17 Malfpie Choire 5 points You invested \( \$ 1500 \) into an account with an annual percentage rate of \( 6 \% \) compounded annually. You plan to ketp the money in the account for the next 10 years Consider the function that relates the time and the amount of money in the account. Which is the most appropriate range? \[ \begin{array}{l}x \geq 0 \\ y \geq 1500 \\ 0 \leq x \leq 10 \\ 1500 \leq y \leq 2686\end{array} \]

To determine the most appropriate **range** for the function that models your investment, let's analyze the given information: 1. **Initial Investment (Principal), \( P \):** \$1500 2. **Annual Interest Rate, \( r \):** 6% (or 0.06 as a decimal) 3. **Time, \( t \):** 10 years 4. **Compounding:** Annually The compound interest formula is: \[ y = P \times (1 + r)^x \] Where: - \( y \) is the amount of money after \( x \) years. - \( x \) is the time in years. Plugging in the known values: \[ y = 1500 \times (1 + 0.06)^x = 1500 \times 1.06^x \] **Calculating the Amount After 10 Years:** \[ y = 1500 \times 1.06^{10} \approx 1500 \times 1.79085 \approx 2686.28 \] **Range Analysis:** - **Initial Amount:** The smallest value \( y \) can be is the initial investment, \$1500. - **After 10 Years:** The largest value \( y \) can reach is approximately \$2686. Therefore, the **range** of the function \( y \) over the 10-year period is: \[ 1500 \leq y \leq 2686 \] **Conclusion:** The most appropriate range for the function, considering the 10-year investment period, is: **\( 1500 \leq y \leq 2686 \)** **Answer:** The range is all y with 1500 ≤ y ≤ 2686