Q1: 14
Q3: 37
Interquartile: 37-14 = 23
THE ANSWER IS 23
(Brainliest please?)
Answer:
Set A's standard deviation is larger than Set B's
Step-by-step explanation:
Standard deviation is a measure of variation. One way to judge the value of standard deviation is by looking at the range of the data. In general, a dataset with a smaller range will have a smaller standard deviation.
The range of data Set A is 25-1 = 24.
The range of data Set B is 18-8 = 10.
Set A's range is larger, so we expect its standard deviation to be larger.
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The standard deviation is the root of the mean of the squares of the differences from the mean. In Set A, the differences are ±12, ±11, ±10. In Set B, the differences are ±5, ±3, ±1. We don't actually need to compute the RMS difference to see that it is larger for Set A.
Set A's standard deviation is larger than Set B's.
For domain 2x sqrt(2+x)>0
x>0,2+x>0,x>-2 combining
we get x>2
f'(x)=[1/{2x sqrt(2+x)}][{2x/(2 sqrt(2+x))}+2 sqrt(2+x)]
That is so confusing the way theyou word it sometimes is like what the...what?
