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.
Determine the minimum cost of transporting all 450 students.

Step-by-step explanation:

From the given information;

Let consider p to be the number of large buses and q to be the number of small buses.

The company has 18 large buses which can carry up to 30 students and 19 small buses which can carry up to 15 students.

The linear inequality represent the students with respect to the total students s:

30p + 15q   ≥  450   -----(1)

They are only 20 drivers available on the day of the field trip, therefore only 20 buses can be used. So,

p  +   q    ≤  20      ----- (2)

Also , There  are only 19 small buses and 18 large buses:

∴

0 ≤  p ≤ 18

0 ≤  q ≤ 19

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

In order to minimize the cost of transporting all 450 students, let z be the minimal cost. i.e

z = 225p + 100q

From equation 1 and 2 ; if we plot them graphically on the graph, the following points of intersection were obtained as shown in the sketch below, in which the shaded region lies the answer.

The point of intersection between equation (1) and (2) is (p,q) = (10,10)

From the critical point in the sketch of our graph attached below, the following values of z can be determined.

Point(s)              Value for Z

(15,0)                 15 ×  225 = 3375

(18,0)                 18 ×  225 =  4050

(18,2)                 (18 × 225 )+ (2 × 100 ) = 4250

(10,10)                (10 × 225) + (10 × 100) = 3250

Thus ; the minimum cost of transporting all 450 students = $3250

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

all it means is function 3