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3 years ago

Please help desperate trying to solve

Mathematics

2 answers:

nata0808 [166]3 years ago

8 0

The sequence needs to be rearranged so the negative numbers come before the positive. The correct sequence would be:
-9, -2, 0, 1, 3, 5

Andrew [12]3 years ago

7 0

You have to put the negatives first, because they are lower than the positives. All the student did was put them from least to greatest, and ignored the negatives, basically. This is correct: -9,-2, 0, 1, 3, and five

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First we need to find the slope of the function

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So the slope is -3/2

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3 years ago

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3 years ago

1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample:

melamori03 [73]

Answer:

The 95 % confidence interval for the mean is between 28.09 and 35.97.

Step-by-step explanation:

Mean of the sample:

30 values.

So we sum all the values and divide by 30.

The sum of all the values(43 + 52 + 18 + ... + 20 + 41) is 961.

961/30 = 32.03

Confidence interval:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{11}{30} = 3.94$

The lower end of the interval is the sample mean subtracted by M. So it is 32.03 - 3.94 = 28.09

The upper end of the interval is the sample mean added to M. So it is 32.03 + 3.94 = 35.97

The 95 % confidence interval for the mean is between 28.09 and 35.97.

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3 years ago