973° Вправа 973Алгебра 8 клас Тарасенкова 2025 · § 24 · стор. 291

973. Для кожного ряду чисел знайдіть обсяг, середнє значення та моду:

1) Упорядкуємо: ${\displaystyle 2}$; ${\displaystyle 10}$; ${\displaystyle 12}$; ${\displaystyle 18}$; ${\displaystyle 18}$; ${\displaystyle 20}$;

${\displaystyle n=}$${\displaystyle 6}$;

${\displaystyle x̄=}$${\displaystyle \frac{2+10+12+18+18+20}{6}=}$${\displaystyle \frac{\stackrel{40}{\cancel{80}}}{\underset{3}{\cancel{6}}}=}$${\displaystyle \frac{40}{3}=}$${\displaystyle 13\frac{1}{3}}$;

${\displaystyle Mo=}$${\displaystyle 18}$;

2) Упорядкуємо: ${\displaystyle -7,1}$; ${\displaystyle -4,3}$; ${\displaystyle 2,5}$; ${\displaystyle 4,3}$; ${\displaystyle 6,2}$; ${\displaystyle 7,1}$; ${\displaystyle 7,1}$;

${\displaystyle n=}$${\displaystyle 7}$;

${\displaystyle x̄=}$${\displaystyle \frac{\cancel{-7,1}+\left(\cancel{-4,3}\right)+2,5+\cancel{4,3}+6,2+\cancel{7,1}+7,1}{7}=}$${\displaystyle \frac{15,8}{7}\approx 2,26}$;

${\displaystyle Mo=}$${\displaystyle 7,1}$;

3) Упорядкуємо: ${\displaystyle -7}$; ${\displaystyle -\sqrt{2}}$; ${\displaystyle 0}$; ${\displaystyle \sqrt{2}}$; ${\displaystyle 2}$; ${\displaystyle 1+\sqrt{2}}$; ${\displaystyle 1+\sqrt{2}}$; ${\displaystyle 5}$;

${\displaystyle n=}$${\displaystyle 8}$;

${\displaystyle x̄=}$${\displaystyle \frac{-7\cancel{-\sqrt{2}}+0\cancel{+\sqrt{2}}+2+1+\sqrt{2}+1+\sqrt{2}+5}{8}=}$${\displaystyle \frac{2+2\sqrt{2}}{8}=}$${\displaystyle \frac{\stackrel{1}{\cancel{2}}(1+\sqrt{2})}{\underset{4}{\cancel{8}}}=}$${\displaystyle \frac{1+\sqrt{2}}{4}}$;

${\displaystyle Mo=}$${\displaystyle 1+\sqrt{2}}$.