Question Number 147158 by puissant last updated on 18/Jul/21 Answered by Olaf_Thorendsen last updated on 18/Jul/21 $$\left.\mathrm{A}\right) \\ $$$$\mathrm{I}. \\ $$$$\mathrm{1}.\:\mathrm{Quel}\:\mathrm{est}\:\mathrm{le}\:\mathrm{nombre}\:\mathrm{de}\:\mathrm{tirages} \\ $$$$\mathrm{possibles}\:? \\ $$$$\mathrm{Dans}\:\mathrm{chaque}\:\mathrm{triplet},\:\mathrm{le}\:\mathrm{chiffre}\:\mathrm{de}…
Question Number 147130 by puissant last updated on 18/Jul/21 Commented by puissant last updated on 18/Jul/21 $$\mathrm{merci}\:\mathrm{prof} \\ $$ Commented by Olaf_Thorendsen last updated on…
Question Number 147018 by puissant last updated on 17/Jul/21 Commented by Tinku Tara last updated on 17/Jul/21 $$\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{rotate}\:\mathrm{image}\:\mathrm{before} \\ $$$$\mathrm{uploading}. \\ $$$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{for}\:\mathrm{your}\:\mathrm{cooperation} \\ $$ Answered…
Question Number 146927 by puissant last updated on 16/Jul/21 Answered by Olaf_Thorendsen last updated on 16/Jul/21 $$\lambda\:=\:\frac{{P}_{\overset{−} {{V}}} \left({M}\right)}{{P}_{{V}} \left({M}\right)} \\ $$$$\lambda\:=\:\frac{{P}_{\overset{−} {{V}}} \left({M}\right){P}\left(\overset{−} {{V}}\right){P}\left({V}\right)}{{P}_{{V}}…
Question Number 146910 by puissant last updated on 16/Jul/21 Answered by Olaf_Thorendsen last updated on 16/Jul/21 $$\mathrm{Soit}\:\mathrm{V}\:\mathrm{l}'\mathrm{evenement}\:“\mathrm{etre}\:\mathrm{vaccine}'' \\ $$$$\mathrm{et}\:\mathrm{M}\:\mathrm{celui}\:“\mathrm{etre}\:\mathrm{malade}''.\:\mathrm{Les} \\ $$$$\mathrm{evenements}\:\mathrm{contraires}\:\mathrm{seront}\:\mathrm{notes}\:\mathrm{par} \\ $$$$\overset{−} {\mathrm{V}}\:\mathrm{et}\:\overset{−} {\mathrm{M}}.…
Question Number 146799 by puissant last updated on 15/Jul/21 Answered by Olaf_Thorendsen last updated on 15/Jul/21 $$\mathrm{Il}\:\mathrm{y}\:\mathrm{a}\:\mathrm{2}\:\mathrm{photocopieuses}\:\mathrm{couleur}\:\mathrm{sur}\:\mathrm{5}\:: \\ $$$${p}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{Il}\:\mathrm{y}\:\mathrm{a}\:\mathrm{3}\:\mathrm{photocopieuses}\:\mathrm{N\&B}\:\mathrm{sur}\:\mathrm{5}\:: \\ $$$$\:{q}\:=\:\frac{\mathrm{3}}{\mathrm{5}}\:=\:\mathrm{1}−{p} \\ $$$$\mathrm{On}\:\mathrm{note}\:\mathrm{X}\:\mathrm{la}\:\mathrm{variable}\:\mathrm{aleatoire}\:\mathrm{qui}…
Question Number 81219 by jagoll last updated on 10/Feb/20 $${given}\:{a}\:{probability}\: \\ $$$${function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4}\:{and}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${in}\:{other}\:{x}.\:{find}\:{the}\:{value}\: \\ $$$${of}\:\sigma^{\mathrm{2}\:} \:? \\ $$ Commented by jagoll last…
Question Number 146613 by puissant last updated on 14/Jul/21 Answered by Olaf_Thorendsen last updated on 14/Jul/21 $$\mathrm{Il}\:\mathrm{suffit}\:\mathrm{de}\:\mathrm{demander}\:\mathrm{a}\:\mathrm{n}'\mathrm{importe} \\ $$$$\mathrm{quel}\:\mathrm{gardien}\:“\mathrm{Si}\:\mathrm{je}\:\mathrm{demandais}\:\mathrm{a} \\ $$$$\mathrm{l}'\mathrm{autre}\:\mathrm{gardien}\:\mathrm{quelle}\:\mathrm{est}\:\mathrm{la}\:\mathrm{porte}\:\mathrm{du} \\ $$$$\mathrm{paradis},\:\mathrm{que}\:\mathrm{me}\:\mathrm{repondrait}−\mathrm{il}\:?'' \\ $$$$…
Question Number 146328 by puissant last updated on 12/Jul/21 Answered by Olaf_Thorendsen last updated on 13/Jul/21 $$\mathrm{Pour}\:\mathrm{realiser}\:\mathrm{l}'\mathrm{evenement}\:\mathrm{E}\::\:“{aucun} \\ $$$${anniversaire}\:{le}\:{meme}\:{jour}''\:\mathrm{prenons} \\ $$$$\mathrm{une}\:\mathrm{premiere}\:\mathrm{personne}.\:\mathrm{Son} \\ $$$$\mathrm{anniversaire}\:\mathrm{peut}\:\mathrm{etre}\:\mathrm{choisi}\:\mathrm{parmi}\:\mathrm{365}. \\ $$$$\mathrm{Puis}\:\mathrm{prenons}\:\mathrm{une}\:\mathrm{deuxieme}\:\mathrm{personne}.…
Question Number 80746 by jagoll last updated on 06/Feb/20 $${what}\:{is}\:{constan}\:{term}\:{in}\:{expansion} \\ $$$$\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\frac{\mathrm{3}}{{x}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$ Commented by jagoll last updated on 06/Feb/20 $${c}\left({x}\right)\:=\:\frac{\left(\mathrm{1}+\mathrm{3}{x}\right)^{\mathrm{5}} \left(\mathrm{3}+{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}}…