Answer:
-384x - 120y
Step-by-step explanation:
General Concepts and Formulas:
Brackets
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
Steps:
1) Distribute
We must distribute the expression above by multiplying -12 by the coefficients and then adding the variables.
So first, multiple -12 by 32 and add the variable 'x' when getting the product:
-12(-384x - 10y)
Now that we have one part, let's solve the other part.
We now must multiple -12 by -10, which gets us 120, now we must plug in:
(-384x - 120y)
(Remove parenthesis)
⇒ -384x - 120y
Your inequality is (x being number of people):
175≥35+10x+3x
First, combine like terms so your new inequality is:
175≥35+13x
Then subtract 35 from both sides which makes it:
140≥13x
then divide 140 by 13 making it:
<span>10.769≥x
</span>But because you can't have .769 of a person the actual answer is 10
then you have to subtract 1 because he isn't including himself and there is the answer!:
9≥x
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.
