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:
Volume = 12 *6 *8 , Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 )
Step-by-step explanation:
Given : Cuboid with length 12 , width 6 and height 8 units.
To find : Drag each expression to show whether it can be used to find the volume, surface area, or neither.
Solution : We have given Cuboid with
Length = 12 units ,
Width = 6 units
Height = 8 units.
Volume of cuboid = length * width * height .
Volume = 12 *6 *8.
Surface area = 2 ( l *w + h *+w *h)
Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 ).
None = 12 +6 +8.
Therefore, Volume = 12 *6 *8 , Surface area = 2 ( 12 *6 + 8 *12 + 6* 8 ) .
Hello! I believe she would have 58,000 at the end of her fourth year. Hope this helps. :)
Answer: Choice A) mean, there are no outliers
Have a look at the image attached below. I made two dotplots for the data points. The blue points represent bakery A. The red points represent bakery B. For any bakery, the points are fairly close together. There is no point that is off on its own. So there are no outliers, making the mean a good choice for the center. If there were outliers, then the median is a better choice. The mean is greatly affected by outliers, while the median is not.
The first, fourth and fifth ones are linear.
The second, third and sixth are all nonlinear.
You can tell this by the fact that their lead exponents are not 1.
