Question Number 118455 by Ravin1989 last updated on 17/Oct/20
Commented by benjo_mathlover last updated on 17/Oct/20
$$\left(\mathrm{4}\right)\:\mid\mathrm{3}{A}\mid\:=\:\mathrm{3}^{\mathrm{3}} ×\mid{A}\mid\:=\:\mathrm{27}×\mathrm{8}=\mathrm{216} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\left.\mathrm{9}\right){d}=\mid\frac{−\mathrm{2}×\mathrm{0}+\mathrm{6}×\mathrm{0}−\mathrm{3}×\mathrm{0}+\mathrm{7}}{\:\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }}\mid=\frac{\mathrm{7}}{\:\sqrt{\mathrm{49}}}=\mathrm{1} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\left.\mathrm{6}\right){y}={lnx}^{\mathrm{2}} −{lne}^{\mathrm{2}} \rightarrow{y}=\mathrm{2}{lnx}−\mathrm{2} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{{x}}\rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{−\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\mathrm{5}\int{x}^{\mathrm{2}} {e}^{{x}^{\mathrm{3}} } {dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int{e}^{{x}^{\mathrm{3}} } {d}\left({x}^{\mathrm{3}} \right)=\frac{\mathrm{1}}{\mathrm{3}}×{e}^{{x}^{\mathrm{3}} } +{c} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\left.\mathrm{7}\right){p}\left({A}\right)\:{sample}\:{space}\:\left\{\mathrm{4},\mathrm{5},\mathrm{6}\right\}\:\:{p}\left({A}\right)=\frac{\mathrm{3}}{\mathrm{6}} \\ $$$${p}\left({B}\right)\:{sample}\:{space}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\}\:{p}\left({B}\right)=\frac{\mathrm{4}}{\mathrm{6}} \\ $$$${p}\left({A}\cap{B}\right)\:\:{sample}\:{space}\:\left\{\mathrm{4}\right\}\:{p}\left({A}\cap{B}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${p}\left({A}\cup{B}\right)={p}\left({A}\right)+{p}\left({B}\right)−{p}\left({A}\cap{B}\right)=\frac{\mathrm{3}}{\mathrm{6}}+\frac{\mathrm{4}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{1} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\left.\mathrm{3}\right)\boldsymbol{{pls}}\:\boldsymbol{{check}}\:… \\ $$$$\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{one}}\:\boldsymbol{{queen}}\:\boldsymbol{{of}}\:\boldsymbol{{spade}}\: \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{probablity}}=\frac{\mathrm{1}}{\mathrm{52}} \\ $$$${other}\:{sirs}\:{pls}\:{comment} \\ $$
Commented by prakash jain last updated on 17/Oct/20
$${I}\:{think}\:{conditional}\:{probability} \\ $$$${should}\:{be}\:{used}. \\ $$$$\mathrm{P}\left({A}\cap{B}\right)={P}\left({A}\mid{B}\right){P}\left({B}\right) \\ $$$${A}\:{Queen}\:{of}\:{spade} \\ $$$${B}\:{Queen} \\ $$$${given}\:{that}\:{it}\:{is}\:{queen}\:{the}\:{probibility} \\ $$$${that}\:{it}\:{is}\:{queen}\:{of}\:{spade}\:{is}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${please}\:{check}\:{and}\:{comment} \\ $$
Answered by TANMAY PANACEA last updated on 17/Oct/20
$$\mid{E}\overset{\rightarrow} {{A}}\mid=\mid{EC}\mid \\ $$$${E}\overset{\rightarrow} {{C}}=−{E}\overset{\rightarrow} {{A}} \\ $$$${E}\overset{\rightarrow} {{D}}=−{E}\overset{\rightarrow} {{B}} \\ $$$${E}\overset{\rightarrow} {{A}}+{E}\overset{\rightarrow} {{B}}+{E}\overset{\rightarrow} {{C}}+{E}\overset{\rightarrow} {{D}}=\mathrm{0} \\ $$