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3 years ago

What is the answer to -2x+3=x-6 ?

Mathematics

1 answer:

saveliy_v [14]3 years ago

5 0

Answer:

x=3

Step-by-step explanation:

-2x+3=x-6

-3x+3=-6

-3x=-9

-x=-3

x=3

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What’s the product (5r+2)(3r-4)

Marrrta [24]

$(5r+2)(3r-4) =15r^2-20r+6r-8=15r^2-14r-8$

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3 years ago

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What's 36% as an irreducible fraction? Choices: 4/25 17/25 9/25 11/37

masha68 [24]

36% = 36/100 divide both numerator and denominator by 4 gives you 9/25

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<img src="https://tex.z-dn.net/?f=3.4%20%5Ctimes%204.37" id="TexFormula1" title="3.4 \times 4.37" alt="3.4 \times 4.37" align="a

shusha [124]

Hello!

Answer: =14.858

Explanation:

Multiply by the numbers. Product it can also multiply by the numbers, fractions, decimals, factors, and multiples.

3.4 it has one decimal place.

4.37 it has two decimal place.

So, the answer it should be have a three decimal places.

$3.4*4.37=14.858$

Hope this helps!

Thanks :)

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abraham has 16 blue marbles and 18 rwd marbles. if he wamts to place them in identical groups without any marbles left over

VladimirAG [237]

Answer:

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Step-by-step explanation:

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3 years ago

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

2 years ago