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3 years ago

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

2 answers:

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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What is the m∠B? Please help this is due soon.

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Answer:

add all given angle

sum of all angles is equal to 180 degree

5 0

3 years ago

Which number is an irrational number?

RSB [31]

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)

=> 1/8 (Rational as it is in the form p/q)

=> -2.2675 (Rational because it is an integer)

6 0

3 years ago

Read 2 more answers

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.

This would have been the same amount contributed by all X women. Thus, we can set the average (480000/X) equal to 5000

(480000/X) = 5000

(480000) = 5000X

(480000)/5000 = X

X = 96

<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

Phoenix [80]

Since the vertex form of a parabola is y=a(x-h)^2+k with h and k being the vertex (the x and y values, respectively), we get a(x-0)^2+20=ax^2+20. Plugging (10, 8) in to find a, we get 8=a(10)^2+20=a*100+20. Subtracting 20 from both sides, we get -12=100*a. Next, we divide 100 from both sides to get a=-12/100. Since h=0 and k=20, we have your answer!

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.

Next,

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

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