gmatophobia wrote:
\( -(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad-(\frac{1}{4})^{-\frac{1}{3}}\)
What is the median of the list of numbers above?
A.\(-(\frac{1}{2})^{-\frac{1}{3}}\)
B.\(-(\frac{1}{4})^{-\frac{1}{2}}\)
C.\(-(\frac{1}{4})^{-\frac{2}{3}}\)
D.\(-(\frac{1}{3})^{-\frac{1}{2}}\)
E.\(-(\frac{1}{4})^{-\frac{1}{3}}\)
Attachment:
Screenshot 2024-01-01 181734.png
\( -(\frac{1}{2})^{-\frac{1}{3}}, \quad -(\frac{1}{4})^{-\frac{1}{2}}, \quad -(\frac{1}{4})^{-\frac{2}{3}},\quad -(\frac{1}{3})^{-\frac{1}{2}}, \quad-(\frac{1}{4})^{-\frac{1}{3}}\)
let us convert the negative power to positivepower
\( -(2)^{\frac{1}{3}}, \quad -(4)^{\frac{1}{2}}, \quad -(4)^{\frac{2}{3}},\quad -(3)^{\frac{1}{2}}, \quad-(4)^{\frac{1}{3}}\)
Now let us raise all to same power1/6
\( -(2^2)^{\frac{1}{6}}, \quad -(4^3)^{\frac{1}{6}}, \quad -(4^4)^{\frac{1}{6}},\quad -(3^3)^{\frac{1}{6}}, \quad-(4^2)^{\frac{1}{6}}\)
[m] -(4)^{\frac{1}{6}},
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Statistics : Posted by chetan2u • on 01 Jan 2024, 04:55 • Replies 1 • Views 140
