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A customer deposits [tex]$500 in an account that pays 4% annual interest. What is the balance after 3 years if the interest is compounded annually?

Compound interest formula: \( V(t) = P \left(1 + \frac{r}{n}\right)^{nt} \)

Where:
- \( t \) = years since initial deposit
- \( n \) = number of times compounded per year
- \( r \) = annual interest rate (as a decimal)
- \( P \) = initial (principal) investment
- \( V(t) \) = value of investment after \( t \) years

A. $[/tex]500.12
B. [tex]$512.00
C. $[/tex]560.00
D. $562.43


Sagot :

Certainly! Let's go through the compound interest formula step-by-step to determine the balance after 3 years for a deposit of [tex]$500 at an annual interest rate of 4%, compounded annually. The formula for compound interest is: \[ V(t) = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( P \) is the initial (principal) investment. - \( r \) is the annual interest rate (as a decimal). - \( n \) is the number of times the interest is compounded per year. - \( t \) is the number of years. - \( V(t) \) is the value of the investment after \( t \) years. Given the parameters: - \( P = 500 \) (the initial deposit), - \( r = 0.04 \) (4% annual interest rate), - \( n = 1 \) (compounded annually), - \( t = 3 \) (3 years), We can plug these values into the formula: \[ V(3) = 500 \left(1 + \frac{0.04}{1}\right)^{1 \cdot 3} \] First, simplify inside the parenthesis: \[ 1 + \frac{0.04}{1} = 1.04 \] Now raise \( 1.04 \) to the power of \( 3 \): \[ 1.04^3 \] Calculate \( 1.04^3 \): \[ 1.04^3 = 1.124864 \] Now multiply this result by the initial principal \( P \): \[ 500 \times 1.124864 \] This gives: \[ 562.432 \] Rounding to two decimal places (as is standard for currency), we get: \[ 562.43 \] Therefore, the balance after 3 years, when the interest is compounded annually, is: \[ \boxed{562.43} \] So, based on the given options, the correct answer is: \[ \$[/tex] 562.43 \]
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