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tatiyna
3 years ago
8

The sum of the deviations about the mean always equals _________.

Mathematics
1 answer:
Cloud [144]3 years ago
5 0
It always equal 0 
hope this helps :)                                                            
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PLEASE HELP ASAP<br><br>Solve. <br><br>y &gt; 2x + 2<br>y ≤ -3/2 x - 5
Fynjy0 [20]
2x + 2 < - \frac{3}{2} x - 5 \\ 2x + \frac{3}{2} x < - 5 - 2 \\ 3.5x < - 7 \\ x< - 2
x < - 2 \\ x < \frac{y - 2}{2}
-2 = \frac{y - 2}{2} \\ y=-2
7 0
2 years ago
If you know the answers pls put this <br> 1<br> 2<br> 3<br> 4<br> 5
Hitman42 [59]

Answer:

1.6 kids

2.6

3.6

4.6

5.6

Step-by-step explanation:

if you count by 6 then you will see that im right

4 0
2 years ago
I need help please only help if you know the answer
vlada-n [284]

Answer:

D

Step-by-step explanation:

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%20%20%5Cdiv%203x%20-%202%20%3D%201%20%5Cdiv%203x%20-%204" id="TexFormula1" title="x \div 3x
Maurinko [17]

Hello from MrBillDoesMath!

Answer:

x = 1/2 (1 +\- i sqrt(23))

Discussion:

x \3x - 2 =   (x/3)*x - 2   =  (x^2)/3 - 2     (*)

1 \3x - 4   =  (1/3)x - 4                               (**)

(*) = (**)   =>

(x^2)/3 -2 = (1/3)x - 4        => multiply both sides by 3

x^2 - 6 = x - 12                 => subtract x from both sides

x^2 -x -6 = -12                   => add 12 to both sides

x^2-x +6  = 0

Using the quadratic formula gives:

x = 1/2 (1 +\- i sqrt(23))

Thank you,

MrB

4 0
3 years ago
Write a rational function that meets the following criteria:
Sliva [168]
You can get a vertical asymptote at x=1 using y = 1/(x-1)
You can generate a hole at x=3 by multiplying by (x - 3/(x - 3) which is undefined at x=3 but otherwise equals 1
You can move the horizontal asymptote up to y=2 by adding 2

y = (x - 3)/((x - 1)(x - 3)) + 2
5 0
3 years ago
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