Question

media aritmetica a nr 1 pe 2; 7 pe 12; 8 pe3; 11 pe 6

Answer 1

$m_{a}= \frac{ \frac{1}{2}+ \frac{7}{12}+ \frac{8}{3}+ \frac{11}{6} }{4}$
$m_{a}= \frac{ \frac{6+7+32+22}{12} }{4}$
$m_{a}= \frac{67}{48}$