- anonymous

Given the following,
x1=2 x2=-1 x3=-2 x4=5 x5=-4 x6=8
Find:
3
∑ ( x subi - 1 ) 2
i=1

- schrodinger

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- anonymous

I think I can handle this ;)

- anonymous

\[\sum_{i=1}^{i=3} (x_i - 1) ^2 \]

- amistre64

the suspense is killing me :)

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## More answers

- anonymous

You given 6 x_i values. The equation only wants you to consider x_i for (i=1 until i=3), that would be x1, x2, x3. The sigma is the symbol for sum. So the problem is to find the sum for the following events: plugging in each x_i subtracting 1 then squaring that. Hence you'll have 3 values that you then add to together to get the final value.
\[ i=1, x_1 = 2, (2-1)^2 = 1\]
\[i=2, x_2 = -1, ((-1) -1)^2 = ... \]
... and so on

- amistre64

could we expand the binomial and pick it apart sum for sum?

- anonymous

OH! THANK YOU. DPFLAN, you are the best <3

- anonymous

@amistre64 for sure, that would create even more suspense!

- amistre64

the butler did it, im sure of it ....

- anonymous

with the candlestick?

- anonymous

@sleestak No problem, glad I could help!

- amistre64

yes, with the candlestick lol ... get a clue

- anonymous

@sleestak I imagine you may have some more problems that are similar to this that you need to work on, so good luck

- anonymous

thanks. I'll tell you what I come up with

- anonymous

9

- anonymous

oh no. sorry one sec

- anonymous

14. OK. I think I can do the other problem. THANKS :D

- anonymous

x1 -> 1
x2 -> 4
x3 -> 9
SUM = 14, yeah, you're now an expert it seems

- anonymous

you guys need to list finance as the next subject :D

- anonymous

It actually had/has a pretty strong following/support from a professor at NYU - http://pages.stern.nyu.edu/~adamodar/

- anonymous

Math is the most active subject

- anonymous

oh. excellent.

- anonymous

So if your questions are mathematical, you'd probably be best served to ask them here because there is a greater mass of users in this group, so you're more likely get an answer.

- anonymous

yeah totally. this is awesome! I am going to use this all the time in school

- anonymous

awesome, tell your friends too! Also, I've the final question you asked about solved

- anonymous

*got

- anonymous

oh okay.
let me post

- anonymous

should I now 'close'
this question?

- anonymous

Yep

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