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.
The area of a triangle is 180
A direct variation is a mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. In other words, a direct variation is where

.
In answer a,

, so it is a direct variation.
In answer c,

, so it is a direct variation.
In answer d,

, so it is a direct variation,
Only answer b is left, which means the answer must be 'b'.
We also know 'b' is the answer because it cannot be expressed as <span>

. Instead, it is expressed as </span>

, which is not the same thing and is therefore not a direct variation.
Hope I helped, and let me know if you have any questions :)
The number in the blank box should be 44 because it said that "3 times x and 4 times y is 44".
You need to multiply the first equation by 3:
3x + 3y = 39.
Then, subtract the first equation from the second equation:
3x - 3x = 0
4y - 3y = y
44 - 39 = 5
Your new equation should look like this: y = 5
Now, substitute y = 5 into the first original equation:
x + 5 = 13
x = 8
Ans. (x, y) = (8, 5)