Um, I don't know if I actually answered the question, but I think that the second one is cheaper to rent.
Answer:
x < -1.5 U x > 9.6
Step-by-step explanation:
12x + 7 < -11
x < -1.5
5x - 8 > 40
5x > 48
x > 9.6
1 1/2 I think is the answer
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.
b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>
They both have an answer that is bigger than the numbers being added or multiplied together <span />
