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s2008m [1.1K]
3 years ago
7

Determine which values of p the following integrals converge. Give your answer in each case by selecting the appropriate inequal

ity and then entering a numerical value in the box to define a range of p values for which the integral converges. If the integral never converges, select = and enter none for the numerical value. In either case be sure that you can rigorously prove convergence and divergence for different values of p.
a) integral from 1 to 2 of (dx/(x(ln(x))^p)

b) integral from 0 to 1 of (ln(x)dx/x^p)
Mathematics
1 answer:
sashaice [31]3 years ago
7 0

Answer:

Step-by-step explanation:

a)

\int\limits^2_1 {\frac{1}{x(lnx)^p} } \, dx

this can be done by substitute lnx = u

dx/x = du

When x =1, u =0 and when x =2, u = ln 2

So integral = \int\limits^{ln2} _0 {du/u^p} \\\=\frac{u^{-p+1} }{-p+1}

We find that this integral value is not definid for p =1

Hence for values of p other than 1, this converges.

When we substitute limits

\frac{1}{1-p} ((ln2)^{1-p} -1)

and converges for p ≠1

b) \int\limits^1_0 {lnx}/x^p \, dx \\\int \frac{\ln \left(x\right)}{x^p}dx=\frac{1}{-p+1}x^{-p+1}\ln \left(x\right)-\frac{x^{-p+1}}{\left(-p+1\right)^2}+C

So not converging for p =1

But ln x is defined only for x >0

So integral 0 to 1 makes this integral not valid and hence not convergent.

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Stanley has a collection of seashells. He found 35% of his collection on Florida beaches. If Stanley has 49 seashells from Flori
m_a_m_a [10]

Answer:

The number of seashells he have in his collection all together is <u>140</u>.

Step-by-step explanation:

Given:

Stanley has a collection of seashells. He found 35% of his collection on Florida beaches.

Stanley has 49 seashells from Florida.

Now, to find the number of seashells of his collection altogether.

Let the number of seashells all together be x.

Percentage of seashells found on Florida beaches = 35%.

Number of seashells found on Florida beaches = 49.

Now, to get the number of seashells altogether we put an equation:

35\%\ of\ x=49.

⇒ \frac{35}{100} \times x=49

⇒ 0.35\times x=49

⇒ 0.35x=49

Dividing both sides by 0.35 we get:

⇒ x=140.

Therefore, the number of seashells he have in his collection all together is 140.

4 0
3 years ago
Andy was given data to draw 3 scatter diagrams.
vredina [299]

Answer:

A) Negative Correlation. B) C. C) B

Step-by-step explanation:

<h2>A:</h2>

As the line of best fit is travelling downwards, it is a negative correlation, if the line of best fit is travelling upwards, it is a positive correlation, if the points on the graph are all over the place, there is no correlation.

<h2>B:</h2>

As there are only three types of correlation, the strongest is always the positive.

<h2>C:</h2>

The answer is B because there is no correlation and that means that there shouldn't be a line of best fit.

6 0
2 years ago
Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi
Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0
3 years ago
Plzzz help with this
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The volume of the cylinder is 1,286.1.
7 0
2 years ago
I need help with this can you?
Reptile [31]

Answer:

A and  D

Step-by-step explanation:

pls make me brainliest

7 0
3 years ago
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