Answer:
5x-y-36=0
Step-by-step explanation:
y+4=5x-40
5x-y=36
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
Hey there,
To solve this problem, let us first define what is mean and median. Mean is the average of all the numbers in the data set while the median is the number in the middle of the data set in ascending order. If we create a box plot for the data of Rome and New York, we can see that there is an outlier in the data for New York. Since New York has an outlier, so the mean is not a good representation on the central tendency of the data. The mean is skewed (distorted) by the outlier. So in this case it is better to use the median. While the Rome data is nice and symmetrical, it does not seem to have an outlier, so we can use the mean for this data set.
Therefore the answer is:
The Rome data center is best described by the mean. The New York data center is best described by the median
Hoped I Helped
Answer:
<h3>
Sample Answer:<em> </em></h3><h3><em>A linear function has a constant rate of change, while a nonlinear function does not. For a table of values to be linear, the outputs must have a constant rate of change as the inputs increase by 1. On a graph, the function must be a straight line to be linear.</em></h3>
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