Acatering service offers 7 appetizers, 3 main courses, and 5 desserts. a customer is to select 4 appetizers, 2 main courses, and 4 desserts for a banquet. in how many ways can this be done?

This question is based on combinations

Out of 7 appetizers, 4 are to be selected, this becomes = 7C4

Out of 3 main courses, 2 are to be selected , this is = 3C2

Out of 5 desserts, 4 are to be selected, this is = 5C4

So, total ways become =

7C4 * 3C2 * 5C4

7!/4!(7-4)! * 3!/2!(3-2)! * 5!/4!(5-4)!

= 35*3*5 =525 ways

Hence, this can be done in 525 ways.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.