Question 185058
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If <i>m</i> is the mid-term score and <i>f</i> is the score on the final exam, then Professor Williamson will give a final grade, <i>g</i>, of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{2}{3}(m) + \frac{1}{3}f = g]


But we are given that <i>m</i> = 48 and the minimum <i>g</i> we are looking for is 70, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{2}{3}(48) + \frac{1}{3}f = 70]


Just solve for <i>f</i> to get the lowest score that will put you in the stated range, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{2}{3}(48) + \frac{1}{3}f = 79]


Solve for <i>f</i> again to get the highest score that will keep the final grade in the stated range.


I think you will find that, unless the good Professor gives a final with more than 100 as a possible score, you can't get there from here.  If the maximum possible score on the final is 100, the best Wendy can do is a 65.33 -- and the moral of the story is study hard; study early.


If you are also asking about the previous problem, the equations are the same if you reverse the fractional coefficients.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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