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2 years ago

Pppppppppppppppppppppppppppppppppppp

Mathematics

2 answers:

Salsk061 [2.6K]2 years ago

6 0

Answer:

Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaàaaaaaaaaaaaaaaa

Step-by-step explanation:

à isn't the same ️️✨

IrinaK [193]2 years ago

4 0

Answer:

Step-by-step explanation:

Findings are not the same

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Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

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3 years ago

Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of t

salantis [7]

Answer:

X density = fXpxq and

Y" =InpXq

Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

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And PXq =pxq

Thus replacing x with y,

PXq = pyq

(a) Hence the density of Y is FYpyq

(b) at p0, fYpyq =fYp0q= 0

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