What point in the feasible region maximizes the objective function?

constraints:

⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
x
≥
0
y
≥
0
−
x
+
3
≥
y
y
≤
1
3
x
+
1
objective function:
c

=

5
x

–

4
y

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.

theres no picture: /

step-by-step explanation:

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

happy to

pls mark as brainliest

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8