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3 years ago

Compute |1-3+5-7+...+13-15|

Mathematics

1 answer:

Allushta [10]3 years ago

4 0

Answer:

Step-by-step explanation:

There are a couple different ways you can solve this, but this is generally the quickest way.

Identify a pattern in the problem. We can see that $1-3=-2$ , $5-7=-2$ , and so on. Subtracting every two odd numbers (without repeating any numbers) in that problem will always give us the same answer: -2.

So, we see how many pairs of odd numbers there are.

1 and 3, 5 and 7, 9 and 11, 13 and 15. We have <u>four</u> pairs of odd numbers. We can just multiply -2 by 4, which leaves us with an answer of -8.

Finally, since this is also an absolute value equation, the answer will be positive, so the final answer is 8.

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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".

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Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

$\bar X = \frac{740+920}{2} = 830$

and the margin of error is given by :

$ME = \frac{920-740}{2}= 90$

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 =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (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:

$n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11$

So the answer for this case would be n=11 rounded up to the nearest integer

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