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.
You just have to multiply the numerators and the denominators.
The answer is 14 over 56. Then, you need to simplify the question.
So, since 56 is divisible by 14, divide the numerator and the denominator by 14.
Then you get your answer, which is one-fourth.
Hope that helped!
Answer:
i would recomend tiger algeabra
Step-by-step explanation:
Step-by-step explanation:
As < A and < B are vertical angles so
<A = < B
5x + 12 = 6x - 11
6x - 5x = 12 + 11
x = 23
Hope it will help :)❤
Answer:
a
Step-by-step explanation:
Using Pythagoras' identity on the right triangle
let x be the other leg to be found, then
155² = 124² + x²
24025 = 15376 + x² ( subtract 15376 from both sides )
8649 = x² ( take the square root of both sides )
= x
Hence x = 93 cm ← length of other leg → a