Zero is less than negative fifteen decreased by y, solve for y.
1 answer:
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Average rate of the function: 3
Step-by-step explanation:
The function in this problem is
The average rate of change of a function over an interval is given by
where
are the values of the function calculated in
respectively
For the function in this problem we have:
, so
, so
So, the average rate of change is
Learn more about rate of change:
#LearnwithBrainly
The box-and-whisker plot is attached.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.
We first order the data from least to greatest:
67,68,68,69,70,72,73,74,75,75,76,78,79,79,80,83,84,85,85,88
Now we find the median. Since there are 20 data values, the median is between 75 and 76:
(75+76)/2 = 151/2 = 75.5
Now we find the Lower Quartile (LQ). We do this by finding the median of the lower half of data (from the median down). There are 10 values here; the median is between 70 and 72:
(70+72)/2 = 142/2 = 71
Find the Upper Quartile (UQ). Do this by finding the median of the upper half of data (from the median up). There are 10 values here; the median is between 80 and 83:
(80+83)/2 = 163/2 = 81.5
The middle bar of the box is over the median. The left hand side of the box is over LQ, and the right hand side of the box is over UQ. Now we draw a whisker from the box to the highest value, 88, and from the box to the lowest value, 67.