Question Number 207387 by sniper237 last updated on 13/May/24 $${Let}\:\:{cardE}={n}\:,\:{and}\:\:{the}\:{set}\:{of}\:{parts} \\ $$$${S}=\left\{\left({A},{B}\right)\in{P}\left({E}\right)×{P}\left({E}\right)\:/\:\:{A}\cap{B}=\varnothing\right\} \\ $$$${Show}\:{that}\:\:{cardS}=\:\mathrm{3}^{{n}} \\ $$ Answered by Berbere last updated on 13/May/24 $${if}\:{card}\left({A}\right)={k};{E}={A}\cup\overset{−} {{A}}\:\:{the}\:{number}\:{of}\:{subset}\:{of}\:{card}={k}…
Question Number 207374 by sniper237 last updated on 13/May/24 $${Show}\:{that}\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$ Answered by mr W last updated on…
Question Number 206514 by cortano21 last updated on 17/Apr/24 $$\:{In}\:{this}\:{covid}\:−\mathrm{19}\:{pandemic},\:{it}\:{is}\: \\ $$$$\:{known}\:{that}\:{are}\:\mathrm{5},\mathrm{667},\mathrm{355}\:{confirmed}\: \\ $$$$\:{cases}\:{out}\:{of}\:\mathrm{273},\mathrm{500},\mathrm{000}\:{in}\:{X}\:{country} \\ $$$$\:{population}\:{based}\:{WHO}.\: \\ $$$$\:{One}\:{of}\:{the}\:{equipment}\:{to}\:{test}\:{the} \\ $$$$\:{covid}−\mathrm{19}\:{is}\:{GeNose}\:{C}\mathrm{19}−{S}\:{developed} \\ $$$$\:{by}\:{UGM}.\:{GeNose}\:{C}\mathrm{19}−{S}\:{is}\:{a}\:{rapid} \\ $$$$\:{screening}\:{equipment}\:{for}\:{Sars}−{CoV}\mathrm{2} \\…
Question Number 205409 by Red1ight last updated on 20/Mar/24 $$\mathrm{let}\:{x},\:{y},\:{z}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y},{z}\in\mathbb{R} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\left.{a}\right)\:\mathrm{all}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{satisfied} \\ $$$$\mid{x}−{y}\mid\geqslant\mathrm{2} \\ $$$$\mid{x}−{z}\mid\geqslant\mathrm{2} \\ $$$$\mid{y}−{z}\mid\geqslant\mathrm{2} \\ $$$$\left.{b}\right)\:\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\…
Question Number 205394 by Red1ight last updated on 19/Mar/24 $$\mathrm{let}\:{x}\:\mathrm{and}\:{y}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y}\in\mathbb{R} \\ $$$$\mid{x}−{y}\mid\geqslant{d} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\left.{a}\right)\:{d}=\mathrm{0} \\ $$$$\left.{b}\right)\:{d}=\mathrm{1} \\ $$$$\left.{c}\right)\:{d}=\mathrm{2} \\…
Question Number 205230 by louisyanni last updated on 13/Mar/24 $$ \\ $$ Commented by Rasheed.Sindhi last updated on 13/Mar/24 $$\mathcal{R}{eserved}\:{space}\:{for}\:{a}\:{question}? \\ $$ Commented by Frix…
Question Number 201972 by Lekhraj last updated on 17/Dec/23 Answered by mr W last updated on 18/Dec/23 $$\boldsymbol{\Phi}_{\leqslant\boldsymbol{{x}}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\boldsymbol{{erf}}\left(\frac{\boldsymbol{{x}}−\boldsymbol{\mu}}{\:\sqrt{\mathrm{2}}\:\boldsymbol{\sigma}}\right)\right] \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{erf}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}}{\:\sqrt{\boldsymbol{\pi}}}\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } \boldsymbol{{dt}}…
Question Number 201969 by BaliramKumar last updated on 17/Dec/23 $$\mathrm{A}\:\mathrm{dice}\:\mathrm{is}\:\mathrm{cast}\:\mathrm{twice},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{appearing}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{10}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{5}\:\mathrm{has}\: \\ $$$$\mathrm{appeared}\:\mathrm{at}\:\mathrm{least}\:\mathrm{once}\:\mathrm{is}. \\ $$ Answered by Frix last updated on 17/Dec/23…
Question Number 201527 by cortano12 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $$#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=#\left(\mathrm{6}\right)+#\left(\mathrm{8}\right)−#\left(\mathrm{6}{n}\mathrm{8}\right) \\ $$$$#\left(\mathrm{6}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{6}}\rfloor=\mathrm{333};#\left(\mathrm{8}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{8}}\rfloor=\mathrm{250} \\ $$$$#\left(\mathrm{6}{n}\mathrm{8}\right)=#\left(\mathrm{24}\right)=\lfloor\frac{\mathrm{2000}}{\mathrm{24}}\rfloor=\mathrm{83} \\ $$$$\Rightarrow#\left(\mathrm{6}\:{or}\:\mathrm{8}\right)=\mathrm{500} \\ $$$$\Rightarrow{Probability}=\frac{\mathrm{2000}−\mathrm{500}}{\mathrm{2000}}=\frac{\mathrm{3}}{\mathrm{4}}…
Question Number 201545 by Calculusboy last updated on 08/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com