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As long as I'm within 10% or so reinout I'm happy!!

And Rosala is quite right.

Have a look at this;

shift log?

Do you mean a logarithm in general?

Have a look at this page;

I think your square is 63/256th Alan 

This paper essentially plots what I've plotted above (but with more data points).  Where do your DO values come from?  The dissolved oxygen values in the paper are plotted against temperature; the CO₂ values against time. I can't find your DO values in the paper anywhere.

Ah! I think CO₂ is actually C_O2, for concentration of oxygen.  

Since your values of DO are different from those in the paper, does that mean you have measured them yourself?  If so, is your problem that your times are different from those in the paper or what?  I'm still puzzled as to your specific difficulty.  The data in the paper (that in the last table) is easy to plot in the same way that I've done in the graph in my earlier answer.

All you need to know are some of the basic rules in algebra;

$$(x^a)^b = x^{a \times b}$$

$$x^a \times x^b = x^{a+b}$$

$$\frac{x^a}{x^b} = x^{a-b}$$

So for d

$$\frac{(x^5)^2 \times x^8}{(x^4)^4} = \frac{x^{10} \times x^8}{x^{16}} = \frac{x^{18}}{x^{16}} = x^2$$

for e

$$\frac{(x^2)^3 \times (y^4)^2}{xy^3 \times x^5} = \frac{x^6 \times y^8}{x \times y^3 \times x^5} = \frac{x^6 \times y^8}{y^3 \times x^6} = \frac{x^6}{x^6}\times \frac{y^8}{y^3} = 1 \times y^{8-3} = y^5$$

I can't see f because the 'top answers' part is blocking my view. 

Reinout 

i think its 1/16 alan 

Perhaps this picture will help.  A quarter of the largest square has been divided into four.  What fraction of the whole do you think the black square is?

one fourth divided by four = 1/4 divide 4

reciprocate = 1/4 * 1/4 

= 1/16 is the answer

Thank you very much for helping me! That's the link but it's in greek. sorry...

See graph, but can't be sure from this!  Can you supply the URL that points to the place on the internet where you found this data?

 It's the etc. that's puzzling!  Other than that you have only three values to plot:

The oxygen concentration (in mg/L) is plotted against time (in seconds) here (the third column is just the total time in seconds).

Thanks a lot for answering! If i give you this table could you maybe explain to me how are the measurements of time depicted like this ( i found it as an example for cod on the internet but i can't get it)

Cool, Melody!!.....I like that compact "box" diagram.....very nice....

$$\frac{1}{2}+\frac{3}{8}=\frac{4}{8}+\frac{3}{8}=\frac{7}{8}$$   

So the 9 deer must be   $$1-\frac{7}{8}=\frac{1}{8}$$      of the total 

If 1/8 of the herd is 9 then ALL of the herd must be 9*8= 72 deer.

Does that make sense?

OK, Rosala...let's examine these statements one-by-one and find a solution.

First. let the total number of deer = x

And half of them are grazing, so that just must be !/2 of x = (1/2)x

And "the remainder" must just be the other half = (1/2)x

And (3/4) of the remainder - (1/2)x -  are playing..... "of" usually means "multiply"...so we have..... (3/4) x (1/2)x = (3/8)x ..

....and 9 others are drinking

So we have

(1/2)x + (3/4)(1/2)x + 9 = x

(1/2)x + (3/8)x + 9 = x     ........ Note that..... (1/2)x = (4/8)x 

(4/8)x + (3/8)x + 9 = x         Add the fractions

(7/8)x + 9 = x        Subtract (7/8)x from both sides

9 =  x - (7/8)x         Note that..... x = 1x = (8/8)x

9 = (8/8)x - (7/8)x

9 = (1/8)x                  Multiply by the reciprocal of (1/8)  =  (8/1)  on both sides

(8/1)*9 = (8/1)* (1/8)x   

72 = x

And there's our answer.....72 deer

hi Rosala, I have referenced this puzzle from "Puzzles" in the sticky notes so it can be found again easily at a later date.

Brain Teaser posted by Rosala (25/5/14)

@@ End of Day Wrap - Sunday 25/5/14   Sydney, Australia   Time 19:50

Hi Everyone,

Today was a quiet Saturday/Sunday.  It's nice to have a quiet day occasionally.  Still there were some excellent answers given by Alan, reinout-g, CPhill and Bertie.  Thank you.

I would like to introduce you to Rosala.  Rosala has only been a member for a few days and she has already greatly impressed me.  Rosala asks good questions, she is quick to say thank you, and gives 'thumbs up' generously on her own questions and also on other people's questions.  She also answers questions well when she has the knowledge.  Rosala is also very prepared to ask for clarification and to say when she does not understand.  This is very important and something not many question askers really do. She may be a very important role model for others to follow!

Now that I have told you why Rosala has impressed me, I'd like to tell you a little about who she is. She is a 13 year old girl from Delhi, India.   I don't know that we have any other regular members from India. She is very proud of her excellent Engish, it is her favourite subject.  She said 'we' are lucky because we come from counties where girls can do anything.  There are much less opportunities for girls in India.  

Thank you Rosala and welcome to web2.0calc forum.  I hope that you continue to learn lots and enjoy this site!

(Incidentally, I did ask Rosala if it would be alright before I repeated all this on the forum.)

##Here are a few posts that may be of interest to people.  

This one demonstrates Rosala's great attitude.

(Note: Rosala did not give herself 5 points - they came from other people)

Is there any limit on the number of books that can fit on a shelf? Is n>k? n<k? n=k?  Are all these allowed?

I assume cod is chemical oxygen demand.

You don't say what you mean by "per 15 seconds" etc.  Do you mean you take a measurement of dissolved oxygen, in mg/L, every 15 seconds and then average the values you collected during the two minute period? Then, for the next four minutes measure the oxygen every 30 seconds and average these values, etc?  If so, then I think I would plot the values against the middle of each time period.  That is, plot the first value at 1 minute, the second value at 4 minutes (the middle of the period from 2 minutes to (2+4) minutes) etc.

However, this might be a completely incorrect interpretation, so you need to provide more details of exactly what you are doing.

Sunday 25/5/14

This one demonstrates Rosala's great attitude.

Your interpretation of the question makes sense, Reinout.  I'm glad I didn't think of that interpretation or it would have been my head that was spinning!