Christina is packing different books into boxes. The paperback books weigh 0.8 pounds each. The hardcover books weigh 1.3 pounds each. Christina needs to pack the boxes so that there are at least 20 books in each box, but the boxes cannot weigh more than 30 pounds. Identify Christina's set of constraints. Check all that apply.

and .

Step-by-step explanation:

Let x and y be the numbers of the paperback book and the hardcover book respectively.

as x and y are the counting numbers, so it will be always positive. So,

and

Given that the weight of 1 paperback book=0.8 pounds and

the weight of 1 paperback book=1.3 pounds.

There are at least 20 books in each box, so

As one box cannot weigh more than 30 pounds, so

Now, from the equations (i),(ii),(iii), and (iv), all the sets of constraints as

and .

Moreover, the sets of all the integral values of (x,y) in the shaded region are the feasible numbers of the paperback book and hardcover book.

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.