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kvv77 [185]
4 years ago
15

Solve this equation5(-7x-3)- 8x=-230​

Mathematics
1 answer:
Brut [27]4 years ago
5 0

Simplifying

5(7x + -3) = 230

Reorder the terms:

5(-3 + 7x) = 230

(-3 * 5 + 7x * 5) = 230

(-15 + 35x) = 230

Solving

-15 + 35x = 230

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '15' to each side of the equation.

-15 + 15 + 35x = 230 + 15

Combine like terms: -15 + 15 = 0

0 + 35x = 230 + 15

35x = 230 + 15

Combine like terms: 230 + 15 = 245

35x = 245

Divide each side by '35'.

x = 7

Simplifying

x = 7

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Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

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Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

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And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

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E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

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and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

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And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

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