12) Mr Mogi borrowed \( \$ 9,000 \) for 10 years to make home improvements. If 1 point he paid \( \$ 11,000 \) in interest, at what interest rate did he borrow the money?

To find the interest rate at which Mr. Mogi borrowed the money, we can use the formula for simple interest: \[ \text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time} \] Given: - Principal (\( P \)) = $9,000 - Time (\( T \)) = 10 years - Interest (\( I \)) = $11,000 We need to find the interest rate (\( R \)). Rearranging the formula to solve for the interest rate: \[ R = \frac{I}{P \times T} \] Substitute the given values into the formula: \[ R = \frac{11000}{9000 \times 10} \] Now, we can calculate the interest rate. Calculate the value by following steps: - step0: Calculate: \(\frac{11000}{\left(9000\times 10\right)}\) - step1: Remove the parentheses: \(\frac{11000}{9000\times 10}\) - step2: Multiply the numbers: \(\frac{11000}{90000}\) - step3: Reduce the fraction: \(\frac{11}{90}\) The interest rate at which Mr. Mogi borrowed the money is 0.122 or 12.2%.