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 to 4.
About 106.66
If they ask if a rounded answer it would be 107. You just divide 2560 by 24 because you know 24 balls is 3 pounds.
Answer:
(10/3, 2/3)
Step-by-step explanation:
You may find it easier to write these two equations x+y=4 2x-y=6 in a column:
x+y=4
2x-y=6
Adding these together will eliminate y:
3x = 10. Then x = 10/3.
Substituting 10/3 for x in the first equation results in:
10/3 + y = 4
Clear fractions by multiplying all three terms by 3:
10 + 3y = 12
Then 3y = 2, and y = 2/3.
The solution is (10/3, 2/3)
Answer:
c
Step-by-step explanation:
Answer:
36 Miles
Step-by-step explanation: 15 + 9 + 12 = 36 miles
