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3 years ago

Cot^2x•sec^2x convert each of these into sine and cosine and simplify

Mathematics

1 answer:

Nana76 [90]3 years ago

8 0

$\cot(x)=\frac{\cos x}{\sin x}$

$\sec x=\frac1{\cos x}$

$\therefore \cot^2x\cdot\sec^2x= \frac{\cos^2x}{\sin^2x}\frac{1}{\cos^2x}=\frac{1}{\sin^2x}$

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The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds

Bumek [7]

Answer: 11.5%

Explanation:

Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds.

Now, we get the z-score for 720 seconds by the following formula:

$\text{z-score} = \frac{x - \mu}{\sigma}$

where

$t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}$

So, the z-score of 720 seconds is given by:

$\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}$

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

$P(t \ \textgreater \ 720) 
\\ = P(z \ \textgreater \ 1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\ \boxed{P(t \ \textgreater \ 720) = 0.115}$

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice.

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