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11Alexandr11 [23.1K]
3 years ago
12

What is the mean for 2,7,3,7,9

Mathematics
2 answers:
ale4655 [162]3 years ago
6 0
Simple add them all 2+7+3+7+9 is 28 then divide 28 by the amount of numbers (5) to get 5.6 or if you are rounding to the nearest number 6...  Hope this helped ;)
mote1985 [20]3 years ago
4 0
2+7+3+7+9= 28 28/5=5.6 The mean is 5.6
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Answer:

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Which number is an irrational number?
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Answer:

\boxed{\sqrt[3]{16} }

Step-by-step explanation:

=> \sqrt[3]{16} is an irrational number because it cannot be written as the form \frac{p}{q} which is the basic requirement of being a rational number.

=> \sqrt{100} = 10 (A rational number because it's a whole number)

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a group of young women decided to raise 480000 to start a business after some time 4 women pulled outand they had to pay additio
Debora [2.8K]

Answer:

96 is the number of original women investors.

Step-by-step explanation:

Let X be the number of young women in the initial group.  They raised 480000, so the average payment per person was (480000/X).  

When 4 pull out, the new average is (480000/(X-4)).  We are told that this new average required the remaining women (X-4) to add another 20000.

The four women therefore had contributed 20000 in total, making their average 20000/4 = 5000 each.  

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(480000/X) = 5000

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<u><em>The original number of women was 96.</em></u>

===

Check

(96 Women)(5000/Woman) = 480000  CHECKS

(4 Women pull out)*(5000) = 20000 that needs to be added to stay at 480000.  CHECKS

5 0
2 years ago
A coordinate grid is mapped onto a video game screen, with the origin at the lower left corner. The game designer programs a tur
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3 0
3 years ago
Read 2 more answers
How to find the limit
Ludmilka [50]
\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}

By the Stolz-Cesaro theorem, this limit exists if

\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}

also exists, and the limits would be equal. The theorem requires that b_n be strictly monotone and divergent, which is the case since k\in\mathbb N.

You have

a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}

so we're left with computing

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}

This can be done with the help of Stirling's approximation, which says that for large n, n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n. By this reasoning our limit is

\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}

Let's examine this limit in parts. First,

\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}

As n\to\infty, this term approaches 1.

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\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}

The term on the right approaches e^k, cancelling the e^{-k}. So we're left with

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}

Divide through the numerator and denominator by n^k:

\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}

So you can see that, by comparison, we have

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}

so this is the value of the limit.
8 0
3 years ago
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