Question 465040
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If the mean of *[tex \Large n] numbers is *[tex \Large \mu], then the sum of the *[tex \Large n] numbers is *[tex \Large n\cdot\mu]


Next you have the mean of *[tex \Large n\ -\ 1] numbers is *[tex \Large \mu_1], hence the sum of the *[tex \Large n\ -\ 1] numbers is *[tex \Large (n\ -\ 1)\mu_1]


The difference in the two sums is the number removed.  So *[tex \Large (5\,\cdot\,6)\ -\ (4\,\cdot\,7)] is the value you seek -- given correctly performed arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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