Answer:
I, II, and IV
Step-by-step explanation:
(I) the probability of a Type I error.
(II) known as the alpha risk.
(IV) the sum of probabilities in the two tails of the normal distribution.
Given function f(x) which is described by "<span>Dawn writes 800 pages in 80 days".
From this function we can notice that Dawn writes 0 pages in 0 days, (i.e. the initial point of the function is point (0, 0).
Recall that the slope, m, of a line passing through points:

and

is given by

Function f(x) passes through points (0, 0) and (80, 800).
Thus the slope of function f(x) is given by

Give function g(x), passing through points (2, -6) and (4, 12), the slope is given by

It can be seen that the slope of f(x) and the slope of g(x) are close.
</span>
Answer: 11, 3, and 17
so C E and F
Step-by-step explanation:
Answer:
a) 0.9
b) Mean = 1.58
Standard Deviation = 0.89
Step-by-step explanation:
We are given the following in the question:
A marketing firm is considering making up to three new hires.
Let X be the variable describing the number of hiring in the company.
Thus, x can take values 0,1 ,2 and 3.

a) P(firm will make at least one hire)

Also,


b) expected value and the standard deviation of the number of hires.
![E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Cdisplaystyle%5Csum%20x_i%5E2P%28x_i%29%5C%5C%3D0%280.1%29%20%2B%201%280.4%29%20%2B%204%280.32%29%20%2B9%280.18%29%20%3D%203.3%5C%5CV%28x%29%20%3D%20E%28x%5E2%29-%5BE%28x%29%5D%5E2%20%3D%203.3-%281.58%29%5E2%20%3D%200.80%5C%5C%5Ctext%7BStandard%20Deviation%7D%20%3D%20%5Csqrt%7BV%28x%29%7D%20%3D%20%5Csqrt%7B0.8036%7D%20%3D%200.89)