Question Number 59099 by Huygens last updated on 04/May/19 $${A}\:{sample}\:{of}\:\mathrm{20}\:{cigarettes}\:{is}\:{tested}\:{to}\:{determine}\:{nicotine}\:{vomtent} \\ $$$${and}\:{the}\:{average}\:{value}\:{observed}\:{was}\:\mathrm{1}.\mathrm{2}{mg}.{Compute} \\ $$$${a}\:\mathrm{99}\:{percentage}\:{two}−{sided}\:{confidence}\:{interval} \\ $$$${or}\:{the}\:{mean}\:{nicotine}\:{content}\:{if}\:{it}\:{is}\:{known}\:{that}\:{the}\:{standard}\:{deviation}\:{of}\:{a}\:{cigarette}'{s}\: \\ $$$${nicotine}\:{content}\:{is}\:\mathrm{0}.\mathrm{2}{mg}. \\ $$ Terms of Service Privacy Policy…
Question Number 59098 by Huygens last updated on 04/May/19 $${A}\:{sample}\:{of}\:\mathrm{20}\:{cigarettes}\:{is}\:{tested}\:{to}\:{determine}\:{nicotine}\:{vomtent} \\ $$$${and}\:{the}\:{average}\:{value}\:{observed}\:{was}\:\mathrm{1}.\mathrm{2}{mg}.{Compute} \\ $$$${a}\:\mathrm{99}\:{percentage}\:{two}−{sided}\:{confidence}\:{interval} \\ $$$${or}\:{the}\:{mean}\:{nicotine}\:{content}\:{if}\:{it}\:{is}\:{known}\:{that}\:{the}\:{standard}\:{deviation}\:{of}\:{a}\:{cigarette}'{s}\: \\ $$$${nicotine}\:{content}\:{is}\:\mathrm{0}.\mathrm{2}{mg}. \\ $$ Terms of Service Privacy Policy…
Question Number 59097 by Huygens last updated on 04/May/19 $${A}\:{sample}\:{of}\:\mathrm{20}\:{cigarettes}\:{is}\:{tested}\:{to}\:{determine}\:{nicotine}\:{vomtent} \\ $$$${and}\:{the}\:{average}\:{value}\:{observed}\:{was}\:\mathrm{1}.\mathrm{2}{mg}.{Compute} \\ $$$${a}\:\mathrm{99}\:{percentage}\:{two}−{sided}\:{confidence}\:{interval} \\ $$$${or}\:{the}\:{mean}\:{nicotine}\:{content}\:{if}\:{it}\:{is}\:{known}\:{that}\:{the}\:{standard}\:{deviation}\:{of}\:{a}\:{cigarette}'{s}\: \\ $$$${nicotine}\:{content}\:{is}\:\mathrm{0}.\mathrm{2}{mg}. \\ $$ Terms of Service Privacy Policy…
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Question Number 59075 by Gorgeous1995 last updated on 04/May/19 Commented by Gorgeous1995 last updated on 05/May/19 $${please}\:{help}\:{me}\:{solve}\:{it} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 124376 by bemath last updated on 02/Dec/20 Answered by liberty last updated on 02/Dec/20 $${n}\left({A}\right)=\:{C}_{\mathrm{1}} ^{\:\mathrm{10}} \:=\:\mathrm{10} \\ $$$${n}\left({S}\right)\:=\:{C}\:_{\mathrm{2}} ^{\mathrm{20}} \:=\:\frac{\mathrm{20}×\mathrm{19}}{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{190}\: \\ $$$${p}\left({A}\right)\:=\:\frac{\mathrm{10}}{\mathrm{190}}\:=\:\frac{\mathrm{1}}{\mathrm{19}}.…
Question Number 124282 by bramlexs22 last updated on 02/Dec/20 Answered by liberty last updated on 02/Dec/20 Commented by malwan last updated on 02/Dec/20 $$\frac{\begin{pmatrix}{\mathrm{5}}\\{\mathrm{5}}\end{pmatrix}}{\begin{pmatrix}{\mathrm{25}}\\{\mathrm{5}}\end{pmatrix}}\:=\:\frac{\mathrm{1}}{\frac{\mathrm{25}×\mathrm{24}×\mathrm{23}×\mathrm{22}×\mathrm{21}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}}\: \\…