B

What is the total count of all integer solutions of the all following equations? \[ \begin{aligned} \log_{\frac12}\!\left(\log_2⁡x\right)&=-1 \\ \log_5\!\left(\log_{\frac15}⁡x\right)&=0 \\ -\log_{\frac13}\!\left(\log_{\frac12}⁡x\right)&=1 \end{aligned}\]

The number \( \left(\frac{27^{-4}\cdot8^{-4}}{16^{-2}\cdot9^{-5}}\right)^{-3} \) equals:

Express \( \frac{\sqrt[3]5\sqrt[6]5}{3\cdot25^3+2\cdot125^2} \) as a power of \( 5 \).

In the tables below the absent hours in lessons of \( 16 \) boys and \( 14 \) girls from one class for half of a year are listed. Use the variance of the number of absent hours to find out the students of which gender had the absence more balanced. I.e. choose the group with more balanced absence and the correct variance of the number of absent hours. The variance is rounded to two decimal places. \[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Girl's ID} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text{Absent hours} & 27 & 61 & 38 & 61 & 17 & 39 & 61 \\\hline \\\hline \text{Girl's ID} & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\\hline \text{Absent hours} & 25 & 21 & 52 & 16 & 34 & 9 & 25 \\\hline \end{array} \] \[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Boy's ID} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline \text{Absent hours} & 67 & 56 & 26 & 36 & 27 & 55 & 17 & 34 \\\hline \\\hline \text{Boy's ID} & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\\hline \text{Absent hours} & 54 & 46 & 13 & 48 & 21 & 49 & 18 & 14 \\\hline \end{array} \]