Question Number 69226 by peter frank last updated on 21/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134760 by rs4089 last updated on 07/Mar/21 Answered by mathmax by abdo last updated on 07/Mar/21 $$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left[\frac{\left(\mathrm{n}!\right)}{\mathrm{n}}\right]^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\frac{\mathrm{n}!}{\mathrm{n}}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \mathrm{e}^{−\mathrm{n}}…
Question Number 134763 by bramlexs22 last updated on 07/Mar/21 $$\mathrm{Given}\:{a}=\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{a}^{\mathrm{3}} −\mathrm{9}{a}+\mathrm{1}. \\ $$ Answered by EDWIN88 last updated on 07/Mar/21 $$=\:\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{3}}\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{9}}−\frac{\mathrm{26}\:\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\frac{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\sqrt[{\mathrm{6}}]{\mathrm{243}}}{\mathrm{3}}−\frac{\mathrm{26}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}}{\mathrm{3}}+\mathrm{4} \\ $$…
Question Number 69222 by mustakim420 last updated on 21/Sep/19 Commented by mr W last updated on 21/Sep/19 $${n}=\mathrm{2}\:{or}\:\mathrm{4} \\ $$$$\mathrm{2}+\mathrm{4}=\mathrm{6} \\ $$ Terms of Service…
Question Number 3685 by Filup last updated on 19/Dec/15 $$\mathrm{let}\:{p}_{{i}} \:\mathrm{be}\:\mathrm{the}\:{i}^{\mathrm{th}} \:\mathrm{prime} \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{fillowing}\:\mathrm{sum}\:\mathrm{converge}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{p}_{{i}} }{{p}_{{i}+\mathrm{1}} } \\ $$$$\left({p}_{\mathrm{1}} =\mathrm{2},\:{p}_{\mathrm{2}}…
Question Number 134758 by rs4089 last updated on 07/Mar/21 Commented by rs4089 last updated on 07/Mar/21 $${prove}\:{it} \\ $$ Answered by mnjuly1970 last updated on…
Question Number 134753 by mathocean1 last updated on 06/Mar/21 $${f}\:{is}\:{defined}\:{in}\:\left[\mathrm{0};\:+\infty\left[.\right.\right. \\ $$$$\begin{cases}{\:}\\{}\\{{f}\left(\mathrm{0}\right)={ln}\mathrm{2}}\end{cases}{f}\left({x}\right)=\int_{{x}} ^{\mathrm{2}{x}} \:\frac{{e}^{−{t}} }{{t}}{dt}\:\:{for}\:{x}>\mathrm{0} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:{Given}\:\mathrm{0}\leqslant{f}\left({x}\leqslant\frac{{e}^{−{x}} −{e}^{−\mathrm{2}{x}} }{{x}}.\right. \\ $$$${Calcule}\:{the}\:{lim}\:{f}\left({x}\right)\:{at}\:\mathrm{0}\:{and}\:+\infty. \\ $$$$\left.\mathrm{2}\right)\:{Calculate}\:{f}\:'\left({x}\right)\:,\:{give}\:{its}\:{variation}…
Question Number 134752 by mathocean1 last updated on 06/Mar/21 $${In}\:{a}\:{locality},\:\mathrm{20\%}\:{of}\:{population}\:{have} \\ $$$${a}\:{chronic}\:{disease}.\:{Among}\:{these}\: \\ $$$${people}\:{who}\:{has}\:{a}\:{chronic}\:{disease},\:\mathrm{2}.\mathrm{5}\:\% \\ $$$${have}\:{COVID}−\mathrm{19}.\:{Among}\:{the}\:{people} \\ $$$${who}\:{don}'{t}\:{have}\:{a}\:{chronic}\:{disease},\:\mathrm{99\%} \\ $$$${have}\:{not}\:{COVID}−\mathrm{19}. \\ $$$$\boldsymbol{{C}}{alculate}\:{the}\:{probability}\:{that}\:{one}\:{person} \\ $$$${of}\:{this}\:{locality}\:{has}\:{COVID}−\mathrm{19}\:{and}\:{a} \\…
Question Number 3682 by Yozzii last updated on 19/Dec/15 $$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}\mathrm{2}^{{r}} }=? \\ $$ Commented by RasheedSindhi last updated on 19/Dec/15 $$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}\mathrm{2}^{{r}}…
Question Number 134751 by mathocean1 last updated on 06/Mar/21 $$\:{In}\:{my}\:{house},\:{there}\:{is}\:\mathrm{250}\:{laptops}:\: \\ $$$$\mathrm{40}\:{are}\:{new},\:\mathrm{100}\:{are}\:{recent}\:{and}\:{the}\: \\ $$$${others}\:{are}\:{old}.\:{A}\:{statistic}\:{showed}\:{that} \\ $$$$\mathrm{4\%}\:{of}\:{new}\:{laptops}\:{are}\:{faulty},\:\mathrm{12\%}\:{of} \\ $$$${recent}\:{ones}\:{are}\:{faulty}\:{and}\:\mathrm{25\%}\:{of}\:{old} \\ $$$${ones}\:{are}\:{faulty}. \\ $$$${Calculate}\:{the}\:{probability}\:{that}\:\mathrm{1}\:{laptop}\:{be} \\ $$$${new}\:,\:{knowing}\:{that}\:{it}\:{is}\:{faulty}. \\…