Question

Determine o valor das médias quadrática, aritmética, geométrica e harmônica dos valores abaixo :

a) 2,4 e 8
b) 6 e 8

Answer 1

Resposta:

Explicação passo-a-passo:

Podemos calcular as médias usando as seguintes fórmulas (vou colocar as formulas para 3 números a,b,c):

Média aritmética:

$\textrm{MA} = \dfrac{a+b+c}3$

Média geométrica:

$\textrm{MG} = \sqrt[3]{abc}$

Média quadrática:

$\textrm{MQ} =\sqrt{ \dfrac{a^2 + b^2 + c^2}{3}$

Média harmônica:

$\textrm{MH} = \dfrac{3}{\dfrac 1a +\dfrac 1b + \dfrac 1c}$

Assim no seu problema fica:

a)

$\textrm{MA} = \dfrac{2+4+8}3 = \dfrac{14}3$

$\textrm{MG} = \sqrt[3]{2\times4\times 8} = 4$

$\textrm{MQ} =\sqrt{ \dfrac{2^2 + 4^2 + 8^2}{3}} = 2 \sqrt 7$

$\textrm{MH} = \dfrac{3}{\dfrac 12 +\dfrac 14 + \dfrac 18} = \dfrac{24}7$

b)

$\textrm{MA} = \dfrac{6+8}2 = 7$

$\textrm{MG} = \sqrt[2]{6\times 8} = 4\sqrt 3$

$\textrm{MQ} =\sqrt{ \dfrac{6^2 + 8^2}{2}} = 5 \sqrt 2$

$\textrm{MH} = \dfrac{2}{\dfrac 16 + \dfrac 18} = \dfrac{48}7$

Answer 2

Resposta:

a)

$M_{ar} = \dfrac{2+4+8}{3} = \dfrac{14}{3} \approx 4,7$

$M_{geo} = \sqrt[3]{2.4.8} = \sqrt[3]{64} = 4$

$M_{qua} = \sqrt{\dfrac{2^2+4^2+8^2}{3}} = \sqrt{\dfrac{4+16+64}{3}} = \sqrt{\dfrac{84}{3}} = \sqrt{\dfrac{4.3.7}{3}} = 2\sqrt{\dfrac{3.7}{3}} = 2\sqrt{7}$

$M_{har} = \dfrac{3}{\dfrac{1}{2} +\dfrac{1}{4}+ \dfrac{1}{8}} = \dfrac{3}{\dfrac{4+2+1}{8}} = \dfrac{3}{\dfrac{7}{8}} = \dfrac{24}{7}$

b)

$M_{ar} = \dfrac{6+8}{2} = \dfrac{14}{3} = 7$

$M_{geo} = \sqrt{6.8} = \sqrt{48} = \sqrt{4.12} = 2\sqrt{12}$

$M_{qua} = \sqrt{\dfrac{6^2+8^2}{2}} = \sqrt{\dfrac{36+64}{2}} = \sqrt{\dfrac{100}{2}} = \sqrt{50} = \sqrt{25.2} = 5\sqrt{2}$

$M_{har} = \dfrac{2}{\dfrac{1}{6} +\dfrac{1}{8}} = \dfrac{2}{\dfrac{4+3}{24}} = \dfrac{2}{\dfrac{7}{24}} = \dfrac{48}{7}$