X // Y, Y // Z,
By inference X // Z
The third option.
For any distribution, the sum of the probabilities of all possible outcomes must be 1. In this case, we have to have
We're told that 
, and we're given other probabilities, so we have
The expected number of calls would be
![E[X]=\displaystyle\sum_xx\,P(X=x)](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2CP%28X%3Dx%29)
![E[X]=0\,P(X=0)+1\,P(X=1)+\cdots+4\,P(X=4)](https://tex.z-dn.net/?f=E%5BX%5D%3D0%5C%2CP%28X%3D0%29%2B1%5C%2CP%28X%3D1%29%2B%5Ccdots%2B4%5C%2CP%28X%3D4%29)
![E[X]=1.4](https://tex.z-dn.net/?f=E%5BX%5D%3D1.4)
Answer:
1.9 inches
Step-by-step explanation:
We need to utilise one important formula in this question which is the volume of a cylinder formula. We need to work out the height of the cylinder given the following information that the radius is 8 inches and the volume is 384 cubic inches. We can set up an equation to find the value of the height so,
→ π × r² × h = 384
⇒ Substitute in 8 for 'r'
→ π × 8² × h = 384
⇒ Simplify
→ π × 64 × h = 384
⇒ Divide both sides by 64 to isolate π and h
→ π × h = 6
⇒ Divide both sides by π to isolate 'h' and find the value of the height
→ h = 1.9098593171
The height of a cylinder with a volume of 384 cubic inches and a radius of 8 inches is 1.9 inches
Answer:
the third one
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
For each box, there is 1 defective lightbulb. Therefore, a delivery of 4 boxes will have 4 defective lightbulbs. A delivery of 2 boxes has 2 defective lightbulbs. 4 defective lightbulbs subtracted by 2 defective lightbulbs equals 2 more defective lightbulbs in 4 boxes than 2. Therefore, it is B.
