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:
212 bp^3 (base pairs cubed) - 121 bp^3 (base pairs cubed) + 222 bp^3 (base pairs cubed) = 313 bp^3 (base pairs cubed
Step-by-step explanation:
Answer:
8 miles away
Step-by-step explanation:
Add the 6 miles and the 2 miles together.
Answer:
No solution
Explanation
−x^2+5x−10=0
Step 1: Use quadratic formula with a=-1, b=5, c=-10.
you plug it in as shown below
and simplify
when you simplify
you get
as shown below
This is no solution....
