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!
F(t) = -5t^2 + 20t + 60
-5(t^2 - 4t - 12)
-5(t + 2) (t - 6)
t = -2 or t = 6
6 seconds
Steps
I’m using elimination so I’m trying to get rid of y
To do that I multiply the first equation by 2 and the second by -3. Then solve
4x + 6x = 12
15x + -6x = -12
—————————
19x + 0 = 0
Divid by 19 to get x to one side. 0/19 is 0 so
X= 0
Plug that into an equation so 0 + 3y = 6
Divid by 3 and y = 2
Answer x = 0 y = 2
<span>
(1/6)/ 1.5 =
(1/6)/(15/10)=
1/6*10/15=10/90=1/9 width</span>
Answer:
Literally use any answer which has a higher slope so . . .
y = 10x + 1
y = 1000000000000000000x + 1
and so on.
