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nalin [4]
3 years ago
10

The ABX Company is interested in conducting a study of the factors that affect absenteeism among its≈α= 0.05

Mathematics
1 answer:
musickatia [10]3 years ago
4 0

Answer:

import pandas as pd

import import statsmodels.api as sm

dataframe = pd.read_csv(Your model)

mod = sm.OLS(formula = ABSENT ~ COMPLX + SENING+FS1+FS2+FS3+FS4)

res = mod.fit()

print(res.summary())

Step-by-step explanation:

Using python you can load the dataframe using pandas library. Once you have your pandas library imported to the system you can also import the statsmodels.api  module. What you do is this. You fit the model using the variables mentioned "COMPLX" ,"SENINV","FS1","FS2","FS3", once you fit the model you use .summary() and that will give you a summary of each coefficient and the level of significance,  the level of significance must be less than 5% in order to be significant. The code would look like this.

import pandas as pd

import import statsmodels.api as sm

dataframe = pd.read_csv(Your model)

mod = sm.OLS(formula = ABSENT ~ COMPLX + SENING+FS1+FS2+FS3+FS4)

res = mod.fit()

print(res.summary())

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2 years ago
Match the parabolas represented by the equations with their foci.
Elenna [48]

Function 1 f(x)=- x^{2} +4x+8


First step: Finding when f(x) is minimum/maximum
The function has a negative value x^{2} hence the f(x) has a maximum value which happens when x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2. The foci of this parabola lies on x=2.

Second step: Find the value of y-coordinate by substituting x=2 into f(x) which give y=- (2)^{2} +4(2)+8=12

Third step: Find the distance of the foci from the y-coordinate
y=- x^{2} +4x+8 - Multiply all term by -1 to get a positive x^{2}
-y= x^{2} -4x-8 - then manipulate the constant of y to get a multiply of 4
4(- \frac{1}{4})y= x^{2} -4x-8
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is 12- \frac{1}{4}=11.75

Hence the coordinate of the foci is (2, 11.75)

Function 2: f(x)= 2x^{2}+16x+18

The function has a positive x^{2} so it has a minimum

First step - x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4
Second step - y=2(-4)^{2}+16(-4)+18=-14
Third step - Manipulating f(x) to leave x^{2} with constant of 1
y=2 x^{2} +16x+18 - Divide all terms by 2
\frac{1}{2}y= x^{2} +8x+9 - Manipulate the constant of y to get a multiply of 4
4( \frac{1}{8}y= x^{2} +8x+9

So the distance of focus from y-coordinate is \frac{1}{8} to the north of y=-14
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: f(x)=-2 x^{2} +5x+14

First step: the function's maximum value happens when x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25
Second step: y=-2(1.25)^{2}+5(1.25)+14=17.125
Third step: Manipulating f(x)
y=-2 x^{2} +5x+14 - Divide all terms by -2
-2y= x^{2} -2.5x-7 - Manipulate coefficient of y to get a multiply of 4
4(- \frac{1}{8})y= x^{2} -2.5x-7
So the distance of the foci from the y-coordinate is -\frac{1}{8} south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when x=8.5 and y=79.25. The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when x=-2.75 and y=-10.125. Manipulating coefficient of y, the distance of foci from y-coordinate is \frac{1}{8} to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when x=1.5 and y=9.5. The distance of the foci from the y-coordinate is \frac{1}{8} to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

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