The probability density function of a continuous random variable $X$ that ranges from 1 to $e^{2}$ is given by $f(x)=\frac{1}{2x}.$ Find the value of $k$ such that $P(X>k)=\frac{1}{10}.$

\ln{\frac{9}{5}}

e^{\frac{9}{5}}

1+\ln{\frac{5}{9}}

1+e^{\frac{5}{9}}

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by Brilliant Staff

If the probability density function of a continuous random variable $X$ is given by $f(x)=\begin{cases} ax^2\ &(0\le x\le1) \\-\frac{a}{9}(x-1)+a\ &(1<x\le10),\end{cases}$ what is the value of $a?$

\frac{6}{29}

\frac{3}{31}

\frac{3}{29}

\frac{6}{31}

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If the probability density function of a continuous random variable $X\in[0,\frac{\pi}{16}]$ is given by $f(x)=a\sin8x,$ what is the variance of $X$ divided by $\pi-3 ?$

\frac{1}{32}

\frac{1}{64}

\frac{1}{8}

\frac{1}{4}

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If the probability density function of a continuous random variable $X$ is given by $f(x)=11x^{10}\ (0\le x\le1),$ what is $E[X]?$

\frac{10}{11}

\frac{11}{12}

\frac{12}{11}

\frac{11}{10}

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