A woman invested \( \$ 10,000 \) in a savings account paying \( 4 \% \) interest compounded quarterly. How much money will she have in her account at the end of 3 years? The woman will have \( \$ \square \), (Type an integer or decimal rounded to the nearest hundredth as needed.)

To find out how much money the woman will have in her account at the end of 3 years, we can use the compound interest formula: \( A = P \left(1 + \frac{r}{n}\right)^{nt} \) Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial investment). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested or borrowed. Plugging in the given values: - \( P = 10,000 \) - \( r = 0.04 \) - \( n = 4 \) (since the interest is compounded quarterly) - \( t = 3 \) Now substituting the values into the formula: \( A = 10,000 \left(1 + \frac{0.04}{4}\right)^{4 \times 3} \) Simplifying it step by step: 1. Calculate \( \frac{0.04}{4} = 0.01 \) 2. Add 1: \( 1 + 0.01 = 1.01 \) 3. Calculate \( 4 \times 3 = 12 \) 4. Raise \( 1.01 \) to the 12th power: \( 1.01^{12} \approx 1.126825 \) 5. Multiply by the principal: \( A \approx 10,000 \times 1.126825 \approx 11,268.25 \) Therefore, she will have approximately \( \$ 11,268.25 \) in her account at the end of 3 years. The woman will have \( \$ 11,268.25 \).