[-/2 Points] DETAILS MY NOTES The owner of a skating rink rents the rink for parties at \( \$ 360 \) if 30 or fewer skaters attend. For each 5 skaters above 30 , she reduces the price per skater by \( \$ 0.50 \). (a) Write the equation that describes the revenue for parties with \( x \) groups of five skaters more than 30 skaters. \( R(x)= \) (b) Find the revenue when 130 skaters attend. \( \$ \).

To solve this problem, let's break it down step by step. (a) Let \( x \) represent the number of groups of five skaters above 30 skaters. Therefore, if there are \( 30 + 5x \) skaters, the rental price decreases by \( 0.50x \) dollars per skater. The total number of skaters can be expressed as \( 30 + 5x \). The revenue \( R(x) \) can be calculated as: \[ R(x) = \text{Number of skaters} \times \text{Price per skater} \] First, determine the price per skater at 30 skaters: \[ \text{Initial price per skater} = \frac{360}{30} = 12 \text{ dollars} \] Now, the price per skater decreases by \( 0.50x \) dollars for each additional group of 5 skaters: \[ \text{Price per skater} = 12 - 0.50x \] Thus, the equation representing the revenue is: \[ R(x) = (30 + 5x) \times (12 - 0.50x) \] (b) To find the revenue when 130 skaters attend, first determine how many groups of five skaters this represents above 30 skaters. Start with the total of 130 skaters: \[ 130 - 30 = 100 \text{ skaters above 30} \] Now, divide by 5 to find \( x \): \[ x = \frac{100}{5} = 20 \] Now, substitute \( x = 20 \) back into the revenue equation: \[ R(20) = (30 + 5 \times 20) \times (12 - 0.50 \times 20) \] Calculating this step-by-step: \[ R(20) = (30 + 100) \times (12 - 10) = 130 \times 2 = 260 \] So, the revenue when 130 skaters attend is: \[ \boxed{260} \]