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k0ka [10]
3 years ago
11

g find all the values of p for which the following functions are improperly integrable on the indicated domain. be sure to prove

your claims, ie show that it is improperly integrable for those numbers p, but is not for other numbers 1.f(x)=1/x^p on i =(0,1) 2. f(x) = 1/(x(lnx)^p) on I = (e,inf)
Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

1. In order to make the integral improper  p  must be 1.

2. In order to make the integral improper  p  must be 1.  

Step-by-step explanation:

Using the rules of integration we get that for

f(x)= \frac{1}{x^p}

\int\limits_{0}^{1}  \frac{1}{x^p}  \, dx  = \frac{x^{1-p}}{(1-p)}  \, |\limits_{0}^{1}   = \frac{1}{1-p} - 0 = \frac{1}{1-p}

Therefore in order to make that integral improper  p  must be 1.

If  p = 1     then you would have a 1/0  indeterminate form.

2.   Using the of integration, specifically substitution we get that for

f(x) = \frac{1}{x(ln(x)^p)}

\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx  = \frac{(ln(x))^{1-p}}{1-p} \, |\limits_{e}^{\infty}

For   p \geq 1  we would have

\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx  = \frac{1}{p-1}

And the problem is the same.  If  p=1   we would have a 1/0 indeterminate form.

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Suppose that 2,4166666 is just 2,416. So the fraction will be 2 in front of a fraction 416 (in nominator) and 1000 ( in denominator) I hooope you understood
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Elena has $0.90. Kia has 10 times as much money
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Answer:

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Troy has one-tenth of the money Elena has. Troy has less money than Elena.

Step-by-step explanation:

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2 years ago
Let n1equals50​, Upper X 1equals30​, n2equals50​, and Upper X 2equals10. Complete parts​ (a) and​ (b) below. a. At the 0.05 leve
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Answer:

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082    

p_v =2*P(Z>4.082)=4.46x10^{-5}  

So the p value is a very low value and using any significance level for example \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation  

X_{1}=30 represent the number of people with a characteristic in 1

X_{2}=10 represent the number of people with a characteristic in 2

n_{1}=50 sample of 1 selected  

n_{2}=50 sample of 2 selected  

p_{1}=\frac{30}{50}=0.6 represent the proportion of people with a characteristic in 1

p_{2}=\frac{10}{50}=0.2 represent the proportion of people with a characteristic in 2

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion 1 is different from proportion 2 , the system of hypothesis would be:  

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{30+10}{50+50}=0.4  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082    

4) Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test.    

Since is a two sided test the p value would be:  

p_v =2*P(Z>4.082)=4.46x10^{-5}  

So the p value is a very low value and using any significance level for example \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.  

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