Найдите среднее арифметическое чисел 1) 10,1; 9,4 и 9,9;

Решите задачу:

$11 \frac{1}{2} + 9.7 + 9 \frac{4}{5} + 10.2 = 20 \frac{5 + 8}{10} + 19.9 = 20 \frac{13}{10} + 19.9 = 21.3 + 19.9 = 41.2 \div 4 = 10.3$

$\frac{9}{20} + 0.65 + \frac{2}{5} = \frac{9 + 8}{20} + \frac{65}{100} = \frac{17}{20} + \frac{13}{20} = \frac{30}{20} = \frac{3}{2} \div 3 = \frac{3}{2 \times 3} = \frac{1}{2} = 0.5$
$5.9 + 6.1 + 6 + 4.8 = 22.8 \div 4 = 5.7$
$3.6 + 3 \frac{3}{4} + 3.66 = 7.26 + 3.75 = 11.01 \div 3 = 3.67$

$5.21 + 4.98 + 5.14 + 8.67 = 24 \div 4 = 6$