Noah is going to invest in an account paying an interest rate of \( 7 \% \) compounded continuously. How much would Noah need to invest, to the nearest hundred dollars, for the value of the account to reach \( \$ 12,900 \) in 10 years?

To determine how much Noah needs to invest today to reach \$12,900 in 10 years with an annual interest rate of \(7\%\) compounded continuously, we can use the formula for continuous compounding: \[ A = P e^{rt} \] 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). - \(t\) is the time the money is invested for in years. - \(e\) is the base of the natural logarithm, approximately equal to 2.71828. **Given:** - \(A = \$12,900\) - \(r = 0.07\) - \(t = 10\) years **Step 1: Rearrange the Formula to Solve for \(P\)** \[ P = \frac{A}{e^{rt}} = \frac{12900}{e^{0.07 \times 10}} = \frac{12900}{e^{0.7}} \] **Step 2: Calculate \(e^{0.7}\)** \[ e^{0.7} \approx 2.01375 \] **Step 3: Compute \(P\)** \[ P = \frac{12900}{2.01375} \approx 6405.69 \] **Step 4: Round to the Nearest Hundred Dollars** \[ P \approx \$6,400 \] **Conclusion:** Noah needs to invest approximately **\$6,400** today to reach \$12,900 in 10 years with a 7% continuous compounding interest rate. **Answer:** Noah needs to invest approximately $6,400 today to reach $12,900 in 10 years.