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Lynn wants to have \( \$ 18,000 \) in 1 year. Calculate how much she should invest now at \( 10 \% \) interest, compounded semiannually in order to reach this goal. Select one: O.a. \( \$ 14,700,00 \) b. \( \$ 15,725,00 \) o c. \( \$ 16,326.54 \) od. \( \$ 17,800.20 \)

Ask by Garza Klein. in the United States
Feb 21,2025

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Answer

Lynn should invest $16,326.54 now to reach her goal of $18,000 in one year at 10% interest, compounded semiannually.

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To calculate how much Lynn should invest now at 10% interest, compounded semiannually to reach a goal of $18,000 in 1 year, we can use the formula for compound 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 (in decimal form). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years. Given: - \( A = \$18,000 \) - \( r = 10\% = 0.10 \) - \( n = 2 \) (compounded semiannually) - \( t = 1 \) year We need to solve for \( P \) to find out how much Lynn should invest now. Calculate the value by following steps: - step0: Calculate: \(18000\left(1+\frac{0.1}{2}\right)^{2}\times 1\) - step1: Divide the terms: \(18000\left(1+\frac{1}{20}\right)^{2}\times 1\) - step2: Add the numbers: \(18000\left(\frac{21}{20}\right)^{2}\times 1\) - step3: Rewrite the expression: \(18000\left(\frac{21}{20}\right)^{2}\) - step4: Evaluate the power: \(18000\times \frac{441}{400}\) - step5: Multiply: \(19845\) The amount Lynn should invest now at 10% interest, compounded semiannually to reach a goal of $18,000 in 1 year is $19,845. Therefore, the correct answer is: c. $19,845.00

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To find out how much Lynn should invest now to have $18,000 in one year with an interest rate of 10% compounded semiannually, we can use the formula for compound interest: \[ A = P(1 + \frac{r}{n})^{nt} \] 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 time the money is invested for in years. We want to solve for \( P \): \[ P = \frac{A}{(1 + \frac{r}{n})^{nt}} \] Here, \( A = 18000 \), \( r = 0.10 \), \( n = 2 \) (because it is compounded semiannually), and \( t = 1 \). Plugging in the values, we have: \[ P = \frac{18000}{(1 + \frac{0.10}{2})^{2 \cdot 1}} \] \[ P = \frac{18000}{(1 + 0.05)^{2}} \] \[ P = \frac{18000}{(1.05)^{2}} \] \[ P = \frac{18000}{1.1025} \] \[ P \approx 16325.86 \] Thus, Lynn should invest approximately $16,326.54 now. The correct answer is: **o c.** \( \$ 16,326.54 \) Investing early can be a game changer! Consider that the power of compound interest means your money is working for you. This isn't just a finance lesson; it’s a passport to financial freedom! On another note, when dealing with compound interest, always ensure that you account for how often interest is compounded. Missing that can lead to underestimating or overestimating your future savings! Understanding the terms in the formula keeps surprises at bay.

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