HELPPPPPP PLEASEEEEE!! I have no idea!

Mathematics

2 answers:

scZoUnD [109]3 years ago

8 0

K. should be your answer.

If t = 0.

Note that the equation has a t in all variables except for c, which will affect the outcome of h (or the first one, as t can change at all times).

If the time is set to 0, the maximum height will only be the height seen at the place where x = 0, which, instead of the true maximum height.

The t intercept will also change, and would not project as far, seeing that it can only 0, instead of t > 0.

hope this helps

mel-nik [20]3 years ago

7 0

I think it's H 3 only

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A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the

olchik [2.2K]

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:

$F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}$

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.

$F(X) = P ( 3=$

- The Expected Value can be determined by sum to infinity of CDF:

E(X) = Σ ( 1 - F(X) )

$E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\= \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\= \frac{1}{(n)} - \frac{1}{(n+1)}\\\\= \frac{1}{(n+1)} - \frac{1}{(n+ 2)}$