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3 years ago

The partial pressure of carbon dioxide in a mixture is 0.31 atm.

Mathematics

1 answer:

juin [17]3 years ago

4 0

Answer:

$P_{Co2}= X_{Co2} P_{tot}$

Where $P_{tot}$ represent the total pressure and $X_{Co2}$ the fraction of carbon dioxide is 0.46 and we can find the total pressure with this formula:

$P_{tot}= \frac{P_{Co2}}{X_{Co2}}$

And replacing we got:

$P_{tot}= \frac{0.31 atm}{0.46}= 0.674 atm$

Step-by-step explanation:

For this case the partial presure of carbon dioxide is given by:

$P_{Co2}= X_{Co2} P_{tot}$

Where $P_{tot}$ represent the total pressure and $x_{Co2}$ the fraction of carbon dioxide is 0.46 and we can find the total pressure with this formula:

$P_{tot}= \frac{P_{Co2}}{X_{Co2}}$

And replacing we got:

$P_{tot}= \frac{0.31 atm}{0.46}= 0.674 atm$

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Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

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The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

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Answer:

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Step-by-step explanation:

We can substitute 6 as x to find out:

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first, because of PEMDAS we'll do 6-3

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multiply 5 by 3 and add 6 and 13

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Since the statement is not true, that means that x=6 is not a solution

Hope this helps!

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Step-by-step explanation:

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