Answer:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:

And replacing we got:

Now we can find the second moment given by:

And replacing we got:

And the variance would be given by:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Answer:
1 minute and 8 seconds or 1:08
Step-by-step explanation:
by subtracting both at starting with 9:17-8:49 you will get a uneven decimal than you gonna have to give as much 60 seconds you can get out of it the 60 seconds equals a minute
Answer:
B. 1/6
Step-by-step explanation:
72 what in inches? please clarify
Answer:
the question is incomplete, so I looked for a similar one:
<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>
iodine's half life (we are given x, we need to find b):
0.5A₀ = A₀eᵇˣ
x = 8 days
we eliminate A₀ from both sides
0.5 = eᵇ⁸
ln 0.5 = ln eᵇ⁸
-0.69315 = b8
b = -0.69315 / 8 = -0.08664
since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):
0.1A₀ = A₀eᵇˣ
0.1 = eᵇˣ
ln 0.1 = bx
where b = -0.08664
x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days