Question Number 209030 by Spillover last updated on 30/Jun/24 Answered by aleks041103 last updated on 30/Jun/24 $${moment}\:{generating}\:{function}\:{is} \\ $$$${m}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}_{{k}} }{{k}!}{t}^{{k}} ={m}_{\mathrm{0}} +{m}_{\mathrm{1}} {t}+……
Question Number 209026 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 07/Jul/24 $${f}\left({x}\right)=\left(\frac{\mathrm{4}}{{x}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−{x}} \:\:\: \\ $$$${x}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$${f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−\mathrm{1}} =\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}}…
Question Number 209027 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 05/Jul/24 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},…..{n} \\ $$$${from}\:{Moment}\:{generating}\:{function}\left({MGF}\right) \\ $$$${M}_{{x}} \left({t}\right)={E}\left({e}^{{tx}} \right) \\…
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Question Number 208969 by Spillover last updated on 29/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by…
Question Number 208968 by MWSuSon last updated on 29/Jun/24 $$ \\ $$$$\mathrm{hello}\:\mathrm{everyone}.\:\mathrm{Im}\:\mathrm{writing}\:\mathrm{a}\:\mathrm{project}\:\:\mathrm{on}\:\mathrm{the}\:\mathrm{topic}“\:\mathrm{Solution}\:\mathrm{of} \\ $$$$\mathrm{Nonlinear}\:\mathrm{partial}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{using}\:\mathrm{charpits}\:\mathrm{methods}'' \\ $$$$\mathrm{curently}\:\mathrm{writing}\:\mathrm{chapter}\:\mathrm{2}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{little}\:\mathrm{bit}\:\mathrm{confused}\:\mathrm{about}\:\mathrm{what}\:\mathrm{ih} \\ $$$$\mathrm{should}\:\mathrm{include}\:\mathrm{in}\:\mathrm{my}\:\mathrm{theoretical}\:\mathrm{framework}.\:\mathrm{Any}\:\mathrm{ideas}? \\ $$ Terms of Service Privacy Policy…
Question Number 208970 by Spillover last updated on 29/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by…
Question Number 208943 by MWSuSon last updated on 28/Jun/24 Answered by A5T last updated on 28/Jun/24 $${a}.\:\frac{\left[{ADM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×{AD}×{DM}}{{AD}×\left({DM}×\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}.\:{Let}\:{AM}\:{and}\:{DN}\:{intersect}\:{at}\:{E} \\ $$$$\frac{\left[{DEM}\right]}{\left[{ABCD}\right]}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{{AD}}{\mathrm{2}}×{DM}}{{AD}×{DM}×\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${c}.\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$…
Question Number 208866 by Adeyemi889 last updated on 26/Jun/24 Commented by Adeyemi889 last updated on 26/Jun/24 $${partial}\:{F}\boldsymbol{{Raction}} \\ $$$$ \\ $$ Answered by Sutrisno last…
Question Number 208836 by Adeyemi889 last updated on 24/Jun/24 Answered by Rasheed.Sindhi last updated on 25/Jun/24 $${Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$={Ax}^{\mathrm{3}} −{A}+{Bx}^{\mathrm{2}} +{Bx}+{B}+{Cx}^{\mathrm{3}} +{Dx}^{\mathrm{2}}…