Your friend makes a comment about how expensive the baseball card pack was that he purchased. He says 10 years ago he could buy a pack of baseball cards for \( \$ 5 \). Assuming inflation is \( 3 \% \), compounded continuously, how much did he pay for the pack today? Round your answer to the nearest cent (hundredth).

To find out how much the friend paid for the pack of baseball cards today, considering the inflation rate of 3% compounded continuously, we can use the formula for continuous compounding: \[ A = P \times e^{rt} \] where: - \( A \) is the amount after the compounding period, - \( P \) is the principal amount (initial amount), - \( r \) is the annual interest rate (inflation rate in this case), - \( t \) is the time period in years. Given: - Initial amount (\( P \)) = $5 - Inflation rate (\( r \)) = 3% = 0.03 - Time period (\( t \)) = 10 years Substitute the values into the formula: \[ A = 5 \times e^{0.03 \times 10} \] Now, we can calculate the amount the friend paid for the pack of baseball cards today. Calculate the value by following steps: - step0: Calculate: \(5e^{0.03\times 10}\) - step1: Multiply the numbers: \(5e^{0.3}\) - step2: Convert the expressions: \(5e^{\frac{3}{10}}\) - step3: Rewrite the expression: \(5\sqrt[10]{e^{3}}\) The friend paid approximately $6.75 for the pack of baseball cards today, considering the inflation rate of 3% compounded continuously over 10 years.