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.
The designer must sell 200 hats for breaking even.
Step-by-step explanation:
Given,
Cost for material for each hat = $5.50
Selling price of each hat = $12.50
Cost spend on advertising = $1400
Let,
x represents the number of hats.
Revenue = 12.50x
Costs = 5.50x+1400
For breaking even,
Revenue = Cost
Dividing both sides by 7
The designer must sell 200 hats for breaking even.
Keywords: addition, division
Learn more about division at:
#LearnwithBrainly
If a function is even, then f(-x) = x.
If a function is odd, then f(-x) = -x.
y = x³ + x² → f(x) = x³ + x² → -f(x) = -(x³ + x²) = -x³ - x²
f(-x) = (-x)³ + (-x)² = [(-1)(x)]³ + [(-1)(x)]² = (-1)³x³ + (-1)²x²
= -1x³ + 1x² =-x³ + x²
f(-x) ≠ f(x) and f(-x) ≠ -f(x)
y = x³ + x² is not odd and not even
Answer: neither
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<u>Here are the steps to solve:</u>
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<em>Distribute negative sign:</em>
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<em>Add like terms:</em>
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Have a great day! Hope it helped!
