Dana wants to make two types of dog treats. she has 10 cups of peanut butter and 12 cups of flour. her dog bone treat recipe uses 3 cups of peanut butter and 2 cups of flour to make one tray. a tray of her oatmeal dog treat recipe uses 1 cup of peanut butter and 4 cups of flour. she plans to sell trays of dog treats at the town festival and charge $6 for a tray of dog bone treats and $7 for a tray of oatmeal treats. dana wants to maximize her income from selling the dog treats. when writing constraints for the problem, what is the most reasonable definition for the variables x and y? a) let x represent the number of cups of peanut butter used and y represent the number of cups of flour used. b) let x represent the number of cups of peanut butter used and y represent the number of trays of dog bone treats made. c) let x represent the number of dog bone treats made and y represents the number of cups of flour used. d) let x represent the number of trays of dog bone treats made and y represent the number of trays of oatmeal dog treats made.

Answer with explanation:

Number of cups of peanut that Dana has =10 cups

Number of cups of flour that Dana has =12 cups

Let x represent the number of trays of dog bone treats made and y represent the number of trays of oatmeal dog treats made.

Constraints

⇒x>0, y>0

⇒3 x+y ≤10

⇒2 x+4 y ≤ 12

Maximize(Objective Function):

Z=6 x +7 y

⇒⇒ The most reasonable definition for the variables x and y as Dana wants to maximize her income from selling the dog treats

Option D: Let x represent the number of trays of dog bone treats made and y represent the number of trays of oatmeal dog treats made.

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.