Answer:
3) C.
4) B.
5) C.
Step-by-step explanation:
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:

1000000.6 because I need sum points cuhhhhh
Answer:
It is -1/196 for those who want the quick answer and not the one below me calling the person dumb which I reported BTW because it is unacceptable.
Step-by-step explanation:
I know i dont work for Brainly but i apologize for the person who was called dumb.
Answer:
I guess that you want to model the elevation of Lake Sam Rayburn.
During the summer, it is 165 ft above the sea level (the sea level is our position 0ft).
If it does not rain, the elevation of the lake decreases by 0.5ft each week.
So if we assume that there is no rain, we can write the elevation fo the lake as a linear relationship with slope equal to -0.5ft and y-intercept equal to 165ft.
L(w) = 165ft - 0.5ft*w
Where w is the number of weeks without rain, if we have 0 weeks without rain, then the level of the lake remains constant at 165ft above sea level,
L(0) = 165ft - 0.