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3 years ago

I NEED HELP PLEASE, THANKS! :)

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

Answer:

First option is the right choice.

Step-by-step explanation:

$\frac{1}{\cos \left(x\right)+1}+\frac{1}{\cos \left(x\right)-1}\\\\\frac{\cos \left(x\right)-1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}+\frac{\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{\cos \left(x\right)-1+\cos \left(x\right)+1}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}\\\\=\frac{2cos\left(x\right)}{cos^2\left(x\right)-1}\\\\=\frac{2cos\left(x\right)}{sin^2\left(x\right)}$

$=2\cdot \frac{cos\left(x\right)}{sin\left(x\right)}\cdot \frac{1}{sin\left(x\right)}\\\\=-2cot(x)csc(x)$

Best Regards!

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The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

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$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$