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:
18x - 3
Step-by-step explanation:
Add all the X values
8x + -2x + 12x = 18x
Put the final numbers on the side.
-3
18x-3
Answer:
the last one(y=2(2/3)^x) is the correct answer
Step-by-step explanation:
I identify two coordinate on the graph (0,2) and (1,3) and I noticed only the last one gives you a appropriate output if you plug the correspond input value
Answer:
try 1 and 2 for supplementary
and 7 and 6 for vertical
Step-by-step explanation:
Answer:
53
Step-by-step explanation:
