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pashok25 [27]
2 years ago
7

The value of

n%20%5E%7B-1%7D%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29" id="TexFormula1" title="\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\sin ^{-1}\left(\frac{1}{2}\right)" alt="\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\sin ^{-1}\left(\frac{1}{2}\right)" align="absmiddle" class="latex-formula"> is​
Mathematics
1 answer:
Mademuasel [1]2 years ago
6 0

Answer:

\displaystyle \frac{\pi}{6}

Explanation:

According to the unit circle, \displaystyle \frac{\pi}{3} and \displaystyle \frac{\pi}{6} are the radian measures you are looking for because all <em>y-coordinates</em> are associated with the <em>sine</em> function, so when evaluating, you will arrive at this:

\displaystyle \boxed{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6}

I am joyous to assist you at any time.

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For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.

For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.
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Is the following a function? (plz help ;-;)
pogonyaev

Answer:

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