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Connor invests a sum of money at an inte...

Connor invests a sum of money at an interest rate of 3% per year compounded annually. The value of the investment, y, after n years can be calculated using the equation `y=A(x)^(n)`. If Connor invests $250 and wants to calculate the value of his investment after 10 years, assuming no further deposits or withdrawals, which of the following equations should be use?

A

`y=250(0.97)^(10)`

B

`y=(250xx0.03)^(10)`

C

`y=250(0.03)^(10)`

D

`y=250(1.03)^(10)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of calculating the future value of Connor's investment after 10 years with an interest rate of 3% compounded annually, we can follow these steps: ### Step 1: Identify the variables in the compound interest formula The formula for compound interest is given by: \[ y = P \times (1 + r)^n \] where: - \( y \) is the final amount of the investment, - \( P \) is the principal amount (the initial investment), - \( r \) is the annual interest rate (as a decimal), - \( n \) is the number of years the money is invested. ### Step 2: Substitute the known values into the formula In this case: - \( P = 250 \) (the initial investment), - \( r = 3\% = \frac{3}{100} = 0.03 \) (the interest rate as a decimal), - \( n = 10 \) (the number of years). Now substituting these values into the formula: \[ y = 250 \times (1 + 0.03)^{10} \] ### Step 3: Simplify the equation Now, simplify the expression inside the parentheses: \[ y = 250 \times (1.03)^{10} \] ### Step 4: Calculate \( (1.03)^{10} \) Using a calculator or exponentiation: \[ (1.03)^{10} \approx 1.3439 \] Now substitute this back into the equation: \[ y \approx 250 \times 1.3439 \] ### Step 5: Calculate the final amount Now perform the multiplication: \[ y \approx 250 \times 1.3439 \approx 335.975 \] ### Step 6: Round to the nearest cent Since we are dealing with currency, we round to two decimal places: \[ y \approx 335.98 \] ### Final Answer Thus, the value of Connor's investment after 10 years will be approximately **$335.98**.
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