Question Number 207813 by Davidtim last updated on 27/May/24 $${the}\:{word}\:{of}\:{atom}\:{is}\:{meant}\:{no}\:{dividable} \\ $$$${however}\:{atom}\:{is}\:{dividable},\:{why}\:{we} \\ $$$${should}\:{use}\:{the}\:{atom}\:{word}\:{nowadays}? \\ $$ Commented by Davidtim last updated on 28/May/24 $${I}\:{need}\:{your}\:{helping} \\…
Question Number 207814 by Davidtim last updated on 27/May/24 $${Some}\:{guys}\:{are}\:{no}\:{caught}\:{by}\:{electricity} \\ $$$${what}'{s}\:{the}\:{reason}? \\ $$ Commented by Davidtim last updated on 28/May/24 $${I}\:{need}\:{your}\:{helping} \\ $$ Terms…
Question Number 207810 by Davidtim last updated on 27/May/24 $${what}\:{is}\:{the}\:{difference}\:{between}\: \\ $$$${Domain}\:{and}\:{Codomain}? \\ $$ Commented by Davidtim last updated on 28/May/24 $${I}\:{need}\:{your}\:{helping} \\ $$ Commented…
Question Number 207789 by Ghisom last updated on 26/May/24 $$\forall{r}\in\mathbb{R}:\:{H}_{{r}} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}^{{r}} −\mathrm{1}}{{t}−\mathrm{1}}{dt} \\ $$$${H}_{{r}+\mathrm{2}} −{H}_{{r}} =\mathrm{1} \\ $$$${r}=? \\ $$ Answered by MM42…
Question Number 207787 by mr W last updated on 26/May/24 $$\boldsymbol{{n}}\:{married}\:{couples}\:{are}\:{invited}\:{to} \\ $$$${a}\:{dance}\:{party}.\:{for}\:{the}\:{first}\:{dance} \\ $$$$\boldsymbol{{n}}\:{paires}\:{are}\:{radomly}\:{selected}.\: \\ $$$${what}'{s}\:{the}\:{probability}\:{that}\:{no}\:{woman} \\ $$$${dances}\:{with}\:{her}\:{own}\:{husband}? \\ $$$$\left.\mathrm{1}\right)\:{if}\:{a}\:{pair}\:{must}\:{be}\:{of}\:{different} \\ $$$$\:\:\:\:\:{genders}. \\ $$$$\left.\mathrm{2}\right)\:{if}\:{a}\:{pair}\:{can}\:{also}\:{be}\:{of}\:{the}\:{same}\:…
Question Number 207779 by Tawa11 last updated on 26/May/24 Answered by mr W last updated on 26/May/24 Commented by Tawa11 last updated on 26/May/24 $$\mathrm{Thanks}\:\mathrm{sir}.…
Question Number 207774 by Ari last updated on 25/May/24 Commented by Rasheed.Sindhi last updated on 26/May/24 $$\mathrm{No}\:\mathrm{unique}\:\mathrm{answer}. \\ $$ Commented by Ari last updated on…
Question Number 207752 by efronzo1 last updated on 25/May/24 $$\:\mathrm{Two}\:\mathrm{ships}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{berth}\: \\ $$$$\:\mathrm{in}\:\mathrm{a}\:\mathrm{port}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\:\mathrm{arrival}\:\mathrm{times}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{ships}\: \\ $$$$\:\mathrm{are}\:\mathrm{independent}\:\mathrm{and}\:\mathrm{have}\:\mathrm{the}\: \\ $$$$\:\mathrm{same}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{docking}\: \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday}\:\left(\mathrm{00}.\mathrm{00}−\mathrm{24}.\mathrm{00}\right) \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{ship} \\ $$$$\:\mathrm{is}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{and}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time} \\…
Question Number 207769 by 073 last updated on 25/May/24 Commented by 073 last updated on 26/May/24 $$???? \\ $$ Answered by Berbere last updated on…
Question Number 207753 by efronzo1 last updated on 25/May/24 Answered by Berbere last updated on 25/May/24 $${A}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\frac{{k}+\mathrm{1}−{k}}{{k}\left({k}+\mathrm{1}\right)}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2024}}=\frac{\mathrm{2023}}{\mathrm{2024}}…