Question

media aritmetică a trei numere nqturale x , y , z , este 28 . Determinati sumq lor

Answer 1

(a+b+c):3= 28
a+b+c= 28*3=84

Answer 2

$m.a. ( a_{1}, a_{2}, a_{3} ... a_{n} ) = \frac{ a_{1} + a_{2} + a_{3}+...+ a_{n} }{n} \\ m.a. (x,y,z) = \frac{x+y+z}{3} = 28. \\ =\ \textgreater \ x+y+z = 28*3 \\ \boxed {x+y+z= 84}$