You manage a restaurant. on any given day you sell on average 20 dinners: 3 fish dinners (0.15),6 chicken dinners (0.30), 6 pork dinners (0.30), and 5 steak dinners (0.25).what is the probability that today you sell 2 fish dinners, 9 chicken dinners, 5 pork dinners, and 4 steak dinners? what probability distribution did you use?

answer: options?

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

2160ft3

step-by-step explanation: