Questions: Select the correct answer. A new amusement park presold discounted tickets for the opening day as well as upon arrival at the park. The opening day ticket sales for the park, including presales, is represented by this function, where T(h) is the number of tickets sold h hours after opening at 7:00 a.m. T(h)=(38+110 h)/h What is the approximate rate of change in the number of tickets sold between 10:00 a.m. and 1:00 p.m.? A. -64 tickets per hour B. -21 tickets per hour C. -39 tickets per hour D. - 1 1 0 tickets per hour

Solution Steps

The function representing the number of tickets sold \( T(h) \) hours after opening at 7:00 a.m. is given by: \[ T(h) = \frac{38 + 110h}{h} \]

We need to find the average rate of change of the function \( T(h) \) between 10:00 a.m. and 1:00 p.m. This corresponds to the interval:

• \( h = 3 \) (10:00 a.m.)
• \( h = 6 \) (1:00 p.m.)

The average rate of change of the function \( T(h) \) over the interval \([3, 6]\) is calculated using the formula: \[ \text{Average Rate of Change} = \frac{T(6) - T(3)}{6 - 3} \]

We evaluate \( T(3) \) and \( T(6) \): \[ T(3) = \frac{38 + 110 \cdot 3}{3} = \frac{38 + 330}{3} = \frac{368}{3} \] \[ T(6) = \frac{38 + 110 \cdot 6}{6} = \frac{38 + 660}{6} = \frac{698}{6} \]

Substituting the values of \( T(3) \) and \( T(6) \) into the average rate of change formula gives: \[ \text{Average Rate of Change} = \frac{\frac{698}{6} - \frac{368}{3}}{3} \]

To simplify, we find a common denominator for the fractions: \[ \frac{368}{3} = \frac{736}{6} \] Thus, the expression becomes: \[ \text{Average Rate of Change} = \frac{\frac{698}{6} - \frac{736}{6}}{3} = \frac{\frac{-38}{6}}{3} = \frac{-38}{18} = -\frac{19}{9} \]

The approximate average rate of change in the number of tickets sold between 10:00 a.m. and 1:00 p.m. is: \[ -\frac{19}{9} \text{ tickets per hour} \approx -2.11 \text{ tickets per hour} \]

Final Answer