robert invested $71,000 in an account paying an interest rate of 6.6% compounded quarterly. assuming no…

robert invested $71,000 in an account paying an interest rate of 6.6% compounded quarterly. assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 18 years?
Answer
Explanation:
Step1: Identify the compound - interest formula
The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $n$ is the number of times interest is compounded per year, and $t$ is the number of years.
Step2: Convert the given values to the appropriate form
Given $P=$71000$, $r = 6.6%=0.066$, $n = 4$ (compounded quarterly), and $t = 18$ years.
Step3: Substitute the values into the formula
$A=71000(1 +\frac{0.066}{4})^{4\times18}$ First, calculate the value inside the parentheses: $\frac{0.066}{4}=0.0165$, and $1 + 0.0165=1.0165$. Then, calculate the exponent: $4\times18 = 72$. So, $A = 71000\times(1.0165)^{72}$.
Step4: Calculate $(1.0165)^{72}$
Using a calculator, $(1.0165)^{72}\approx3.2177$.
Step5: Calculate the final amount $A$
$A=71000\times3.2177 = 228456.7$. Rounding to the nearest ten dollars, $A\approx228460$.
Answer:
$228460$