Category: Set Theory

Question Number 190573 by sami123 last updated on 06/Apr/23 $$\hat {{a}}\mathrm{2}+\mathrm{2}{ab}+\hat {{b}}\mathrm{2} \\ $$ Commented by a.lgnaoui last updated on 06/Apr/23 $$\mathrm{1}\bullet{pour}\:{editer}\:\:{exposant}\left({a}^{\mathrm{2}} \right) \\ $$$${dans}\:{clavier}\:{ecrire}\:\:{a}\:\:{puis}\:\:\:\:…

Question Number 124409 by oustmuchiya@gmail.com last updated on 03/Dec/20 $${write}\:\mathrm{A}'\cup\:{B}'\:{in}\:{a}\:{disjoint}\:{set} \\ $$ Answered by Ar Brandon last updated on 03/Dec/20 $$\left(\mathrm{A}\cup\mathrm{B}\right)−\left(\mathrm{A}\cap\mathrm{B}\right) \\ $$ Terms of…

Question Number 189091 by mnjuly1970 last updated on 12/Mar/23 $$ \\ $$$$\:\:\:\:\mathrm{1}\::\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\:\mathrm{1}\:\right)^{\:{n}} \mathrm{H}_{\:{n}} }{{n}^{\:\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\mathrm{2}\::\:\:\:\:\:\eta\:\left(−\mathrm{1}\:\right)=\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$…

Question Number 122436 by sahiljakhar04 last updated on 17/Nov/20 $$\boldsymbol{\mathrm{Mean}}\::\:\mathrm{It}\:\mathrm{is}\:\mathrm{found}\:\mathrm{by}\:\mathrm{adding}\:\mathrm{all}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{observation}\:\mathrm{and}\:\mathrm{dividing}\:\mathrm{it}\:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{observations}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{denoted}\:\mathrm{by}\:\bar {{x}}. \\ $$$$\mathrm{So},\:\bar {{x}}\:=\:\frac{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{i}} }{{n}}.\:\mathrm{For}\:\mathrm{an}\:\boldsymbol{\mathrm{ungrouped}}\:\boldsymbol{\mathrm{frequency}}\:\boldsymbol{\mathrm{distribution}},\:\mathrm{it}\:\mathrm{is}\:\bar {\boldsymbol{{x}}}\:=\:\frac{\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\boldsymbol{{f}}_{\boldsymbol{{i}}} \boldsymbol{{x}}_{\boldsymbol{{i}}} }{\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\boldsymbol{{n}}}…

Question Number 56744 by Umar last updated on 22/Mar/19 $$\mathrm{If}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{on}\:\mathrm{a}\:\mathrm{set}\:\mathrm{of}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\mathrm{R}=\left\{\left(\mathrm{x},\mathrm{y}\right):\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{0}\right\}, \\ $$$$\mathrm{find}\: \\ $$$$\:\:\mathrm{i}−\:\mathrm{R}\:\mathrm{in}\:\mathrm{roster}\:\mathrm{form} \\ $$$$\:\:\mathrm{ii}−\mathrm{Domain}\:\mathrm{of}\:\mathrm{R} \\ $$$$\:\:\mathrm{iii}−\mathrm{Range}\:\mathrm{of}\:\mathrm{R}\: \\ $$ Answered…