the equation for the future value of a deposit earning compound interest is $v(t)=p(1 + \frac{r}{n})^{nt}$…

the equation for the future value of a deposit earning compound interest is $v(t)=p(1 + \frac{r}{n})^{nt}$, where: $p =$ the initial deposit $t =$ years invested $r =$ rate at which interest is compounded annually $n =$ number of times the interest is compounded per year after 10 years, a $2,000 - dollar investment compounded annually has grown to $3600. what is the interest rate to the nearest whole - number percent?

the equation for the future value of a deposit earning compound interest is $v(t)=p(1 + \frac{r}{n})^{nt}$, where: $p =$ the initial deposit $t =$ years invested $r =$ rate at which interest is compounded annually $n =$ number of times the interest is compounded per year after 10 years, a $2,000 - dollar investment compounded annually has grown to $3600. what is the interest rate to the nearest whole - number percent?

Answer

Explanation:

Step1: Identify given values

$P = 2000$, $V(t)=3600$, $t = 10$, $n = 1$ (compounded annually)

Step2: Substitute into formula

$3600=2000(1 + r)^{10}$

Step3: Isolate the power - term

$\frac{3600}{2000}=(1 + r)^{10}$ $1.8=(1 + r)^{10}$

Step4: Take the 10th - root of both sides

$1.8^{\frac{1}{10}}=1 + r$

Step5: Calculate 1.8^(1/10)

$1.8^{\frac{1}{10}}\approx1.06054$

Step6: Solve for r

$r=1.06054 - 1$ $r = 0.06054$

Step7: Convert to percentage

$r\approx6%$

Answer:

$6$