Step-by-step explanation:
In the case of a discrete probability distribution of a random variable X, the mean is equal to the sum over every possible value weighted by the probability of that value; that is, it is computed by taking the product of each possible value x of X and its probability p(x), and then adding all these products together
3x - y + z = 5 . . . (1)
x + 3y + 3z = -6 . . . (2)
x + 4y - 2z = 12 . . . (3)
From (2), x = -6 - 3y - 3z . . . (4)
Substituting for x in (1) and (3) gives
3(-6 - 3y - 3z) - y + z = 5 => -18 - 9y - 9z - y + z = 5 => -10y - 8z = 23 . . (5)
-6 - 3y - 3z + 4y - 2z = 12 => y - 5z = 18 . . . (6)
(6) x 10 => 10y - 50z = 180 . . . (7)
(5) + (7) => -58z = 203
z = 203/-58 = -3.5
From (6), y - 5(-3.5) = 18 => y = 18 - 17.5 = 0.5
From (4), x = -6 - 3(0.5) - 3(-3.5) = -6 - 1.5 + 10.5 = 3
x = 3, y = 0.5, z = -3.5
Answer:
No
Reasoning:
If something is a perfect cube, it is able to be put under a cube root (![\sqrt[3]{..}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B..%7D)
) and will result in an integer (a non-decimal number > 0, basically).
So let's calculate ![\sqrt[3]{48}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B48%7D)
, and see if the result is an integer.
= 3.634.......
As you can see, the result is not an integer, therefore 48 is not a perfect cube.
Move -4 to the other side
7x=24+4
7x=28
x=4
a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
