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stellarik [79]
4 years ago
13

Verify the identity

e="csc^{2}x-csc^{2}xcos^{2}x=1" alt="csc^{2}x-csc^{2}xcos^{2}x=1" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Degger [83]4 years ago
3 0

Answer:

The answer to your question is below

Step-by-step explanation:

                               csc²x - csc²x cos²x = 1

Process

1.- Convert the second term to sines

                             csc²x - cos²x/sin²x = 1

2.- Convert cos²x/sin²x to cot²x

                             csc²x - cot²x = 1

3.- Solve for csc²x

                             csc²x = 1 + cot²x

We have verified it, because there is a Pythagorean identity which sets that

                              csc²x = 1 + cot²x

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Read 2 more answers
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Answer:

P(\bar X >80)=P(Z>2.143)=1-P(z

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

X \sim N(\mu=74, \sigma=14)

The sampling distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

The deduction is explained below we have this:

E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu

Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)

Since the variance for each individual observation is Var(x_i)=\sigma^2 then:

Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}

And then for this special case:

\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)

We are interested on this probability:

P(\bar X >80)

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

Applying this we have the following result:

P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)

And using the normal standard distribution, Excel or a calculator we find this:

P(Z>2.143)=1-P(z

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