Prove that triangle ABD is congruent to triangle CBD.

Mathematics

1 answer:

Ede4ka [16]2 years ago

8 0

Answer:

AB=AC, BD is shared side, angle 1 and angle 2 are equal as BD bisects angle ABC

So we have two sides and included angle equal.

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

According to the rule above we can say triangles ΔABD ≅ ΔCBD

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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) )

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