A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students. There are only 20 drivers available on the day of the field trip. The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.

The answer is below

Step-by-step explanation:

Let x represent the big buses and y represent small buses. The large buses can carry 30 students and the small buses can carry 15 students. The total number of students are 450, this can be represented by the inequality:

30x + 15y ≤ 450

They are only 20 drivers, therefore only 20 buses can be used. It is represented by:

x + y ≤ 20

They  are only 19 small buses and 18 large buses:

x ≤ 18

y ≤ 19

After plotting the graph, the minimum solution to the graph are at:

A (15,0), B(18,0), C(10, 10), D(18, 2).

The cost function is given as:

The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.

F(x, y) = 225x + 100y

At point A:

F(x, y) = 225(15) + 100(0) = $3375

At point B:

F(x, y) = 225(18) + 100(0) = $4050

At point C:

F(x, y) = 225(10) + 100(10) = $3250

At point D:

F(x, y) = 225(18) + 100(2) = $4250

The minimum cost is at point C(10, 10) which is $3250

the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation: