Answer:
2/6
Step-by-step explanation:
There are 6 sides, and two are green. That makes it 2/6
Answer:

Step-by-step explanation:
According to Angles of intersecting chords theorem:
∠GAJ = 
Where ∠GAJ = 50 , BD = 24 and GJ = ?
50 = 
Multiplying both sides by 2
100 = 24 + GJ
GJ = 100-24
GJ = 76 degrees
I hope this helps you
Volume =pi.r^2.h
Volume =3,14.8^2.10
Volume =31,4.64
Volume =200,96
Answer:
S = {0,2,3,4}
P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001
Mean = 0.879
Standard Deviation = 1.033
Step-by-step explanation:
Let the number of people having same birth month be = x
The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)
The number of ways of distributing their birthdays = 12⁴
The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)
P(X = 0) = 
P(X=0) = 0.573
P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)
P(X=2) = 
P(X=2) = 0.401
P(X=3) = 
P(x=3) = 0.025
P(X=4) = 
P(X=4) = 0.001
Mean, 

Standard deviation, ![SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}](https://tex.z-dn.net/?f=SD%20%3D%20%5Csqrt%7B%5Csum%20x%5E%7B2%7D%20P%28x%29%20-%20%5Cmu%5E%7B2%7D%7D%20%20%5C%5CSD%20%3D%5Csqrt%7B%20%5B%20%280%5E%7B2%7D%20%2A%200.573%29%20%2B%20%282%5E%7B2%7D%20%20%2A%200.401%29%20%2B%20%283%5E%7B2%7D%20%2A%200.025%29%20%2B%20%284%5E%7B2%7D%20%2A%200.001%29%5D%20-%200.879%5E%7B2%7D%7D)
SD = 1.033