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Vinvika [58]
2 years ago
9

HELP ME OUT PLS!!!!!!!

Mathematics
1 answer:
Anarel [89]2 years ago
8 0
I think it is D
Hope his is right!!!
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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in
Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

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

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

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

\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

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3 years ago
Sunnydale elementary school is having its annual read-a-thon.the third grader have read 573 books so far.their goal is to read m
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Rasek [7]

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Step-by-step explanation:

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Can a triangle be drawn with side lengths 8.4 inches, 9.2 inches, and 17 inches?
sasho [114]

Answer:

yes

Step-by-step explanation:

based on the triangle inequality theorem, add the two smallest numbers together (8.4,9.2) to get 17.6 which is bigger than 17

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Graph the solution to the following system of inequalities: <br> 7x +4y&gt;=-12<br> -3x+4y&lt;4
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