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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