GRE Question of the Day (January 24th)

By - Jan 24, 02:00 AM Comments [0]

Math

\({\frac{{1}}{{x^3}},\frac{{1}}{{x^2}},\frac{{1}}{{x}},x,x^2,x^3}\)

If \({-1<x<0}\) , what is the median for the six numbers in the list above ?

A. \({\frac{{1}}{{x}}}\)

B. \({x^2}\)

C. \({\frac{{x^2(x+1)}}{{2}}}\)

D. \({\frac{{x(x^2+1)}}{{2}}}\)

E. \({\frac{{x^2+1}}{{2x}}}\)

 

Correct Answer - - (click and drag your mouse to see the answer)

Question Discussion & Explanation

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