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:
243 = 3⁵
Step-by-step explanation:
Firstly we will find the factors of 243. On calculating it see that the factors of 243 are 3 x 3 x 3 x 3 x 3.
or
243 = 9 x 9 x 3
We need to write 243 as the product of primes.
The number that divides itself or 1 are prime numbers.
3 is the only prime number here such that,
243 = 3⁵
Hence, this is the required solution.
Simplify both sides then isolate the variable and it gives you y=3.6
You could first solve for y, assuming x is 0, and y = 6. Then solve for x, and that means y = 4.
