- A$7$
- ✓$10$
- C$13$
- D$17$
$ = \left( {\frac{{\log 512}}{{\log 3}} \times \frac{{\log 9}}{{\log 4}} - \frac{{\log 3}}{{\log 4}} \times \frac{{\log 8}}{{\log 3}}} \right)$
$×$ $\left( {\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} - \frac{{\log 3}}{{\log 8}} \times \frac{{\log 4}}{{\log 3}}} \right)$
$= \left( {\frac{{\log {2^9}}}{{\log 3}} \times \frac{{\log {3^2}}}{{\log {2^2}}} - \frac{{\log {2^3}}}{{\log {2^2}}}} \right)$ $×$ $\left( {\frac{{\log {2^2}}}{{\log 2}} - \frac{{\log {2^2}}}{{\log {2^3}}}} \right)$
$= \left( {\frac{{9 \times 2}}{2} - \frac{3}{2}} \right)\,\left( {2 - \frac{2}{3}} \right) = \frac{{15}}{2} \times \frac{4}{3} = 10$.
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| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
$\frac{\sum_{k=1}^{12}\left|\alpha_{k+1}-\alpha_k\right|}{\sum_{k=1}^3\left|\alpha_{4 k-1}-\alpha_{4 k-2}\right|}$ is