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: -5.6
Step-by-step explanation:
9514 1404 393
Answer:
Step-by-step explanation:
Let x and y represent amounts invested at 6% and 9%, respectively.
y = 3x +58 . . . . . . . the amount invested at 9%
0.06x +0.09y = 1097.19 . . . . . . total interest earned
__
Substituting for y, we have ...
0.06x +0.09(3x +58) = 1097.19
0.33x + 5.22 = 1097.19 . . . . . . . . . simplify
0.33x = 1091.97 . . . . . . . . . . . . subtract 5.22
x = 3309 . . . . . . . . . . . . . . . . divide by 0.33
y = 3(3309) +58 = 9985
$3309 is invested at 6%; $9985 is invested at 9%.
Nick has tennis practice every sixth day = 6
Mark has tennis practice every 4th day = 4
They both had tennis practice on = 31st July
We have to find when will be the next time that they both have practice on the same day?
For this, we will find the LCM of 4 and 6
So, LCM of 4 and 6 is 12.
So, it will be 12 days after 31st July, that they will practice on the same day.
So it is 12th August.
Answer:
what is the unit and class
Step-by-step explanation:
