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Ludmilka [50]
4 years ago
11

{29, 29, 29, 28, 28, 27}

Mathematics
2 answers:
spayn [35]4 years ago
7 0
Hello!

You first have to look and see if there is an outlier

An outlier is something that is away from the rest of the data

All the numbers are with 1 from each other so there is no outliers

The answer is the first one

Hope this helps!
son4ous [18]4 years ago
4 0

Answer:

The answer is mean or median because there is no outlier

Step-by-step explanation:   I took the quiz and got it right.

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The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

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