Question and Answers Forum

AllQuestion and Answers: Page 358

Question Number 182695 Answers: 1 Comments: 2

If two events A and B are such that P(A^c )=0.3 ; P(B)=0.4 and P(A∩B^c )= 0.5 then P((B/(A∪B^c )))=? (A) 0.20 (B) 0.25 (C) 0.30 (D) 0.35

$$\:\:\mathrm{If}\:\mathrm{two}\:\mathrm{events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{such}\: \\ $$$$\:\:\mathrm{that}\:\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \right)=\mathrm{0}.\mathrm{3}\:;\:\mathrm{P}\left(\mathrm{B}\right)=\mathrm{0}.\mathrm{4}\:\mathrm{and} \\ $$$$\:\:\mathrm{P}\left(\mathrm{A}\cap\mathrm{B}^{\mathrm{c}} \right)=\:\mathrm{0}.\mathrm{5}\:\mathrm{then}\:\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}\cup\mathrm{B}^{\mathrm{c}} }\right)=? \\ $$$$\:\left(\mathrm{A}\right)\:\mathrm{0}.\mathrm{20}\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{0}.\mathrm{25}\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{0}.\mathrm{30}\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{0}.\mathrm{35} \\ $$$$\:\: \\ $$

Question Number 182694 Answers: 1 Comments: 0

Question Number 182681 Answers: 5 Comments: 6

Solve (√(39− 10 a −a^2 )) + (√(9−a^2 )) = 2 (√(14)) ; a∈ R^(+★) @Mr. Rasheed, yesterday i solved it on a draft paper When i came today to write it down on the notebook i got into a maze, do you have an easy way to deal with? “if you have time” i lost that paper, and this formula makes me laugh at myself, it′s for 13 y/o “confused face” Sure Every friend can share!

$$\:{Solve}\:\sqrt{\mathrm{39}−\:\mathrm{10}\:{a}\:−{a}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{9}−{a}^{\mathrm{2}} }\:=\:\mathrm{2}\:\sqrt{\mathrm{14}}\:\:\:;\:{a}\in\:\mathbb{R}^{+\bigstar} \\ $$$$\:@{Mr}.\:{Rasheed},\:{yesterday}\:{i}\:{solved}\:{it}\:{on}\:{a}\:{draft}\:{paper} \\ $$$$\:{When}\:{i}\:{came}\:{today}\:{to}\:{write}\:{it}\:{down}\:{on} \\ $$$$\:{the}\:{notebook}\:{i}\:{got}\:{into}\:{a}\:{maze},\:{do}\:{you}\:{have} \\ $$$$\:{an}\:{easy}\:{way}\:{to}\:{deal}\:{with}?\:``{if}\:{you}\:{have}\:{time}'' \\ $$$$\:{i}\:{lost}\:{that}\:{paper},\:{and}\:{this}\:{formula}\:{makes}\:{me} \\ $$$$\:{laugh}\:{at}\:{myself},\:{it}'{s}\:{for}\:\mathrm{13}\:{y}/{o}\:``{confused}\:{face}''\: \\ $$$$\:{Sure}\:{Every}\:{friend}\:{can}\:{share}! \\ $$

Question Number 182676 Answers: 2 Comments: 0

Question Number 182675 Answers: 2 Comments: 0

Question Number 182659 Answers: 1 Comments: 0

lim_(n→∞) (((n!))^(1/n) /n)=?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{{n}}]{{n}!}}{{n}}=? \\ $$

Question Number 182657 Answers: 0 Comments: 0

as we can know for Q_1 , set the funvction in question as f(x) and make the first step as: a=1,f(x)=x^2 −7x+3ln x df(x)=2x−7+(3/x)=h(x) set h(x)=0⇒x=(1/2)&x=3 so,the monotonicity of f(x) is f(n)<f((1/2))_(∣n<(1/2)) ,f((1/2))>f(2),f(2)<f(p)_(∣p>2.) Q_1 has been proved finished Q_(2 ) ,set ax^2 −(a+6)x+3ln x>−6 when x∈[2,3e], make the f(x) dive two part as g(x)=ax^2 −(6+a)x&k(x)=−3(ln x+2), the middle value of g(x) is x=((a+6)/(2a)) when x=3e, k_(min) (3e)=−15, x=2, k_(max) (2)=−3(ln 2+2) g(2)=2a−12>k_(max) (x)⇒a>3−(3/2)ln 2 when x=(1/2)+(3/a), g(x)=−(((a+6)^2 )/(4a))>−3(ln (((a+6)/(2a)))+2)&a>0&2<(1/2)+(3/a)<3e

$${as}\:{we}\:{can}\:{know}\:{for}\:{Q}_{\mathrm{1}} ,\:{set}\:{the}\:{funvction}\:{in}\:{question}\:{as}\:{f}\left({x}\right) \\ $$$${and}\:{make}\:{the}\:{first}\:{step}\:{as}: \\ $$$${a}=\mathrm{1},{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{3ln}\:{x} \\ $$$${df}\left({x}\right)=\mathrm{2}{x}−\mathrm{7}+\frac{\mathrm{3}}{{x}}={h}\left({x}\right) \\ $$$${set}\:{h}\left({x}\right)=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\&{x}=\mathrm{3} \\ $$$${so},{the}\:{monotonicity}\:{of}\:{f}\left({x}\right)\:{is}\:{f}\left({n}\right)<{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\mid{n}<\frac{\mathrm{1}}{\mathrm{2}}} ,{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)>{f}\left(\mathrm{2}\right),{f}\left(\mathrm{2}\right)<{f}\left({p}\underset{\mid{p}>\mathrm{2}.} {\right)} \\ $$$${Q}_{\mathrm{1}} \:{has}\:{been}\:{proved}\:{finished} \\ $$$${Q}_{\mathrm{2}\:} \:,{set}\:{ax}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){x}+\mathrm{3ln}\:{x}>−\mathrm{6}\:{when}\:{x}\in\left[\mathrm{2},\mathrm{3}{e}\right], \\ $$$${make}\:{the}\:{f}\left({x}\right)\:{dive}\:{two}\:{part}\:{as} \\ $$$$\:{g}\left({x}\right)={ax}^{\mathrm{2}} −\left(\mathrm{6}+{a}\right){x\&k}\left({x}\right)=−\mathrm{3}\left(\mathrm{ln}\:{x}+\mathrm{2}\right), \\ $$$$\:{the}\:{middle}\:{value}\:{of}\:{g}\left({x}\right)\:{is}\:{x}=\frac{{a}+\mathrm{6}}{\mathrm{2}{a}} \\ $$$${when}\:{x}=\mathrm{3}{e},\:{k}_{{min}} \left(\mathrm{3}{e}\right)=−\mathrm{15},\:\:{x}=\mathrm{2},\:{k}_{{max}} \left(\mathrm{2}\right)=−\mathrm{3}\left(\mathrm{ln}\:\mathrm{2}+\mathrm{2}\right) \\ $$$${g}\left(\mathrm{2}\right)=\mathrm{2}{a}−\mathrm{12}>{k}_{{max}} \left({x}\right)\Rightarrow{a}>\mathrm{3}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}},\:{g}\left({x}\right)=−\frac{\left({a}+\mathrm{6}\right)^{\mathrm{2}} }{\mathrm{4}{a}}>−\mathrm{3}\left(\mathrm{ln}\:\left(\frac{{a}+\mathrm{6}}{\mathrm{2}{a}}\right)+\mathrm{2}\right)\&{a}>\mathrm{0\&2}<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}}<\mathrm{3}{e} \\ $$$$ \\ $$

Question Number 182650 Answers: 0 Comments: 0