Answer:
At a certain pizza parlor,36 % of the customers order a pizza containing onions,35 % of the customers order a pizza containing sausage, and 66% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.
Step-by-step explanation:
Hello!
You have the following possible pizza orders:
Onion ⇒ P(on)= 0.36
Sausage ⇒ P(sa)= 0.35
Onions and Sausages ⇒ P(on∪sa)= 0.66
The events "onion" and "sausage" are not mutually exclusive, since you can order a pizza with both toppings.
If two events are not mutually exclusive, you know that:
P(A∪B)= P(A)+P(B)-P(A∩B)
Using the given information you can use that property to calculate the probability of a customer ordering a pizza with onions and sausage:
P(on∪sa)= P(on)+P(sa)-P(on∩sa)
P(on∪sa)+P(on∩sa)= P(on)+P(sa)
P(on∩sa)= P(on)+P(sa)-P(on∪sa)
P(on∩sa)= 0.36+0.35-0.66= 0.05
I hope it helps!
Answer:
11
Step-by-step explanation:
56.50-7=49.50
49.50divided by 4.50=11
4x+1=6x-2
4x+3=6
3=2x
3/2=2x/2, because you have to divide 2 from both sides
X=3/2
Answer:
x^2 ×1 2x + 36
Step-by-step explanation:
area of a square = L × B
= (x+6) (x+6) since a square has 4 equal sides, it becomes (x+6) (x+6)
= x(x+6) +6(x+6)
= x^2 + 6x + 6x +36
= x^2 + 12x +36 added all the like terms 6x + 6x = 12x
hope this helps ☆☆☆
9/10 = x/13. Solve for x. X = 11.7.
Your answer would be 11.7oz. Hope this helps