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Question Number 93691 by Aniruddha Ghosh last updated on 10/Jun/20

If Σ_(i = 1) ^(10) (x_i +4) = 60  then find the value of x^� .

$$\boldsymbol{\mathrm{If}}\:\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\boldsymbol{\mathrm{x}}_{\mathrm{i}} +\mathrm{4}\right)\:=\:\mathrm{60}\:\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\bar {\boldsymbol{{x}}}. \\ $$

Commented by hknkrc46 last updated on 15/May/20

Σ_(i=k) ^m c=(m−k+1)c  c=x+4 ; m=10 ; k=1  ⇒Σ_(i=1) ^(10) (x+4)=(10−1+1)(x+4)=60  ⇒10(x+4)=60⇒x+4=6⇒x=2

$$\underset{{i}={k}} {\overset{{m}} {\sum}}{c}=\left({m}−{k}+\mathrm{1}\right){c} \\ $$$${c}={x}+\mathrm{4}\:;\:{m}=\mathrm{10}\:;\:{k}=\mathrm{1} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{10}} {\Sigma}}\left({x}+\mathrm{4}\right)=\left(\mathrm{10}−\mathrm{1}+\mathrm{1}\right)\left({x}+\mathrm{4}\right)=\mathrm{60} \\ $$$$\Rightarrow\mathrm{10}\left({x}+\mathrm{4}\right)=\mathrm{60}\Rightarrow{x}+\mathrm{4}=\mathrm{6}\Rightarrow{x}=\mathrm{2} \\ $$

Commented by i jagooll last updated on 14/May/20

Σ_(i = 1) ^(10)  (x_i +4) = 60 .

$$\underset{\mathrm{i}\:=\:\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\left(\mathrm{x}_{\mathrm{i}} +\mathrm{4}\right)\:=\:\mathrm{60}\:.\: \\ $$

Commented by Aniruddha Ghosh last updated on 10/Jun/20

right. it was by mistake

Answered by Aniruddha Ghosh last updated on 14/May/20

Σ_(i=1) ^(10) (x+4) = x_1 +x_2 +x_3 +....+x_(10) +(4×10) = 60  ⇒ x_1 +x_2 +...........+x_(10 ) = 60−40  ⇒ x_1 +x_2 +...........+x_(10)  = 20  ∴ x^�  = ((x_1 +x_2 +x_3 +.....+x_(10) )/(10(frequency)))           = ((20)/(10))           = 2

$$\underset{\boldsymbol{{i}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\boldsymbol{{x}}+\mathrm{4}\right)\:=\:\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +....+\boldsymbol{{x}}_{\mathrm{10}} +\left(\mathrm{4}×\mathrm{10}\right)\:=\:\mathrm{60} \\ $$$$\Rightarrow\:\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +...........+\boldsymbol{{x}}_{\mathrm{10}\:} =\:\mathrm{60}−\mathrm{40} \\ $$$$\Rightarrow\:\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +...........+\boldsymbol{{x}}_{\mathrm{10}} \:=\:\mathrm{20} \\ $$$$\therefore\:\bar {\boldsymbol{{x}}}\:=\:\frac{\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +.....+\boldsymbol{{x}}_{\mathrm{10}} }{\mathrm{10}\left(\mathrm{frequency}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{20}}{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2} \\ $$$$ \\ $$

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