ANSWER CHOICE C
We have been given two box plots representing attendance at a local movie theater and high school basketball games. WE are asked to choose the correct option about the measure the spread of the data represented by box plots.
Since we know that Standard deviation is not a good measure of spread in highly skewed data. IQR is better measure of spread for highly skewed data.
We can see that box plot representing attendance at movies is right skewed, therefore, mean will be greater than median and SD will not be a good measure of spread.
Box plot representing basket-ball games attendance is left skewed left,therefore, mean will be less than median and SD will not be a good measure of spread.
Our both box plots are highly skewed, therefore, the IQR is the best measurement of spread for games and movies and option C is the correct choice.
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<u>Answer:</u>
$6.00
<u>Step-by-step explanation:</u>
To solve this problem, we have to construct two separate equations for each family, and then use any of the methods for solving simultaneous equations.
Let's consider <em>h </em>to represent the cost of 1 hot dog, and <em>w </em>to mean the cost of 1 water bottle.
• For the first family:

We can rearrange the equation to make <em>w</em> the subject:
⇒ 
⇒ 
• For the second family:

Since we have previously obtained an expression for <em>w</em> in terms of <em>h</em>, we can substitute that expression for <em>w</em> in the above equation, and then solve for <em>h</em>:
⇒ 
⇒ 
⇒
[Multiplying both sides of the equation by 3]
⇒ 
⇒ 
⇒ 
∴ The price of one hot dog is $6.00.
Domain is X
Range is Y
Therefore {-3,-1,1}
Remember you don’t repeat numbers if there is more than one