Question

Mr Williams is planning a seventh grade field trip to a math museum. school policy requires a minimum of 2 adults as chaperones for every 9 students on the trip

Answer 1

This is an incomplete question. However, I read the full question in the other site.

Given:
minimum of 2 adults as chaperons for every 9 students.
food budget of $6 per person
educational group pricing: $4.50 per student; $7.25 per adult
bus parking fee that is 4% of the total ticket price

Part A. 171 students. Find minimum number of chaperons required.
171 ÷ 9 = 19 x 2 = 38 chaperons

Part B. Budget for tickets and parking
171 students * 4.50 per student = $769.50
38 chaperons * 7.25 per adult = $275.50
bus parking fee: (769.50 + 275.50) * 4% = 1,045 * 4% = $41.80
Total amount to budget : 769.50 + 275.50 + 41.80 = $1,086.80

Part C. Budget for tickets, parking, and food
Total amount to budget for tickets and parking: $1,086.80 
Food budget: (171 + 38) * $6 per person = 209 * $6 = $1,254
Total budget: 1,086.80 + 1,254 = $2,340.80

Answer 2

1254 + 1086.80 = 2340.80

total = Answer:

this is not a full question but i still found the answer

Step-by-step explanation:

part a :

171 / 9 = 19

19 x 2 =38 chaperones

38 x 7.25 =275.50

171 x 4.50 = 769.50

275.50 + 769.50 = 1045

4% x 1045 = 41.80

1045 + 41.80 =1086.80 = total cost of museum tickets and parking

part b:

171 +38 = 209

209 x 6 = 1254

1254 +1086.80 = 2340.80 = total budget including food, parking tickets for students and chaperones