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Alexeev081
5 days ago
15

You have a ruler of length 1 and you choose a place to break it using a uniform probability distribution. Let random variable X

represent the length of the left piece of the ruler. X is distributed uniformly in [0, 1]. You take the left piece of the ruler and once again choose a place to break it using a uniform probability distribution. Let random variable Y be the length of the left piece from the second break. (a) Find the conditional expectation of Y given X, E(Y\X). (b) Find the unconditional expectation of Y. One way to do this is to apply the law of iterated expectation which states that E(Y) = E(E(Y\X)). The inner expectation is the conditional expectation computed above, which is a function of X. The outer expectation finds the expected value of this function. (c) Compute E(XY). This means that E(XY\X) = XE(Y\X) (d) Using the previous results, compute cov(X, Y).
Mathematics
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2 months ago
In a GP if T3 = 18 and T6 = 486 Find:- T10
lawyer [12517]

Answer:

The 10th term in the geometric progression is 29.

Step-by-step explanation:

Given: In a geometric series, [T3 = 18] and [T6 = 486].

To find: The term [T10]?

Solution:

A geometric sequence takes the form [a, ar, ar^2,...]

Where, a represents the first term, and r denotes the common ratio.

The nth term is expressed as [Tn = a * r^(n-1)]

From the information provided: [T3 = a * r^2 = 18]

And [T6 = a * r^5 = 486]

By dividing the second equation by the first:

[(a * r^5) / (a * r^2)] = 486 / 18

[r^3 = 27]

Taking the cube root provides: r = 3.

Inserting r into one of the equations allows us to solve for a.

Substituting r gives: [T3 = a * r^2 = 18]

Thus, the first term is a = 2, and the common ratio is r = 3.

The 10th term in the geometric progression is computed as:

[T10 = a * r^(10-1)]

[Thus, T10 = 29.]

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Response 2 centimeters per night

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