Aschool district needs to arrange transportation for at least 360 students for a field trip. the district can use large buses or small buses. the large buses hold 60 students each and have an operating expense of $300 for the day. the small buses hold 45 students each and have an operating expense of $200 for the day. the district only has chaperones for, at most, seven buses. what is the minimum cost for the transportation for the field trip? . a.$1,400. b. $1,600. c.$1,700. d.$1,800

$1700

Step-by-step explanation:

Let the number of large buses be x

Let number of small buses be y

Let C be the total cost for the trip

Since w eare given that cost of 1 large bus is 300 and 1 small bus is 200

So, Cost C  = 300 * number of large buses + 200 * number of small buses

So  

Constraints are :

Since we are given that large buses can hold 60 students each

So, total students in x buses = 60 x

We are also given that small buses can hold 45 students each.

So, total students in y buses = 45y

Since we are also given that A school district needs to arrange transportation for at least 360 students

So,

Number of large buses

Number of small buses

We are also given that The district only has chaperones for, at most, seven buses.

So,

Now we graph the constraints and find all the endpoints

Refer the attached Graph

So,End points are (3,4) , (6,0) and (7,0)

Now substitute these points in Cost function to find minimum cost.

with(3,4)

with (6,0)

With (7,0)

Hence the minimum cost is $1700.

4/6 = 2/3

so it should take 3/3 hours per man. so 9 hours is the answer 3*3 =9

step-by-step explanation:

answer: the answer is fre sha voca do

step-by-step explanation: