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.
Answer: 36
Step-by-step explanation:
Answer:
21/2
Step-by-step explanation:
The answer can be determined by substracting the lengrh of the woodland trail from the Marsh trail
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converting the fraction to its simplest form gives 21/2
I think you can find it graphically by points through a table you create like:
let x = 0, and find the value of y
for example for the first equation when we let x = 0 you find y = 6 and be in the form of (0,6) then draw them or you can let x = -2 or -3 , etc as you like
to be more accurate you can find 3 pionts and start graphing them
these two equations are straight lines and they will intersect in a point
The question is somewhat poorly posed because the equation doesn't involve <em>θ</em> at all. I assume the author meant to use <em>x</em>.
sec(<em>x</em>) = csc(<em>x</em>)
By definition of secant and cosecant,
1/cos(<em>x</em>) = 1/sin(<em>x</em>)
Multiply both sides by sin(<em>x</em>) :
sin(<em>x</em>)/cos(<em>x</em>) = sin(<em>x</em>)/sin(<em>x</em>)
As long as sin(<em>x</em>) ≠ 0, this reduces to
sin(<em>x</em>)/cos(<em>x</em>) = 1
By definition of tangent,
tan(<em>x</em>) = 1
Solve for <em>x</em> :
<em>x</em> = arctan(1) + <em>nπ</em>
<em>x</em> = <em>π</em>/4 + <em>nπ</em>
(where <em>n</em> is any integer)
In the interval 0 ≤ <em>x</em> ≤ 2<em>π</em>, you get 2 solutions when <em>n</em> = 0 and <em>n</em> = 1 of
<em>x</em> = <em>π</em>/4 <u>or</u> <em>x</em> = 5<em>π</em>/4