taub invested $38,000 in an account paying an interest rate of 3.9% compounded quarterly. assuming no…

taub invested $38,000 in an account paying an interest rate of 3.9% 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?

taub invested $38,000 in an account paying an interest rate of 3.9% 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 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 values to appropriate form

Given $P=$38000$, $r = 3.9%=0.039$, $n = 4$ (compounded quarterly), and $t = 18$ years.

Step3: Substitute values into the formula

$A=38000(1 +\frac{0.039}{4})^{4\times18}$ First, calculate the value inside the parentheses: $\frac{0.039}{4}=0.00975$, then $1+\frac{0.039}{4}=1 + 0.00975=1.00975$. Next, calculate the exponent: $4\times18 = 72$. So, $A = 38000\times(1.00975)^{72}$.

Step4: Calculate $(1.00975)^{72}$

Using a calculator, $(1.00975)^{72}\approx2.0477$.

Step5: Calculate $A$

$A=38000\times2.0477 = 77812.6$.

Step6: Round to the nearest ten dollars

Rounding $77812.6$ to the nearest ten dollars gives $77810$.

Answer:

$77810$