An agency wants to ship food and clothing to hurricane victims in Louisiana. Commercial carriers have volunteered to transport the packages provided they fit in the available cargo space each 20ft3 box of food weights 40 pounds

300 cartons of food and 400 cartons of clothes

Step-by-step explanation:

Total weight  = 16000lb

Total volume = 18000lb

Let number of food box be x

Let number of cloth box be y

therefore,

Maximize Z = 13x + 6y

Subject to :

                   20x + 30y ≤ 18000 (Divide through by 10)

                   40x + 10y ≤ 16000 (Divide through by 10)

                      2x + 3y ≤ 1800(1)

                      4x + y ≤ 1600 (2)

y ≤ 1600 - 4x

Substitute the value of y in equation (1)

2x +3(1600 - 4x) ≤ 1800

2x + 4800 - 12x ≤1800

-10x ≤ -3000

Divide all through by -10

x ≤ 300

Substitute x = 300 in equation (2)

y ≤ 1600 - 4(300)

y ≤ 1600 - 1200

y ≤ 400

Therefore, lines intersects at 300, 400 means we can 300 cartons of food and 400 cartons of clothes

Total people helped = 13x + 6y

                                   = 13 (300) + 6(400)

                                   = 3900 + 2400

                                    = 6300

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

the answer is. -82562234434584

step-by-step explanation: