Question Number 201982 by Calculusboy last updated on 17/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201977 by Calculusboy last updated on 17/Dec/23 Commented by Calculusboy last updated on 17/Dec/23 $$\boldsymbol{{make}}\:\boldsymbol{{R}}\:\boldsymbol{{subject}}\:\boldsymbol{{of}}\:\boldsymbol{{theformula}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201979 by mr W last updated on 17/Dec/23 Commented by mr W last updated on 19/Dec/23 $${Q}\mathrm{155657}\:{reposted}\:{for}\:{alternative} \\ $$$${solutions}. \\ $$ Answered by…
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 201965 by ajfour last updated on 17/Dec/23 $${m}−{h}=\mathrm{2}{p} \\ $$$${p}\left({m}−{h}\right)={k}−{q} \\ $$$${mk}−{qh}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${k}−\mathrm{2}{q}=\mathrm{1}+{ph} \\ $$$${Assume}\:{one}\:{find}\:{the}\:{rest}! \\ $$$$\checkmark \\ $$ Answered by Frix…
Question Number 201966 by mnjuly1970 last updated on 17/Dec/23 $$ \\ $$$$\:\:\:\:\:\:\:{solve}\: \\ $$$$\:\: \\ $$$$\:\:\:\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\right)=\:\mathrm{1} \\ $$$$ \\ $$ Answered by ajfour last…
Question Number 201949 by sonukgindia last updated on 16/Dec/23 Answered by Frix last updated on 17/Dec/23 $${n}^{{k}} \equiv{n}\mathrm{mod2024};\:{n}\in\left\{\mathrm{529},\:\mathrm{737},\:\mathrm{760},\:\mathrm{1265},\:\mathrm{1288},\:\mathrm{1496}\right\} \\ $$$$\Rightarrow \\ $$$$\Sigma{n}^{{n}} \equiv\mathrm{3mod2024} \\ $$…
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Question Number 201947 by Mingma last updated on 16/Dec/23 Answered by Frix last updated on 18/Dec/23 $$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\left(\pi−\alpha−\beta\right)\:+ \\ $$$$\:\:\:\:\:+\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\pi−\alpha−\beta\right)\:\mathrm{cos}\:\alpha\:+ \\ $$$$\:\:\:\:\:+\mathrm{sin}\:\left(\pi−\alpha−\beta\right)\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:= \\ $$$$=\frac{\mathrm{3}−\left(\mathrm{cos}\:\mathrm{2}\alpha\:+\mathrm{cos}\:\mathrm{2}\beta\:+\mathrm{cos}\:\mathrm{2}\left(\alpha+\beta\right)\right.}{\mathrm{4}}= \\ $$$$\:\:\:\:\:\left[\mathrm{Let}\:\alpha={u}−{v}\wedge\beta={u}+{v}\right]…