Question

arturo invested $630 in an account paying an interest rate of 4.3% compounded continuously. assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 9 years?

Explanation:

Step1: Recall the formula for continuous compounding

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the amount of money accumulated after $t$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (in decimal form), and $t$ is the time the money is invested for in years.

Step2: Identify the values of \( P \), \( r \), and \( t \)

Given that $P = 630$ dollars, $r = 4.3\%=0.043$ (converted to decimal), and $t = 9$ years.

Step3: Substitute the values into the formula

Substitute $P = 630$, $r = 0.043$, and $t = 9$ into the formula $A = Pe^{rt}$. So we have $A=630\times e^{0.043\times9}$.

Step4: Calculate the exponent first

Calculate $0.043\times9 = 0.387$.

Step5: Calculate \( e^{0.387} \)

Using a calculator, $e^{0.387}\approx1.472$.

Step6: Calculate the amount \( A \)

Multiply $630$ by $1.472$: $A = 630\times1.472 = 927.36$.

Step7: Round to the nearest ten dollars

To round to the nearest ten dollars, we look at the digit in the ones place, which is 7. Since $7\geq5$, we round up the tens place. So $927.36$ rounded to the nearest ten dollars is $930$.

Answer:

\( \$930 \)