Select ALL statements that are true about the linear equation.
1 answer:
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Answer:
B.
Step-by-step explanation:
To simplify something that looks like you would multiply the top and bottom by the conjugate of the bottom. So you multiply the top and bottom for this problem I just made by:
.
If you had , then you would multiply top and bottom the conjugate of
which is
.
The conjugate of a+b is a-b.
These have a term for it because when you multiply them something special happens. The middle terms cancel so you only have to really multiply the first terms and the last terms.
Let's see:
(a+b)(a-b)
I'm going to use foil:
First: a(a)=a^2
Outer: a(-b)=-ab
Inner: b(a)=ab
Last: b(-b)=-b^2
--------------------------Adding.
a^2-b^2
See -ab+ab canceled so all you had to do was the "first" and "last" of foil.
This would get rid of square roots if a and b had them because they are being squared.
Anyways the conjugate of is
.
This is the thing we are multiplying and top and bottom.
Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.
The standard deviation is a measure of dispersion obatined from the sample variance and is given by:
Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.