Answer:28
Step-by-step explanation:
hope this helpes
Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.

- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )

E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
If a line's slope is

, the slope of the perpendicular line is

in this case, the given line has a slope 4, which is 4/1, so the perpendicular line has a slope/gradient
Answer:
a) 47
b) 3
c) 1
d) 2
e) 47
f) 3
Step-by-step explanation:
read the textbook
Look at points (0, -1) and (1, 5).
m = (y2 - y1)/(x2 - x1) = (5 - (-1))/(1 - 0) = 6/1 = 6
Slope = 6
Answer: 6