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If [tex]$\$[/tex]360[tex]$ is invested at an interest rate of $[/tex]4 \%[tex]$ per year and is compounded quarterly, how much will the investment be worth in 18 years?

Use the compound interest formula $[/tex]A=P\left(1+\frac{r}{n}\right)^{n t}[tex]$.

A. $[/tex]\[tex]$175.86$[/tex]
B. [tex]$\$[/tex]422.39[tex]$
C. $[/tex]\[tex]$430.61$[/tex]
D. [tex]$\$[/tex]736.96$


Sagot :

Certainly! Let's determine the future value of an investment of \[tex]$360 at an annual interest rate of 4% compounded quarterly over 18 years. The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate, - \( n \) is the number of times interest is compounded per year, - \( t \) is the number of years the money is invested or borrowed for, - \( A \) is the amount of money accumulated after n years, including interest. Given: - \( P = 360 \) (Principal), - \( r = 0.04 \) (4% annual interest rate), - \( n = 4 \) (Compounded quarterly), - \( t = 18 \) (Number of years). Now, substitute these values into the formula: \[ A = 360 \left(1 + \frac{0.04}{4}\right)^{4 \times 18} \] First, calculate the term inside the parentheses: \[ \frac{0.04}{4} = 0.01 \] So, \[ 1 + 0.01 = 1.01 \] Next, calculate the exponent: \[ 4 \times 18 = 72 \] Thus, the expression becomes: \[ A = 360 \left(1.01\right)^{72} \] Using the exponentiation: \[ 1.01^{72} \approx 2.046 \] So now multiply: \[ A = 360 \times 2.046 \] \[ A \approx 736.96 \] So, after 18 years, the investment will be worth approximately \$[/tex]736.96.

Therefore, the correct answer is \$736.96.