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pogonyaev
3 years ago
7

Integrate sec x ( sec x + tan x ) dx

Mathematics
1 answer:
il63 [147K]3 years ago
8 0
<span>  by taking integral we get
integral sec(x) (tan(x)+sec(x)) dx
applying integral we get
 sec(x) (tan(x)+sec(x)) gives sec^2(x)+tan(x) sec(x)
  = integral (sec^2(x)+tan(x) sec(x)) dx Integrate the sum term by term
= integral sec^2(x) dx+ integral tan(x) sec(x) dx For the integrand tan(x) sec(x), now we will use substitution

substitute u = sec(x) and du = tan(x) sec(x) dx
 = integral 1 du+ integral sec^2(x) dx The integral of sec^2(x) is tan(x)
 = integral 1 du+tan(x) The integral of 1 is u
 = u+tan(x)+constant
 Substitute the value of u which is  equal to 
 = sec(x):
 so our conclusion is 
:tan(x)+sec(x)+constant
hope this helps</span>
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Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{182 - 187}{4}

Z = -1.2

Z = -1.2 has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182

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3 years ago
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Answer:

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3 years ago
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