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:
Rent, wages and car insurance don't vary much. Grocery expenses depend upon the particular groceries purchased each time one goes to the grocery store, and are least likely to be steady / constant from week to week.
Answer: -3 and 1/3
Step-by-step explanation:
There’s your answer
Answer:
B. 4t=27
Step-by-step explanation:
let t be the price of one ticket,
since there are 4 people buying tickets in total, the price of 4 tickets would be 4t.
so 
.
Answer:
480 people were unemployed full time
Step-by-step explanation:
First thing to do here is to check the percentage of people that were unemployed full time
Looking at the graph, this is equal to 60%
This means that 60% of the total 800 are unemployed full time
Mathematically, that would be 60/100 * 800 = 480 people
