Answer:
Kindly check explanation
Step-by-step explanation:
Given the data :
Technician __Shutdown
Taylor, T___4
Rousche, R _ 3
Hurley, H__ 3
Huang, Hu___2
Gupta, ___ 5
The Numbe of samples of 2 possible from the 5 technicians :
We use combination :
nCr = n! ÷ (n-r)!r!
5C2 = 5!(3!)2!
5C2 = (5*4)/2 = 10
POSSIBLE COMBINATIONS :
TR, TH, THu, TG, RH, RHu, RG , HHu, HG, HuG
Sample means :
TR = (4+3)/2 = 3.5
TH = (4+3)/2 = 3.5
THu = (4+2) = 6/2 = 3
TG = (4 + 5) = 9/2 = 4.5
RH = (3+3) = 6/2 = 3
RHu = (3+2) /2 = 2.5
RG = (3 + 5) = 8/2 = 4
HHu = (3+2) = 2.5
HG = (3+5) = 8/2 = 4
HuG = (2+5) / 2 = 3.5
Mean of sample mean (3.5+3.5+3+4.5+3+2.5+4+2.5+4+3.5) / 10 = 3.4
Population mean :
(4 + 3 + 3 + 2 + 5) / 5 = 17 /5 = 3.4
Population Mean and mean of sample means are the same.
This distribution should be approximately normal.
The problem can be translated into an equation that is something like 4/5 + 3/x = 1/2
we cannot have x equal to zero because the number can be infinite.
So the LCD here is 10x, so multiply both sides by that to get:
8x + 30 = 5x
Subtract 5x and 30 from both sides:
3x = -30
divide:
x = -10
The solution isn't zero so there is a solution.
the right answer is 7 and 4 because the angles are between the two lines that are not the transversal and are on diagonal opposite sides of the transversal.
Y=mx+b
m=slope
b=yintercept
given
slope=-1/4
yint=5
y=-1/4x+5
