Is useful<span> in </span>proving<span>various theorems about </span>triangles<span> and other polygons.
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Answer:
0.8 Not Outlier
1.1 Not Outlier
10.2 Not Outlier
10.9 Not Outlier
Solution:
Arranging the numbers in ascending order:
0.8 1.1 4.9 5.2 5.8 5.9 6.1 6.1 7.4 10.2 10.9
we can see that the median is 5.9.
We can find the first quartile Q1 by getting the median in the lower half of the data
0.8 1.1 4.9 5.2 5.8
that is, Q1 = 4.9
We can find the third quartile Q3 by getting the median for the upper half of the data
6.1 6.1 7.4 10.2 10.9
that is, Q3 = 7.4
We subtract Q1 from Q3 to find the interquartile range IQR.
IQR = Q3 - Q1 = 7.4 - 4.9 = 2.5
We can now calculate for the upper and lower limits:
upper limit = Q3 + 1.5*IQR = 7.4 + (1.5*2.5) = 11.15
lower limit = Q1 – 1.5*IQR = 0.8 - (1.5*2.5) = -2.95
There is no data point that lies above the upper limit and below the lower limit, therefore, there are no outliers in the data set.
Answer:
6.25 <7
Step-by-step explanation:
Answer:
8
Step-by-step explanation:
4-2=2
2x9=18
18+6=24
24/4=6
6+2=8
=8
Answer:
Dude same
Step-by-step explanation:
