Ask Question
21 August, 22:35

Tonya plans to join a fitness club. She researched the costs to join different clubs in her area and found that there is a linear relationship between the total cost of belonging to a club and the number of months that you are a member. Each club charges monthly dues, and some clubs require new members to also pay a one-time membership fee. The total costs of being a member for 6 months, 12 months, and 24 months at each of three clubs are shown in the table. Total Cost of Fitness Club Membership Length of membership Club 1 Club 2 Club 3 6 months $108 $120 $126 12 months $216 $210 $252 24 months $432 $390 $504 How many of the clubs require new members to pay a one-time membership fee? 0 1 2 3

+5
Answers (1)
  1. 22 August, 00:35
    0
    Club 1 charges $108 for 6 months, $216 for 12 months ($216 = 2 ($108) and 12 months = 2 (6 months)), and $432 for 24 months ($432 = 2 ($216) and 24 months = 2 (12 months)). Since each period of time doubles as well as the cost doubling, we can see that there is no other fee added for membership.

    Club 2 charges $120 for 6 months, $210 for 12 months ($210 ≠ 2 ($120) and 12 months = 2 (6 months)), and $390 for 24 months ($390 ≠ 2 ($≠210) and 24 months = 2 (12 months)). These prices are not doubled each time as the time periods are. Instead, each price is $30 more than double each time. This means there is an additional $30 fee for joining in this club.

    Club 3 charges $126 for 6 months $252 for 12 months ($252 = 2 ($126) and 12 months = 2 (6 months)), and $504 for 24 months ($504 = 2 ($252) and 24 months = 2 (12 months)). Each price doubles along with the time period, so there is no extra membership fee in this club.

    Only 1 club charges the extra membership fee.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Tonya plans to join a fitness club. She researched the costs to join different clubs in her area and found that there is a linear ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers