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:
y=-1/4x+5
Step-by-step explanation:
Use the point-slope form which is y-y1=m(x-x1) where m is the slope. In this case, y is equal to 3, and x is equal to 8, and the slope is equal to -1/4. Therefore, the equation will be equal to y-3=-1/4(x-8). Simplify that to get y-3=-1/4x+2. Simplify this again to get y=-1/4x+5. Therefore, the equation is y=-1/4x+5.
Both sides aren't equal because 1-sin^2x = cos^2x and cos^2x / 1+cosx can never equal cosx
Answer:
3
(
3
)
+
6
(
2
)
+
2
=
2
(
3
)
+
3
(
2
)
+
5
Step-by-step explanation:
Answer:
18
Step-by-step explanation:
