Formula 1 point You deposit \( \$ 112 \) in an account that pays interest with annual compounding. If the account contains \( \$ 145 \) exactly 9 years from today, what is the annual interest rate? Enter your answer as a decimal, and show 4 decimal places. For example, if your answer is \( 6.5 \% \), you should enter .0650 . Type your answer.

To find the annual interest rate, we can use the formula for compound interest: \[ A = P(1 + r)^n \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (initial deposit). - \( r \) is the annual interest rate (in decimal form). - \( n \) is the number of years the money is invested. Given: - \( P = \$112 \) (initial deposit) - \( A = \$145 \) (amount after 9 years) - \( n = 9 \) years We need to solve for \( r \) (annual interest rate). Substitute the given values into the formula: \[ 145 = 112(1 + r)^9 \] Now, we can solve for \( r \) using the formula for compound interest. Solve the equation by following steps: - step0: Solve for \(r\): \(145=112\left(1+r\right)^{9}\) - step1: Swap the sides: \(112\left(1+r\right)^{9}=145\) - step2: Divide both sides: \(\frac{112\left(1+r\right)^{9}}{112}=\frac{145}{112}\) - step3: Divide the numbers: \(\left(1+r\right)^{9}=\frac{145}{112}\) - step4: Simplify the equation: \(\sqrt[9]{\left(1+r\right)^{9}}=\sqrt[9]{\frac{145}{112}}\) - step5: Calculate: \(1+r=\sqrt[9]{\frac{145}{112}}\) - step6: Simplify the root: \(1+r=\frac{\sqrt[9]{145\times 112^{8}}}{112}\) - step7: Move the constant to the right side: \(r=\frac{\sqrt[9]{145\times 112^{8}}}{112}-1\) - step8: Subtract the numbers: \(r=\frac{\sqrt[9]{145\times 112^{8}}-112}{112}\) The annual interest rate \( r \) is approximately 0.0291 or 2.9108%.