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Aleks04 [339]
3 years ago
15

" \frac{1}{8} + \frac{1}{3} = " alt=" \frac{1}{8} + \frac{1}{3} = " align="absmiddle" class="latex-formula">

Mathematics
1 answer:
barxatty [35]3 years ago
6 0

Answer:

It won't do you any good to just copy the answer. You need to know that, for example,

\frac{13}{17}=\frac{13}{17}\times\frac{123457}{123457}=\frac{13\times123457}{17\times123457}=\frac{1604941}{2098769}

The 3 numbers (13,17,123457) are just examples; the equation is true for all such numbers except zero.

Once you know that, you need to know how to add fractions with the same denominator:

\frac{2}{23}+\frac{3}{23}=\frac{5}{23}

Next, you have to know that you can't add fractions with different denominators.

Instead, you have to use the first trick to change the fractions so they have the same denominators without changing the value of either fraction.

\frac{1}{3}+\frac{1}{8}=\frac{1}{3}\times\frac{8}{8}+\frac{1}{8}\times\frac{3}{3}=\frac{1\times8}{3\times8}+\frac{1\times3}{8\times3}=\frac{8+3}{3\times8}

Once you know how to do this, you can add fractions, but you might end up with results that are like

\frac{1604941}{2098769}\text{ which is really just }\frac{13}{17}

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Answer:

y = 6

Step-by-step explanation:

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Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
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Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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The answer is The second one because I took this test before and I got it right
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The answer for this question is C. X=5.41

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