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loris [4]
4 years ago
7

If five lamps cost $200, what is the average price per lamp?

Mathematics
1 answer:
kkurt [141]4 years ago
3 0
$200/5 lamps= $40 per lamp

Final answer: $40
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Answer:

"ALL of the beverages at The Zest Nest ARE LIKELY to contain 46 grams of sugar or less, but NONE of the beverages at Jim's Juice Join ARE LIKELY to contain LESS THAN 46 grams of sugar."

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Suppose we take a random sample of 41 state college students. Then we measure the length of their right foot in centimeters. We
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Answer:

ME = \frac{25.09-21.01}{2}= 1.69

The general formula for the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the width is:

Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}

And if we decrease the confidence level from 95% to 90% then the critical value t_{\alpha/2} would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

Step-by-step explanation:

For this cae we know that the sample size selected is n =41

And we have a confidence interva for the true mean of foot length for students at a college selected.

The confidence interval is given by this formula:

\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the 95% confidence interval is given by: (21.71,25.09)

A point of etimate for the true mean is given by:

\bar X = \frac{21.71+25.09}{2}= 23.4

And the margin of error would be:

ME = \frac{25.09-21.01}{2}= 1.69

The general formula for the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

And for this case the width is:

Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}

And if we decrease the confidence level from 95% to 90% then the critical value t_{\alpha/2} would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

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The answer is below in the picture as well

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साहस को सलाम पाठ/अरुणिमा सिन्हा कहां बैठी थी

Step-by-step explanation:

I hope you understand

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Cos x = 15÷17
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