Apizzeria sells pizza according to size: small pizzas cost 10$, medium pizzas 15$ and large pizzas cost 40$. they usually sell as many small pizzas as medium and large pizzas combined. the number of medium pizzas sold is usually twice as many as large ones. how many of each size pizza must they sell to get 600$.

6 large, 12 medium and 18 small

Step-by-step explanation:

Lets call small pizzas s, medium pizzas m and large pizzas l, with its respective prices $10, $15 and $40.

As they sell as many small as medium and large combined we can say that:

s = m + l [eq 1]

Also, as the number of medium is twice the larges we can say that:

m = 2l [eq 2]

Finally, as the get $600 we know that every amount sold multiplied by its price, and summing all together must bring $600:

10s + 15m + 40l = 600 [eq 3]

Lets replace s by its value in eq 1:

10s + 15m + 40l = 10 (m+l) 15m + 40l = 10m +10l + 15m + 40l = 600

25m + 50l = 600

Now we can replace m by its value in eq 2:

25m + 50l = 25 (2l) + 50 l = 50l + 50l = 100l = 600

100l = 600

Now divide both sides by 100:

l = 6

So, 6 large pizzas were sold.

Replace l=6 in eq 2:

m = 6*2

m = 12

12 medium pizzas were sold.

Finally, replace l=6 and m=12 in eq 1

s = m + l = 12 + 6 = 18

And 18 small pizzas were sold.

Lets verify our results in eq. 3:

10*(18) + 15*(12) + 40*(6) = 600

180 + 180 + 240 = 600

360 + 240 = 600 ---> Verified!

Step-by-step explanation:

Given ;

a) P ( X = 13 ) = C(15,13) x 0.56^13 x (1-0.56)^2 = 0.0108

b) Mean = np = 15 x 0.56= 8.400

Standard deviation = √(np(1-p)) = √(15 x 0.56 x (1-0.56)) = 1.9225

c) P(X>10) = 1 - P(X≤10) = 1 - 0.8633 = 0.1367

d) P(X≥12) = P(X=12) + P(X=13) + P(X=14) + P(X=15) = 0.0498

e) P(Y=3) = C(3,1) x 0.56^1 x (1-0.56)^2 x 0.56 = 0.1821

f) P(Y=2) = C(2,1) 0.56^1 x (1-0.56)^1 x 0.56 = 0.2760