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Komok [63]
3 years ago
5

Verify cotB+tanB=secBcscB

Mathematics
1 answer:
masya89 [10]3 years ago
3 0
Hmm let me stab at this
has. been a while and identifies aren't fresh.
\frac{1}{ \tan( \alpha ) }  +  \tan( \alpha )  \\  =  \frac{1}{ \tan( \alpha ) }  +  \frac{ { \tan( \alpha ) }^{2} }{ \tan( \alpha ) }  \\  =  \frac{1 +  { \tan( \alpha ) }^{2} }{ \tan( \alpha ) }  \\  =  \frac{  { \sec( \alpha ) }^{2} }{ \tan( \alpha ) }
so then we get
\sec( \alpha ) ( \frac{ \sec( \alpha ) }{ \tan( \alpha ) } ) \\  =  \sec( \alpha ) ( \frac{ \frac{1}{  \cos( \alpha )} }{ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } ) \\  =  \sec( \alpha ) ( \frac{1}{ \sin( \alpha ) } ) \\  =  \sec( \alpha )  \csc( \alpha )
vote brainliest if you like my answer!
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2 = b
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ur inequality for this line is : y < = x + 2

(0,6)(3,0)
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(0,6)...x = 0 and y = 6
sub and find b, the y int
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4 0
3 years ago
K^2+k=0 what does k=
FinnZ [79.3K]

Answer:

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6 0
4 years ago
A TV company when purchasing thousands electronic components apply this sampling plan: randomly select 15 of them and then accep
katrin [286]
<h2>Answer with explanation:</h2>

According to the Binomial probability distribution ,

Let x be the binomial variable .

Then the probability of getting success in x trials , is given by :

P(X=x)=^nC_xp^x(1-p)^{n-x} , where n is the total number of trials or the sample size and p is the probability of getting success in each trial.

As per given , we have

n = 15

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Probability of getting defective components = P = 0.03

The whole batch can be accepted if there are at most two defective components. .

The probability that the whole lot is accepted :

P(X\leq 2)=P(x=0)+P(x=1)+P(x=2)\\\\=^{15}C_0(0.03)^0(0.97)^{15}+^{15}C_1(0.03)^1(0.97)^{14}+^{15}C_2(0.03)^2(0.97)^{13}\\\\=(0.97)^{15}+(15)(0.03)^1(0.97)^{14}+\dfrac{15!}{2!13!}(0.03)^2(0.97)^{13}\\\\\approx0.63325+0.29378+0.06360=0.99063

∴The probability that the whole lot is accepted = 0.99063

For sample size n= 2500

Expected value : \mu=np= (2500)(0.03)=75

The expected value = 75

Standard deviation :  \sigma=\sqrt{np(1-p)}=\sqrt{2500(0.03)(0.97)}\approx8.53

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6 0
3 years ago
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weeeeeb [17]

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4 0
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