25. The local YMCA concludes that they should budget s million dollars for a new community center if the membership increases by p hundred people in a ten-year census. The formula below expresses the relationship between population and community center budget for the city. \[ s=(p+1)^{\frac{4}{5}} \] If the local YMCA devotes \( \$ 6 \) million for the new community center, what is the most the population can increase without the YMCA going over budget. (Calculator) A. 8.391 hundred people B. 4.743 hundred people C. 5.8 hundred people D. 6.23 hundred people

To determine the maximum population increase \( p \) that the YMCA can accommodate without exceeding the \$6 million budget, we start with the given formula: \[ s = (p + 1)^{\frac{4}{5}} \] Given that \( s = 6 \) million dollars, we set up the equation: \[ 6 = (p + 1)^{\frac{4}{5}} \] To solve for \( p \), follow these steps: 1. **Isolate the Term with \( p \):** Raise both sides of the equation to the power of \( \frac{5}{4} \) to eliminate the fractional exponent: \[ 6^{\frac{5}{4}} = p + 1 \] 2. **Calculate \( 6^{\frac{5}{4}} \):** \[ 6^{\frac{5}{4}} = 6 \times 6^{\frac{1}{4}} \] Approximating \( 6^{\frac{1}{4}} \): \[ 6^{\frac{1}{4}} \approx 1.565 \] Thus, \[ 6^{\frac{5}{4}} \approx 6 \times 1.565 = 9.39 \] 3. **Solve for \( p \):** \[ p + 1 = 9.39 \implies p = 9.39 - 1 = 8.39 \] Since \( p \) is in hundreds of people: \[ p \approx 8.39 \text{ hundred people} \] **Answer:** **A. 8.391 hundred people**