Question

Calculati media aritmetica ponderata a numerelor 0,25; 1,(3); 0,1(6) cu ponderile 6,2 respectiv 5

Answer 1

Explicație pas cu pas:

$m_{p} = \dfrac{6 \cdot 0,25 + 2 \cdot 1,(3) + 5 \cdot 0,1(6)}{6 + 2 + 5} = \\ = \dfrac{6 \cdot \dfrac{25}{100} + 2 \cdot \dfrac{13 - 1}{9} + 5 \cdot \dfrac{16 - 1}{90}}{13} \\ = \dfrac{6 \cdot \dfrac{1}{4} + 2 \cdot \dfrac{12}{9} + 5 \cdot \dfrac{15}{90}}{13} = \dfrac{ \dfrac{3}{2} + \dfrac{8}{3} + \dfrac{5}{6}}{13} \\ = \dfrac{9 + 16 + 5}{13 \cdot 6} = \dfrac{30}{13 \cdot 6} = \bf \dfrac{5}{13}$

Answer 2

$m _{a _{p}} = \tfrac{0.25\cdot6 + 1.(3)\cdot2 + 0.1(6)\cdot5}{6 + 2 + 5} \\ \\ m _{a _{p} } = \tfrac{ \frac{25}{100}\cdot6 + 1 \frac{3}{9} \cdot2 + \frac{16 - 1}{90}\cdot5 }{13} \\ \\ m _{a _{p} } = \tfrac{ \frac{1}{4}\cdot6 + \frac{12}{9}\cdot2 + \frac{15}{90}\cdot 5 }{13} \\ \\ m _{a _{p}} = \tfrac{ \frac{6}{4} + \frac{24}{9} + \frac{75}{90} }{13} \\ \\ m _{a _{p}} = \tfrac{ {}^{ {}^{3)} } \frac{3}{2} + {}^{ {}^{2)} } \frac{8}{3} + \frac{5}{6} }{13} \\ \\ m _{a_{ p} } = \tfrac{ \frac{9 + 16 + 5}{6} }{13} \\ \\ m _{a _{p} } = \tfrac{ \frac{30}{6} }{13} \\ \\ m _{a _{p}} = \tfrac{5}{13}$