Question

Aflati media aritmetica a numerelor
a)1/6;2/9;3/4
b)3/7;1/5;2;6
c)1,2;0,15;3,4
d)1,(5);0,1(5);1,5

Answer 1

a)(1/6+2/9+3/4)/3=[(6+8+27) /36]/3=41/108 =0,37(962)
b)(3/7+1/5+2+6)/4=(15+7+70+42)/4=134/4 =67/2=33,5
c)(1,2+0,15+3,4)/3 = 4,75/3=1,58(3)
d)[1,(5)+0,1(5)+1,5]/3= [(15-1)/9+(15-1)/90+15/10]/3=(14/9+14/90+15/10)/3=
=(140+14+135)/270=289/270=1,0(703)

Answer 2

$a).m_a= \frac{ \frac{1}{6} + \frac{2}{9}+ \frac{3}{4} }{3} = \frac{ \frac{6}{36} + \frac{8}{36} + \frac{27}{36} }{3} = \frac{ \frac{41}{36} }{3} = \frac{41}{36} :3= \frac{41}{36} * \frac{1}{3} = \frac{41}{108}$

$b).m_a= \frac{ \frac{3}{7} + \frac{1}{5} +2+6}{4} = \frac{ \frac{15}{35} + \frac{7}{35} + \frac{70}{35} + \frac{210}{35} }{4} = \frac{ \frac{302}{35} }{4} = \frac{302}{35} :4= \frac{302}{35} * \frac{1}{4} = \frac{302}{140} = \frac{151}{70}$

$c).m_a= \frac{1,2+0,15+3,4}{3} = \frac{4,75}{3} =1,58(3)$

$d).m_a= \frac{1,(5)+0,1(5)+1,5}{2} = \frac{ \frac{14}{9} + \frac{14}{90} + \frac{15}{10} }{3} = \frac{ \frac{140}{90} + \frac{14}{90} + \frac{135}{90} }{3} = \frac{ \frac{289}{90} }{3} = \frac{289}{90} :3= \\ = \frac{289}{90}* \frac{1}{3} = \frac{289}{270}$