Question

Ajutați-mă la acest exercițiu, va rog!
"Calculați media aritmetică ponderată a numerelor:"​

Answer 1

$M_{p}=\frac{\frac{48}{9} }{9}$$M_{p}=\frac{\frac{6}{3}+\frac{30}{9} }{9}$c $M_{p}=\frac{12*\frac{5}{6}+10*\frac{9}{5} }{22}$

$M_{p}=\frac{\frac{60}{6}+\frac{90}{5} }{22}$

$M_{p}=\frac{10+14}{22}$

$M_{p}=\frac{24}{22}$

$M_{p}= 1,09$

d $M_{p}=\frac{4*2,5+3,4*5}{9}$

$M_{p}=\frac{10+13,5}{9}$

$M_{p}=\frac{23,5}{9}$

$M_{p}=2,61$

d mai intai vom transforma numerele $0,(3)$ si $1,(3)$ in fractii ordinare

$0,(3)=\frac{3}{9}=\frac{1}{3}$

$1,(3)=\frac{13-3}{9} =\frac{10}{9}$

$M_{p}=\frac{6*\frac{1}{3}+3*\frac{10}{9} }{9}$

$M_{p}=\frac{\frac{6}{3}+\frac{30}{9} }{9}$

$M_{p}=\frac{\frac{48}{9} }{9}$

$M_{p}=\frac{48}{9} * \frac{1}{9}$

$M_{p}=\frac{48}{9}$

$M_{p}=5,33$