Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1215

\[\boxed{\mathbf{1215}\mathbf{.}}\]

\[R = 2 - ( - 8) = 2 + 8 = 10;\]

\[Mo_{1} = - 8;\text{\ \ }Mo_{2} = - 5;\]

\[Me = \frac{- 5 + ( - 5)}{2} = - \frac{10}{2} = - 5;\]

\[\overline{X} = \frac{- 8 - 8 - 5 - 5 + 0 + 2}{6} =\]

\[= - \frac{24}{6} = - 4.\]

\[R = 9 - ( - 4) = 9 + 4 = 13;\]

\[Mo_{1} = - 4;\text{\ \ }Mo_{2} = 9;\]

\[Me = \frac{0 + 2}{2} = \frac{2}{2} = 1;\]

\[\overline{X} = \frac{- 4 - 4 + 0 + 2 + 9 + 9}{6} =\]

\[= \frac{12}{6} = 2.\]