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pogonyaev
3 years ago
5

Can someone solve this plz

Mathematics
1 answer:
galina1969 [7]3 years ago
7 0
-1+-6=-7

The small line shows the number -1 going back one. The bigger line shows a continuation from the smaller line “stopping” at -7.
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A random variable X with a probability density function () = {^-x > 0
Sliva [168]

The solutions to the questions are

  • The probability that X is between 2 and 4 is 0.314
  • The probability that X exceeds 3 is 0.199
  • The expected value of X is 2
  • The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

P(x) = \int\limits^a_b {f(x) \, dx

So, we have:

P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx

Using an integral calculator, we have:

P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2

Expand the expression

P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx

Using an integral calculator, we have:

P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3

Expand the expression

P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

E(x) = \int\limits^a_b {x * f(x) \, dx

So, we have:

E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx

This gives

E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx

Using an integral calculator, we have:

E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx

This gives

E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx

Using an integral calculator, we have:

E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

#SPJ1

<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

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2 years ago
How many diagonals can be constructed from one vertex of an n-gon? State your answer in terms of n and, of course, justify your
Rasek [7]

Polygon Diagonals. The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides. For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n(n-3).

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What is the length of AB?
IceJOKER [234]

Answer:

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Step-by-step explanation:

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Which of the following shows the graph of the inequality y &gt; –2x + 9?
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It will be a horizontal line at y=7 and shaded above. If you have pictures of the graph I can show which one it would be.
7 0
3 years ago
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Find x .<br><br> 8<br> 4√2<br> √8<br> 16
love history [14]

Answer:

8

Step-by-step explanation:

(8\sqrt{2}) ^{2} = 128 \\128 / 2 = 64\\x= \sqrt{64} = 8

8 0
3 years ago
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