what are the extraneous variables eliminated by randomly selecting schools into the experiment and control groups?
1 answer:
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Answer:
x = 0; y = -6; z = 1
Step-by-step explanation:
You want to find the solution to the system of equations ...
- 9x+y-3z=-9
- 10x-y+2z=8
- -10x-y+4z=10
This set of equations is conveniently solved using the matrix row-reduction features of a scientific or graphing calculator, spreadsheet, or any of a number of apps or on-line calculators.
The attachment shows the solution to be ...
(x, y, z) = (0, -6, 1)
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<em>Additional comment</em>
Solving a system of three or more equations "by hand" often can be done by an ad hoc process. It can be done in systematic fashion using Gauss-Jordan reduction techniques, but that often gets messy. Similarly, Cramer's Rule can be used, but that math tends to involve more arithmetic operations than are really necessary.
Adding the first equation to the other two eliminates the y-variable and reduces the system to two equations in two unknowns.
(9x +y -3z) +(10x -y +2z) = (-9) +(8) . . . . adding [1] and [2]
19x -z = -1 . . . . simplified
(9x +y -3z) +(-10x -y +4z) = (-9) +(10) . . . . adding [1] and [3]
-x +z = 1 . . . . simplified
At this point, we could graph the two equations, or we can proceed algebraically.
Adding these two equations gives ...
(19x -z) +(-x +z) = (-1) +(1)
18x = 0 ⇒ x = 0
Using the second of the reduced equations, we find ...
-x +z = 1
0 +z = 1 ⇒ z = 1
And using the second of the original equations, gives us ...
10(0) -y +2(1) = 8
-y = 6 . . . . . subtract 2
y = -6 . . . . multiply by -1
Then the solution is (x, y, z) = (0, -6, 1), as above.
