The system is:
i) <span>2x – 3y – 2z = 4
ii) </span><span>x + 3y + 2z = –7
</span>iii) <span>–4x – 4y – 2z = 10
the last equation can be simplified, by dividing by -2,
thus we have:
</span>i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
iii) 2x +2y +z = -5
The procedure to solve the system is as follows:
first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:
i) 2x – 3y – 2z = 4
iii) 2x +2y +z = -5
2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.
Equalize:
3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9
similarly, using i and ii, eliminate x:
i) 2x – 3y – 2z = 4
ii) x + 3y + 2z = –7
multiply the second equation by 2:
i) 2x – 3y – 2z = 4
ii) 2x + 6y + 4z = –14
thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:
3y+2z+4=-6y-4z-14
9y+6z=-18
So we get 2 equations with variables y and z:
a) 5y+3z=-9
b) 9y+6z=-18
now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.
Let's use elimination method, multiply the equation a by -2:
a) -10y-6z=18
b) 9y+6z=-18
------------------------ add the equations:
-10y+9y-6z+6z=18-18
-y=0
y=0,
thus :
9y+6z=-18
0+6z=-18
z=-3
Finally to find x, use any of the equations i, ii or iii:
<span>2x – 3y – 2z = 4
</span>
<span>2x – 3*0 – 2(-3) = 4
2x+6=4
2x=-2
x=-1
Solution: (x, y, z) = (-1, 0, -3 )
Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:
check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
Answer:
We have that:
And we want to find the value of g(4)
Then we are evaluating the function g(x) in x = 4, this means that we need to locate all the "x" in g(x), and replace them by 4.
If we do that, we get:
Now we can just solve this to get:
If 
is the cumulative distribution function for 
, then
Then the probability density function for is 
:
The 
th moment of is
![E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5Enf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bn-1%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Let 
, so that 
and 
:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu%7De%5E%7B-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du)
Complete the square in the exponent:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B%5Cfrac12%28n%5E2-%28u-n%29%5E2%29%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac%7Be%5E%7B%5Cfrac12n%5E2%7D%7D%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du)
But 
is exactly the PDF of a normal distribution with mean and variance 1; in other words, the 0th moment of a random variable 
:
![E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1](https://tex.z-dn.net/?f=E%5BU%5E0%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du%3D1)
so we end up with
![E[Y^n]=e^{\frac12n^2}](https://tex.z-dn.net/?f=E%5BY%5En%5D%3De%5E%7B%5Cfrac12n%5E2%7D)
Answer:
24
Step-by-step explanation:
8 16 24 32 40...
6 12 18 24 30 36....
24 and 24.
