Question

Yolanda invests $30,000 in an account that offers a compound interest rate of 7.9% per year. Which of the following is the correct equation for how much Yolanda will have after 10 years?
Possible answers here: https://imgur.com/M5JkywQ

Answer 1

To solve this, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
tex]P[/tex] is the initial investment 
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

We know form our problem that the initial invest is $30,000 and the period is ten years, so $P=30000$ and $t=10$. We also know that the interest is compounded one time per year, so $n=1$. To convert the interest rate to decimal form, we are going to divide the rate by 100%
$r= \frac{7.9}{100}$
$r=0.079$
Lets replace the values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=30000(1+ \frac{0.079}{1} )^{(1)(10)}$
$A=3000(1+0.079)^{10}$
Replacing $A$ with $P_{10}$:
$P_{10}=3000(1+0.079)^{10}$

We can conclude that the correct answer is D. $P_{10}=3000(1+0.079)^{10}$

Now, to find much Yolanda will have after 10 years, we just need to perform the operations in our equation: 
$P_{10}=3000(1+0.079)^{10}$
$P_{10}=30000(1.079)^{10}$
$P_{10}=64170.54$

We can conclude that Yolanda will have $64,170.54 in her account after ten years of earning a compound interest rate of 7.9% per year.