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:
3.2 km * 1 / 1.609 mi / km * 5280 ft / mi = 10,501 ft
Check: 3280.84 ft / km * 3.2 km = 10,499 ft
If A = (0,2,3,4,9,11), B = {2,3,6,8,9,10) and C= {0,2,3,9), then (A-B) n(A-C) is
timofeeve [1]
Answer:
(A-B)n(A-C) ={ }
<h3><em>hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em>.</em></h3>
<em>Wishing</em><em> </em><em>a</em><em> </em><em>nice</em><em> </em><em>day</em><em>.</em>
Area = 1/2 * base * height
= 1/2 * 20 * 6
= 1/2 * 120
= 60 ft^2
