25 Points! Midterm Today! Can anyone help with this problem?

Mathematics

1 answer:

Andrews [41]3 years ago

6 0

I really do wish I could help But I’m not sure I tried to figure it out and i just can’t get it

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Which one is it plsss help

anzhelika [568]

Answer:

You are correct it is c

Step-by-step explanation:

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2 years ago

12x^2y-32xy+8y factored form

siniylev [52]

Factor the following:
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Hint: | Factor common terms out of 12 x^2 y - 32 x y + 8 y.
Factor 4 y out of 12 x^2 y - 32 x y + 8 y:
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3 years ago

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Step-by-step explanation:

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3 years ago

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A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the populati

Marysya12 [62]

The $100(1-\alpha)\%$ confidence interval of a standard deviation is given by:

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

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval,

$\chi^2_{1- \frac{\alpha}{2} }=43.773 \\ \\ \chi^2_{\frac{\alpha}{2} }=18.493$

Therefore, the 90% confidence interval for the standard deviation is given by

$\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\ \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\ \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59} \\ \\ \Rightarrow41.4\leq\sigma\leq63.7$

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2 years ago