Question and Answers Forum

AllQuestion and Answers: Page 696

Question Number 149532 Answers: 1 Comments: 0

on realise une suite infinie d′epreuves independantes.chaque epreuve resulte en un succes avec la probabilite p∈]0;1[ ou un echec avec la probabilite q=1−p.soit A_n l′evement “obenir au moins un succes au cours des premieres epreuves.” determiner P(A_n )

$${on}\:{realise}\:{une}\:{suite}\:{infinie}\:{d}'{epreuves} \\ $$$${independantes}.{chaque}\:{epreuve}\:{resulte}\:{en} \\ $$$$\left.{un}\:{succes}\:{avec}\:{la}\:{probabilite}\:{p}\in\right]\mathrm{0};\mathrm{1}\left[\:{ou}\:{un}\right. \\ $$$${echec}\:{avec}\:{la}\:{probabilite}\:{q}=\mathrm{1}−{p}.{soit}\:{A}_{{n}} \\ $$$${l}'{evement}\:``{obenir}\:{au}\:{moins}\:{un}\:{succes}\:{au} \\ $$$${cours}\:{des}\:{premieres}\:{epreuves}.''\:{determiner} \\ $$$${P}\left({A}_{{n}} \right) \\ $$

Question Number 149527 Answers: 1 Comments: 1

Question Number 149516 Answers: 1 Comments: 0

Ω = ∫_( 0) ^( 3) ((√((x + 2)^2 - 8x))) dx = ?

$$\Omega\:=\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{3}} {\int}}\:\left(\sqrt{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:-\:\mathrm{8}{x}}\right)\:{dx}\:=\:? \\ $$

Question Number 149510 Answers: 1 Comments: 7

Question Number 149505 Answers: 3 Comments: 2

lim_(n→∞) (((n + 3)/(n + 1)))^n = ?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}\:+\:\mathrm{3}}{\mathrm{n}\:+\:\mathrm{1}}\right)^{\mathrm{n}} =\:? \\ $$

Question Number 149503 Answers: 2 Comments: 0

Calcular: 3sin∙(150 - 𝛂) - 2cos∙(60 - 𝛂) = (1/2)

$$\mathrm{Calcular}: \\ $$$$\mathrm{3sin}\centerdot\left(\mathrm{150}\:-\:\boldsymbol{\alpha}\right)\:-\:\mathrm{2cos}\centerdot\left(\mathrm{60}\:-\:\boldsymbol{\alpha}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 149498 Answers: 2 Comments: 0

∫((px+q)/( (√(x^2 +r^2 ))))

$$ \\ $$$$\int\frac{{px}+{q}}{\:\sqrt{{x}^{\mathrm{2}} +{r}^{\mathrm{2}} }} \\ $$

Question Number 149586 Answers: 0 Comments: 0

$$ \\ $$$$ \\ $$

Question Number 149485 Answers: 2 Comments: 0

proof that 1=2??

$$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}=\mathrm{2}?? \\ $$

Question Number 149481 Answers: 1 Comments: 0

Question Number 149478 Answers: 2 Comments: 2

Solve for real numbers: x^2 −2x−3siny−4cosy + 6 = 0

$${Solve}\:{for}\:{real}\:{numbers}: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}{sin}\boldsymbol{{y}}−\mathrm{4}{cos}\boldsymbol{{y}}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$

Question Number 149474 Answers: 0 Comments: 2

x;y;z≥0 and x+y+z=3 prove that: (x^5 )^(1/(12)) + (y^5 )^(1/(12)) + (z^5 )^(1/(12)) ≥ xy + yz + zx

$${x};{y};{z}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:{x}+{y}+{z}=\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{12}}]{{x}^{\mathrm{5}} }\:+\:\sqrt[{\mathrm{12}}]{{y}^{\mathrm{5}} }\:+\:\sqrt[{\mathrm{12}}]{{z}^{\mathrm{5}} }\:\geqslant\:{xy}\:+\:{yz}\:+\:{zx} \\ $$

Question Number 149471 Answers: 1 Comments: 0

on distribue au hasard 8 boules b_1 ...b_8 dans 6 tiroirs t_1 ...t_6 .soit A_i l′evenement “le tiroir t_i est vide” les evenements A_1 et A_2 sont-ils independants ?

$${on}\:{distribue}\:{au}\:{hasard}\:\mathrm{8}\:{boules}\:{b}_{\mathrm{1}} ...{b}_{\mathrm{8}} \\ $$$${dans}\:\mathrm{6}\:{tiroirs}\:{t}_{\mathrm{1}} ...{t}_{\mathrm{6}} .{soit}\:{A}_{{i}} \:{l}'{evenement} \\ $$$$``{le}\:{tiroir}\:{t}_{{i}} \:{est}\:{vide}''\:{les}\:{evenements}\:{A}_{\mathrm{1}} {et} \\ $$$${A}_{\mathrm{2}} {sont}-{ils}\:{independants}\:? \\ $$

Question Number 149468 Answers: 1 Comments: 0

sin𝛂 ∙ cos𝛂 = (3/8) ⇒ 3∣sin𝛂 - cos𝛂∣=?

$${sin}\boldsymbol{\alpha}\:\centerdot\:{cos}\boldsymbol{\alpha}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:\Rightarrow\:\mathrm{3}\mid{sin}\boldsymbol{\alpha}\:-\:{cos}\boldsymbol{\alpha}\mid=? \\ $$

Question Number 149467 Answers: 1 Comments: 0

Question Number 149462 Answers: 1 Comments: 0

Question Number 149469 Answers: 1 Comments: 0

Compare: 1900^(3/4) + 99^(3/4) and 1999^(3/4)

$$\mathrm{Compare}: \\ $$$$\mathrm{1900}^{\frac{\mathrm{3}}{\mathrm{4}}} \:+\:\:\mathrm{99}^{\frac{\mathrm{3}}{\mathrm{4}}} \:\:\:\boldsymbol{\mathrm{and}}\:\:\:\mathrm{1999}^{\frac{\mathrm{3}}{\mathrm{4}}} \\ $$

Question Number 149463 Answers: 1 Comments: 0

Let the independent random variables X_1 and X_2 have binomial distribution with parameters n_1 =3,p=2/3 and n_2 =4 p=1/2 respectively. Compute P(X_1 =X_2 )

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{independent}\:\mathrm{random}\:\mathrm{variables} \\ $$$$\mathrm{X}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{X}_{\mathrm{2}} \:\mathrm{have}\:\mathrm{binomial}\:\mathrm{distribution} \\ $$$$\mathrm{with}\:\mathrm{parameters}\:\mathrm{n}_{\mathrm{1}} =\mathrm{3},\mathrm{p}=\mathrm{2}/\mathrm{3}\:\mathrm{and}\:\mathrm{n}_{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{p}=\mathrm{1}/\mathrm{2}\:\:\mathrm{respectively}.\: \\ $$$$\mathrm{Compute}\:\mathrm{P}\left(\mathrm{X}_{\mathrm{1}} =\mathrm{X}_{\mathrm{2}} \right) \\ $$

Question Number 149458 Answers: 1 Comments: 3