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sattari [20]
3 years ago
7

Given the following discrete uniform probability distribution, find the expected value and standard deviation of the random vari

able. Round your final answer to three decimal places, if necessary.
Probability Distribution
x 0 1 2 3 4 5 6 7 8 9 10
P(X=x) 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11 1/11
Mathematics
1 answer:
katovenus [111]3 years ago
3 0

The expected value is

E[X]=\displaystyle\sum_xx\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x=\dfrac{0+1+\cdots+9+10}{11}=\dfrac{55}{11}=\boxed{5}

The standard deviation is the square root of the variance, which is

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x^2=\dfrac{0^2+1^2+\cdots+9^2+10^2}{11}=\dfrac{385}{11}=35

so that

V[X]=35-5^2=10

making the standard deviation

\sqrt{V[X]}=\sqrt{10}\approx\boxed{3.16}

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olga nikolaevna [1]

6(3x+5)

6*3x=18x

6*5=30

<u>18x+30</u> is your answer! I hope this helped! :)

5 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Could someone please help me with this?
allsm [11]

Answer:

21.5 mm^2

Step-by-step explanation:

First find the area of the square

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Then find the area of the circle

d = 10 so r = d/2 = 10/2 =5

A = pi r^2

A = 3.14 (5)^2 =78.5

The difference is the shaded region

100-78.5 =21.5

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3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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2 years ago
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