marshall had $65,361.50 in a savings account with simple interest. he had opened the account with $61,000…

marshall had $65,361.50 in a savings account with simple interest. he had opened the account with $61,000 just 6 months earlier. what was the interest rate? use the formula $i = prt$, where $i$ is the interest earned, $p$ is the principal (starting amount), $r$ is the interest rate expressed as a decimal, and $t$ is the time in years.

marshall had $65,361.50 in a savings account with simple interest. he had opened the account with $61,000 just 6 months earlier. what was the interest rate? use the formula $i = prt$, where $i$ is the interest earned, $p$ is the principal (starting amount), $r$ is the interest rate expressed as a decimal, and $t$ is the time in years.

Answer

Explanation:

Step1: Calculate the interest earned

The initial amount $p = 61000$, and the final amount is $65361.50$. So the interest earned $i=65361.50 - 61000=4361.50$. The time $t = 6$ months $=\frac{6}{12}=0.5$ years.

Step2: Rearrange the simple - interest formula to solve for $r$

The simple - interest formula is $i = prt$. Rearranging for $r$, we get $r=\frac{i}{pt}$.

Step3: Substitute the values of $i$, $p$, and $t$ into the formula for $r$

Substitute $i = 4361.50$, $p = 61000$, and $t=0.5$ into $r=\frac{i}{pt}$. Then $r=\frac{4361.50}{61000\times0.5}=\frac{4361.50}{30500}=0.143$.

Answer:

$14.3$