P = \$875 \) (initial investment), - \( r = 6\% = 0.06 \) (annual interest rate), - \( t = 3 \) years. Substitute these values into the formula: \[ A = 875 \times e^{0.06 \times 3} \] Now, we will calculate the value of \( A \) after 3 years. Calculate the value by following steps: - step0: Calculate: \(875e^{0.06\times 3}\) - step1: Multiply the numbers: \(875e^{0.18}\) - step2: Convert the expressions: \(875e^{\frac{9}{50}}\) - step3: Rewrite the expression: \(875\sqrt[50]{e^{9}}\) The amount of money Emma will have in her account after 3 years is approximately $1047.57. Therefore, after 3 years, Emma will have $1047 in her account.

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Beyond the Answer

To find out how much Emma's investment will be worth in 3 years with continuous compounding, we can use the formula for continuous compounding: \[ A = Pe^{rt} \] 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 annual interest rate (decimal). - \( t \) is the time the money is invested or borrowed for, in years. - \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Using the given values: 1. **Substituting the known values:** - \( P = 875 \) - \( r = 0.06 \) (which is \( 6\% \) as a decimal) - \( t = 3 \) The setup through substitution will look like this: \[ A = 875 \cdot e^{(0.06 \cdot 3)} \] 2. **Calculate the final amount:** - First calculate the exponent: \( 0.06 \cdot 3 = 0.18 \). - Now substitute it back: \[ A = 875 \cdot e^{0.18} \] Using \( e^{0.18} \approx 1.1972 \): \[ A \approx 875 \cdot 1.1972 \approx 1041.05 \] So, after rounding to the nearest whole dollar, Emma will have approximately \( \$ 1,041 \) in her account after 3 years.

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