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ziro4ka [17]
3 years ago
15

Find the z score that cuts-off the largest 9% under the standard normal curve. a) -1.34 b) 1.34 c) 0.536 d) 1.28

Mathematics
1 answer:
topjm [15]3 years ago
4 0

Answer:

b) 1.34

Step-by-step explanation:

The z score is used to determine the number of standard deviations by which the raw score is above or below the mean. If the z score is positive then the z score is above the mean while for a negative z score implies that it is below the mean. The z score is given by:

z=\frac{x-\mu}{\sigma}

For the largest 9%, the score is 100% - 9% = 91% = 0.91

From the normal distribution table, the z score that corresponds to 0.91 is 1.34

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Answer:

\rm\displaystyle  0,\pm\pi

Step-by-step explanation:

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===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

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factor out tanγ:

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divide both sides by tanαtanβ-1 and that yields:

\rm\displaystyle   \tan( \gamma ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

multiply both numerator and denominator by-1 which yields:

\rm\displaystyle   \tan( \gamma ) =   -  \bigg(\frac{ \tan( \alpha )  +  \tan( \beta ) }{ 1 - \tan( \alpha )  \tan( \beta )   } \bigg)

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\rm\displaystyle   \tan( \gamma ) =   -  \tan( \alpha  +  \beta )

let α+β be t and transform:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( t)

remember that tan(t)=tan(t±kπ) so

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm k\pi )

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm \pi )

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\rm\displaystyle   \tan( \gamma ) =    \tan(   -\alpha  -\beta\pm \pi )

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\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm \pi

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ \pm \pi  }

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

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isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ 0  }

and we're done!

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