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:
50
Step-by-step explanation:
6/100=3/y
Let y be the total number of items on the test
[cross multiply]
6×y =3×100
6y = 300
divide both sides by 6
y = 50
Answer:
F
Step-by-step explanation:
Your welcome my guy :D
Answer:
The measure of segment AC is 36 units
Step-by-step explanation:
- The mid-point divides the segment into two equal parts in length
- B is the mid point of segment AC
- That means B divides segment AC into two equal parts in length
∴ AB = BC
∵ AC = 5x - 9
∵ AB = 2x
- The two parts AB and BC are equal in length
∴ BC = 2x
∵ AC = AB + BC
- Substitute the values of AB and BC in the expression of AC
∴ AC = 2x + 2x
∴ AC = 4x
∵ AC = 5x - 9
- Equate the two values of AC
∴ 5x - 9 = 4x
- Add 9 to both sides
∴ 5x = 4x + 9
- Subtract 4x from both sides
∴ x = 9
- Substitute the value of x in any expression of AC
∵ AC = 4x
∵ x = 9
∴ AC = 4(9) = 36
* The measure of segment AC is 36 units
