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:
Step-by-step explanation:
<u>Total amount to be paid is the sum of 4 bills:</u>
- 21.47 + 14.43 + 13.27 + 25.37 = $74.54
<u>Divide the total by the number of people:</u>
- 74.54/4 = 18.635 = $18.64 rounded to the nearest cent
If the total to be paid equally, then each person has to pay $18.64
Answer:
x = 2 1/3
y = 14 / (2 1/3) = 14 / (7/3) = 14*3/7 = 42/7 = 6
y = 6
x = 4 1/5
y = 14 / (4 1/5) = 14 / (21/5) = 14*5/21 = 70/21 = 3 7/21 = 3 1/3
y = 3 1/3
x = 7/6
y = 14 / (7/6) = 14*6 / 7 = 12
y = 12
Step-by-step explanation:
