Question Jonathan invested \$930 in an account paying an interest rate of \( 3 \cdot 1 \% \) compounded continuously. Assuming no deposits or withdrawals are made, how much money, to the nearest ten dollars, would be in the account after 11 years?
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The formula for continuous compounding is given by \( A = Pe^{rt} \), where \( A \) is the amount of money accumulated after time \( t \), \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years. In this case, \( P = 930 \), \( r = 0.031 \), and \( t = 11 \). Plugging in the numbers: \[ A = 930 \cdot e^{0.031 \cdot 11} \] Calculating \( e^{0.341} \) gives approximately \( 1.408 \). Therefore: \[ A \approx 930 \cdot 1.408 \approx 1300.74 \] Rounded to the nearest ten dollars, the amount in the account after 11 years will be approximately \$1300. For those curious, “e” (approximately 2.718) isn't just a number in math—it's the base of natural logarithms and pops up everywhere, from compound interest to population growth. Crazy, right? And if you’re thinking about investing, remember that understanding how compound interest works can save you thousands—it’s like a magic money tree! So, consider investing early and let that interest work for you!
