claire deposited $2,500 into an account that accrues interest monthly. she made no additional deposits or…

claire deposited $2,500 into an account that accrues interest monthly. she made no additional deposits or withdrawals. after 2 years, claire had $2,762.35 in the account. what is the annual interest rate of the account? compound interest formula: $v(t)=p(1 + \frac{r}{n})^{nt}$\n$t$ = years since initial deposit\n$n$ = number of times compounded per year\n$r$ = annual interest rate (as a decimal)\n$p$ = initial (principal) investment\n$v(t)$ = value of investment after $t$ years\n11%\n31%\n26%\n5%

claire deposited $2,500 into an account that accrues interest monthly. she made no additional deposits or withdrawals. after 2 years, claire had $2,762.35 in the account. what is the annual interest rate of the account? compound interest formula: $v(t)=p(1 + \frac{r}{n})^{nt}$\n$t$ = years since initial deposit\n$n$ = number of times compounded per year\n$r$ = annual interest rate (as a decimal)\n$p$ = initial (principal) investment\n$v(t)$ = value of investment after $t$ years\n11%\n31%\n26%\n5%

Answer

Explanation:

Step1: Identify the values

$P = 2500$, $V(t)=2762.35$, $t = 2$, $n=12$ (monthly compounding)

Step2: Substitute into the formula

$2762.35=2500(1 + \frac{r}{12})^{12\times2}$

Step3: Isolate the exponential term

$\frac{2762.35}{2500}=(1 + \frac{r}{12})^{24}$ $1.09694=(1 + \frac{r}{12})^{24}$

Step4: Take the 24 - th root of both sides

$1.09694^{\frac{1}{24}}=1+\frac{r}{12}$ $1.00375 = 1+\frac{r}{12}$

Step5: Solve for $r$

$1.00375-1=\frac{r}{12}$ $0.00375=\frac{r}{12}$ $r=0.00375\times12$ $r = 0.045$ or $4.5%$ (closest option is $5%$)

Answer:

$5%$