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blondinia [14]
3 years ago
7

Verify the identity: tan(x+pi/2)=-cotx please!!

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0
First we use sin(a+b)= sinacosb+sinbcosa
and cos(a+b)=cosa cosb -sinasinb

tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)
and sin(x+pi/2) = sinxcospi/2 + sinpi/2cosx =cosx, 
<span>cos(x+pi/2) = cosxcospi/2- sinxsinpi/2= - sinx, 
</span> because <span>cospi/2 =0,  </span>and <span>sinpi/2=1

</span><span>=tan(x+pi/2)= sin(x+pi/2) / cos(x+pi/2)= cosx / -sinx = -1/tanx = -cotx
</span>from where <span>tan(x+pi/2)=-cotx</span> 
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Answer:

Part A:

P(X=x)=n_{C_{x}}*p^x*(1-p)^{n-x}

x is a binomial Random Variable.

Part B:

Value of p=0.8

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Part C:

P(X=15)=0.17456

Part D:

P(X≥15)=0.80417

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Step-by-step explanation:

Part A:

Binomial Distribution is used because the number of tracks and the probability to find the intruding object is constant for all tracks

It means:

P(X=x)=n_{C_{x}}*p^x*(1-p)^{n-x}

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Part B:

Value of p=0.8

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Part C:

x=15

P(X=15)=20_{C_{15}}*0.8^{15}*(1-0.8)^{20-15}\\

By solving above Expression:

P(X=15)=0.17456

Part D:

P(X≥15)=P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)

P(X=15)=20_{C_{15}}*0.8^{15}*(1-0.8)^{20-15}\\

P(X=15)=0.17456

P(X=16)=20_{C_{16}}*0.8^{16}*(1-0.8)^{20-16}\\P(X=16)=0.21819

P(X=17)=20_{C_{17}}*0.8^{17}*(1-0.8)^{20-17}\\P(X=17)=0.20536

P(X=18)=20_{C_{18}}*0.8^{18}*(1-0.8)^{20-18}\\P(X=18)=0.13691\\\\P(X=19)=20_{C_{19}}*0.8^{19}*(1-0.8)^{20-19}\\P(X=19)=0.05765\\\\P(X=20)=20_{C_{20}}*0.8^{20}*(1-0.8)^{20-20}\\P(X=20)=0.01153

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P(X=15)+P(X=17)+P(X=16)+P(X=18)+P(X=19)+P(X=20)\\=0.17456+0.21819+0.20536+0.13691+0.05765+0.01153\\

P(X≥15)=0.80417

Part E:

Mean Distribution:

E(x)=n*p

E(x)=20*0.8

E(x)=16

E(x) tells us that out of 20 tracks the SBIRS will detect from 16 tracks.

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