Question 639990
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If you know the value of *[tex \LARGE n\ -\ 1] items, and you let the value of the *[tex \LARGE n\text{th}] item be represented by *[tex \LARGE x], the average, *[tex \LARGE A], of all *[tex \LARGE n] items is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\sum_{i=1}^{n-1}\,V_i\right) +\ x}{n}\ =\ A]


Solving for *[tex \LARGE x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ nA\ -\ \left(\sum_{i=1}^{n-1}\,V_i\right)]


For your problem you want:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3A\ -\ \left(206\ +\ 176)]


Once you figure out what average you want him to achieve, you can solve for *[tex \LARGE x]


Extra Credit:  What range of values for a 3-game average would cause you to tell the person that they either needed a math tutor or were trying to cheat?


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