Answer:
Outlier therefore could only be values below - 12.75
or could only be values above + 121.125
Step-by-step explanation:
0, 4, 6, 14, 17
inner quartile range of 0 - 17 is 1/2 of 17 subtracted from the higher number = 17 - 1/2 of 8.5 = 8.5 - 4.25 = 4.25 - 4.25 x 3
= 4.25 to 12.75 for inner quartile
inner quartile range is 12.75-4.25 = 8.5
We then 1.5 x 8.5 to show the outlier
= 12.75 meaning there is no outlier if is below.
Lower quartile fences = 4.25 - 1.5 = 2.75
or -1.5 x 8.5 (the range) = -12.75
Upper quartile fence = 12.75 + 1.5 = 14.25 x 8.5 = 121.125 this would be an outlier if it is 12.75 higher than 121.125 or 12.75 lower than 5.50.
Outlier therefore could only be values below - 12.75
or could only be values above + 121.125
An observation is considered an outlier if it exceeds a distance of 1.5 times the interquartile range (IQR) below the lower quartile or above the upper quartile. The values of the lower quartile - 1.5 x IQR and upper quartile + 1.5 x IQR are known as the inner fences.
An observation is an outlier if it falls more than above the upper quartile or more than below the lower quartile. The minimum value is so there are no outliers in the low end of the distribution. The maximum value is so there are no outliers in the high end of the distribution.
135000 meters and 135 kilometers
450 * 300
135000 then you convert -> 135
Parallel to 2x+y=-5
2x+y-2x=-5-2x
y=-2x-5
y=mx+b
Slope: m=-2
As the line is parallel must have the same slope:
m=-2
And it has an x-intercept of 2: when x=2, y=0→Point: P1=(2,0)=(x1,y1)
x1=2, y1=0
y-y1=m(x-x1)
y-0=(-2)(x-2)
y=-2x+4
In standard form:
y+2x=-2x+2x+4
2x+y=4
Answer: The equation of the line, in standard form, that passes through the origin and is parallel to x+y=6 is x+y=0
Not correct.. i should choose 3 .-3x+6
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<em>X</em><em>=</em><em>1</em><em>2</em><em> </em><em>°</em><em>ANSWER</em>
