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Alinara [238K]
1 year ago
7

15. Which rule was used to create the arithmetic sequence below?

Mathematics
1 answer:
Nikolay [14]1 year ago
4 0

The rule used to create the arithmetic sequence -2, -4, -6, -8,.. is an = -2 - 2(n-1)

The arithmetic sequence is

-2, -4, -6, -8,...

The initial term = -2

The common difference = The second term - The first term

= -4 - (-2)

= -4 + 2

= -2

The nth term of the arithmetic sequence is

an = a+ (n-1) d

Where a is the initial term

an is the nth term

d is the common difference

Substitute the values in the equation

an = -2 + (n-1)(-2)

Rearrange the terms

an = -2 -2(n-1)

Hence, the rule used to create the arithmetic sequence -2, -4, -6, -8,.. is an = -2 - 2(n-1)

Learn more about arithmetic sequence here

brainly.com/question/12108818

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<h2>Answer:</h2>

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(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at col
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Answer:

a) n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean

\sigma=16 represent the population standard deviation

n represent the sample size  

Solution to the problem

Part a

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40

So the answer for this case would be n=40 rounded up to the nearest integer

Part b

For this case if we see the formula for the margin of error

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

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Step-by-step explanation:

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