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JulsSmile [24]
1 year ago
10

For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate, o

r if neither method is appropriate. Do not evaluate the integrals.
a. ∫ xSinx dx
i. substitution
ii. neither
iii.integration

b. ∫ x⁴/(1+x³). dx
i. substitution
ii. neither
iii.integration

c. ∫ x⁴. e^x³. dx
i. substitution
ii. neither
iii.integration

d. ∫x⁴ cos(x⁵). dx
i. substitution
ii. neither
iii.integration

e. ∫1/√9x+1 .dx
i. substitution
ii. neither
iii.integration
Mathematics
1 answer:
Andrei [34K]1 year ago
6 0

Answer:

a. ∫ xSinx dx

iii. integration by parts

u =x and dv= sinx

b. ∫ x⁴/(1+x³). dx  

ii. neither

Long division is an option here before integration is done

c. ∫ x⁴. e^x³. dx

i. substitution

where u = x⁵

d. ∫x⁴ cos(x⁵). dx

i. substitution

where u = x⁵

e. ∫1/√9x+1 .dx

i. substitution

where u = 9x+1

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An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
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Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

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Answer:

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