Answer:
Step-by-step explanation:
9.)
z + 3 > 2/3
z > -7/3
You want an unshaded dot on the point -7/3 going towards the right.
10.)
1/2 ≤ c - 3/4
5/4 ≤ c
You want a shaded dot on the point 5/4 going towards the left.
2, 6, 18, 54, 162, 486...
Start with the number 2.
![2*3=6 \\ 6*3=18 \\ 18*3=54 \\ 54*3=162 \\ 162*3=486](https://tex.z-dn.net/?f=2%2A3%3D6%20%5C%5C%206%2A3%3D18%20%5C%5C%2018%2A3%3D54%20%5C%5C%2054%2A3%3D162%20%5C%5C%20162%2A3%3D486)
And continue with that pattern from there.
Answer:
y = -2
Step-by-step explanation:
Put x as -19 and solve for y.
2(-19) - 5y = -28
Multiply.
-38 - 5y = -28
Add 38 on both sides of the equation.
-5y = -28 + 38
-5y = 10
Divide both sides by -5.
y = 10/-5
y = -2
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.
![IQR= Q_3 -Q_1](https://tex.z-dn.net/?f=IQR%3D%20Q_3%20-Q_1)
The standard deviation is a measure of dispersion obatined from the sample variance and is given by:
![s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%3D%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
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.