samuel is going to invest $870 and leave it in an account for 20 years. assuming the interest is compounded…

samuel is going to invest $870 and leave it in an account for 20 years. assuming the interest is compounded monthly, what interest rate, to the nearest hundredth of a percent, would be required in order for samuel to end up with $2,580? answer attempt 1 out of 2
Answer
Explanation:
Step1: Recall compound - interest formula
The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. Given $P = 870$, $A=2580$, $n = 12$ (compounded monthly), and $t = 20$. Substitute these values into the formula: $2580=870(1 +\frac{r}{12})^{12\times20}$.
Step2: Simplify the equation
First, divide both sides of the equation by 870: $\frac{2580}{870}=(1 +\frac{r}{12})^{240}$. $3=(1 +\frac{r}{12})^{240}$.
Step3: Take the 240 - th root of both sides
$3^{\frac{1}{240}}=1+\frac{r}{12}$. Using a calculator, $3^{\frac{1}{240}}\approx1.00456$. So, $1.00456 = 1+\frac{r}{12}$.
Step4: Solve for $r$
Subtract 1 from both sides: $1.00456−1=\frac{r}{12}$. $0.00456=\frac{r}{12}$. Multiply both sides by 12 to solve for $r$: $r = 0.00456\times12$. $r=0.05472$.
Step5: Convert $r$ to a percentage
To convert $r$ to a percentage, multiply by 100: $r = 5.472%$. Rounding to the nearest hundredth of a percent, $r\approx5.47%$.
Answer:
$5.47%$