valentina wanted to buy a new motorcycle and put $4,000.00 into an account which earns interest compounded…

valentina wanted to buy a new motorcycle and put $4,000.00 into an account which earns interest compounded continuously. after 1 year, she withdrew the entire balance of $4,388.00 and bought the motorcycle. what was the interest rate on the account? round your answer to the nearest tenth of a percent.

valentina wanted to buy a new motorcycle and put $4,000.00 into an account which earns interest compounded continuously. after 1 year, she withdrew the entire balance of $4,388.00 and bought the motorcycle. what was the interest rate on the account? round your answer to the nearest tenth of a percent.

Answer

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), and $t$ is the time in years. We know that $P = 4000$, $A = 4388$, and $t = 1$. Substituting these values into the formula gives $4388=4000e^{r\times1}$, or $4388 = 4000e^{r}$.

Step2: Solve for $r$

First, divide both sides of the equation by 4000: $\frac{4388}{4000}=e^{r}$. Simplify $\frac{4388}{4000}$ to get $1.097 = e^{r}$. Then, take the natural logarithm of both sides. Since $\ln(e^{r})=r$, we have $r=\ln(1.097)$. Using a calculator, $\ln(1.097)\approx0.093$.

Step3: Convert $r$ to a percentage

To convert $r$ from a decimal to a percentage, multiply by 100: $r = 0.093\times100 = 9.3%$.

Answer:

$9.3$