jose invested $79,000 in an account paying an interest rate of 6.8% compounded quarterly. assuming no…

jose invested $79,000 in an account paying an interest rate of 6.8% compounded quarterly. assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 9 years?

jose invested $79,000 in an account paying an interest rate of 6.8% compounded quarterly. assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 9 years?

Answer

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal form), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years.

Step2: Convert values to appropriate form

The principal $P=$79000$, the annual interest rate $r = 6.8%=0.068$, the number of times compounded per year $n = 4$ (quarterly compounding), and the time $t = 9$ years.

Step3: Substitute values into the formula

$A=79000(1 +\frac{0.068}{4})^{4\times9}$. First, calculate the value inside the parentheses: $\frac{0.068}{4}=0.017$, and $1 + 0.017=1.017$. Then, calculate the exponent: $4\times9 = 36$. So, $A = 79000\times(1.017)^{36}$.

Step4: Calculate $(1.017)^{36}$

Using a calculator, $(1.017)^{36}\approx1.8777$.

Step5: Calculate the final amount $A$

$A=79000\times1.8777\approx148338$.

Answer:

$148338$