The amusement park has two plans available for the school to choose from. Plan A is listed in the table below. Cost (per vehicle) Cost (per person)
$10 $30

The cost of Plan B can be calculated using the formula c = $15v + $28p, where c represents the cost, v represents the number of vehicles, and p represents the number of people. The school has a budget of $7200 for the cost of tickets and parking. If 241 students are going on the field trip, which plan should the school purchase? Explain your reasoning.

the second plan should be purchased.

Step-by-step explanation:

school budget is $7200

plan 1: 241 students are $30 each is

241 × 30 = $7230 which is over budget with no money for parking for the vehicles.

plan 2: 241 students are $28 each is

241 × 28 = $6748

with a budget of $7200 for the trip the amount left for parking is

$7200 - $6748 = $452

plan 2: Parking is $15 per vehicle

Therefore the remaining $452 can be used to pay for 452 ÷ 15 = 30.13 vehicles

The school should purchase Plan 2 for the field trip.

The budget of $7200 can pay for 241 tickets for students and parking for 30 vehicles

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt