Question

logan opened a savings account and deposited $700.00 as principal. the account earns 11% interest, compounded quarterly. what is the balance after 3 years?use the formula $a = p\left(1 + \frac{r}{n}\
ight)^{nt}$, where $a$ is the balance (final amount), $p$ is the principal (starting amount), $r$ is the interest rate expressed as a decimal, $n$ is the number of times per year that the interest is compounded, and $t$ is the time in years. round your answer to the nearest cent. $

Explanation:

Step1: Identify given values

$P = 700$, $r = 0.11$, $n = 4$, $t = 3$

Step2: Calculate $\frac{r}{n}$

$\frac{0.11}{4} = 0.0275$

Step3: Calculate $nt$

$4 \times 3 = 12$

Step4: Calculate $1+\frac{r}{n}$

$1 + 0.0275 = 1.0275$

Step5: Calculate $(1+\frac{r}{n})^{nt}$

$1.0275^{12} \approx 1.38709$

Step6: Calculate final amount $A$

$A = 700 \times 1.38709$

Answer:

$\$970.96$