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2 years ago

5

Help please! I will give a like and Brainly pls ! Help me !:D

1 answer:

Lina20 [59]2 years ago

5 0

Answer:

C) $t = 2 \frac{1}{3}$

Step-by-step explanation:

Calculation steps:

Turn $2 \frac{7}{12}$ as a importer fraction = $\frac{31}{12}$

$\frac{31}{12} - \frac{1}{4}$

$\frac{31}{12}-\frac{1*3}{4*3}$

$\frac{31}{12}-\frac{3}{12}$

$\frac{31-3}{12}$

$\frac{28}{12}$

$\frac{28/4}{12/4}$

$=\frac{7}{3}=2\frac{1}{3}$

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-1.1.2<br> x(7x + 2)=1 (correct to TWO decimal places)

Angelina_Jolie [31]

Answer:

1.1.2

=1

Step-by-step explanation:

x(7× +2)=1

7X2=14

14÷7

=1

3 0

3 years ago

Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea use a 0.01 significance le

kenny6666 [7]

Answer:

The null hypothesis is $H_o: p \leq 0.2$

The alternate hypothesis is $H_a: p > 0.2$

The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.

Step-by-step explanation:

Test the claim that more than 20% of users develop nausea

This means that the null hypothesis is that 20% or less of the users develop nausea, that is:

$H_o: p \leq 0.2$

And the alternate hypothesis is that more than 20% develop, so:

$H_a: p > 0.2$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that $\mu = 0.2, \sigma = \sqrt{0.2*0.8}$

Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea.

This means that $n = 229, X = \frac{52}{229} = 0.2271$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.2271 - 0.20}{\frac{\sqrt{0.2*0.8}}{\sqrt{229}}}$

$z = 1.03$

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Probability of finding a proportion above 0.2271, which is 1 subtracted by the pvalue of z = 1.03

Looking at the z-table, z = 1.03 has a pvalue of 0.8485

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The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.

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3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

sineoko [7]

Answer:

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Step-by-step explanation:

For this case we need to determine if the following integral converges or not:

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$\int_{-\infty}^{-1} \frac{1}{(x-2)^3} dx$

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$\int_{-\infty}^{-3} \frac{1}{u^3} du=\int_{-\infty}^{-3} u^{-3} du$

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$=\frac{u^{-3+1}}{-3+1} =-\frac{u^{-2}}{2}=-\frac{1}{2}\frac{1}{u^2}$

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$= -\frac{1}{2} [\frac{1}{(-1-2)^2} -\lim_{x\to -\infty} \frac{1}{(x-2)^2}]$

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$\int_{-\infty}^{-1} (x-2)^{-3} dx =-\frac{1}{18}$

And we see that our integral on this case converged to -1/18.

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4 years ago

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vaieri [72.5K]

Answer:

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Step-by-step explanation:

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4 years ago

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4 years ago