The z is 5
−2⋅<em>z</em>=−8+18
-
2
⋅
z
=
-
8
+
18
Simplify :
−2⋅<em>z</em>=10
-
2
⋅
z
=
10
Dividing by the variable coefficient :
<em>z</em>=−
10
2
z
=
-
10
2
Simplify :
<em>z</em>=−5
z
=
-
5
The solution of equation
2⋅<em>z</em>−18−4⋅<em>z</em>=−8
2
⋅
z
-
18
-
4
⋅
z
=
-
8
is
[−5]
To find your answer, you just have to graph the points, then connect them to see what shape they make! :) Hope this helps!
Answer:
Third graph, I took the test
Step-by-step explanation:
In the original graph, R (0,3), S(3,3), T(4, -1) and Q(0, 1)
Reflecting it across y=-1, which is a horizontal line going through the point T, all the x coordinates stay the same,
the new confidantes are : R'(0, -5), S'(3, -5), T'(4, -1), Q(0', -3) (Notice; T stays where it is now, because it is right now the line y=-1)
Locate these new points, you have your graph.
Answer:
The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.
Step-by-step explanation:
We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.
The probability of one point of being over UCL=104 is:

The probability of one point of being under LCL=96 is:

Then, the probability of exhibit lack of control is:

The probability of having at least one point out of control in the first three points is:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.