Why do I have three separate weight ranges according to three separate weight charts?
366081
Weight charts are shiiit.\n\nYou can use the chart of me.\nYou sound like you're at the perfect weight for your height. You could possibly gain a few pounds though.
953641
I can only say so much because those equations are too long for two in the morning. Not that it matters; I just won't be retyping them. But you certainly need a bunch of random things for this. First, you need a variable -- a char, probably, since you don't care about any input beyond that. I don't know what the compare thing is going to output, but that'll come.\n\nI suggest an array for the old and new charts.\n\nGeez, what's the syntax? int oldchart[11], or something? I'm getting too used to Java. Anyway. Since you're going in increments of 5 from 0 to 50, you'll have a 0, 5, 10, ... 45, 50 -- that is, 11 different speeds, right? So you'll need an array for the old chart and an array for the new chart, both with a size of 11. Do you get where I'm going?\n\nCalculating the chart takes a bit more thinking. We'll ignore the actual process of getting input for the moment and just deal with filling it in. I'll suggest a method, and explain afterwards:\n\nint speed;\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   oldchart[i] = <calculate wind chill>;\n}\n\nThe same thing could be applies to the newchart array. The index, i, will go from 0 to 10, matching the indexes of the array, right? Since you're going by increments of 5, the speed for slot i will be i*5. That takes you from 0 to 50.\n\nSo after you pull that off for the old and new, you've got charts saved in the variables. Printing them is kind of similar to what happened above -- you use a loop to output what's in the array, using i*5 for the speed.\n\nNow, granted, you could combine this and the above step so you wouldn't need the arrays at all--you'd just recalculate and print every time. But I find it easier to organize my thoughts by going a way that deals with one issue at a time... It depends on what you prefer at this point.\n\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   cout << "Speed: "<< speed << " Wind Chill: " << oldarray[i] << endl;\n}\n\nLet's see...\n\nDisplaying the menu is easy--just use cout << "your menu stuff" << endl;\nInput is a bit stranger, I think. For something like this, just use a cin and check afterwards if it's right. Um, let's see... I'll give this to you in pseudocode--you should be able to work out the actual stuff.\n\nwhile ( the user doesn't want to exit )\n//this is just a logic thing -- if the user is done, then what?\n{\n     take the user's input;\n\n     if the input is 'o' or 'O'\n     {\n          calculate and print the old chart\n     }\n\n     else if the input is 'n' or 'N'\n     {\n          calculate and print the new chart\n     }\n\n     else if the input is 'c' or 'C'\n     {\n          calculate and print both the old and new charts (?) (whatever compare does)\n     }\n\n     else if the user wants to exit\n     {\n          end the program (usu. with return 0 or something of that nature)\n     }\n\n     else the user's input was bad\n}\n\nMind that for this to go well, you'd have to mind a few issues -- as I wrote above, if the user doesn't enter what you expect them to. Also, you'd have to deal with a case where the user decided to print charts before entering data into the old or new charts. For that, you should give the temperature a default value, like 0, so the charts can still be calculated.\n\nYou can see that I wrote in a calculation for every print, which is what I was saying before with it not being necessary. You could technically use the for loop I gave for printing but just replace the reference to the array with the calculation for the wind chill.\n\n...Don't know if I covered all of it, but I should be getting to bed. ^^;
799374
ok. i think i know what you're asking.\n\nDo you have a chart from whatever you're doing filled with information? Put that in your Results and Analysis section.\n\nBasically, just put anything with numbers into that section.\n\nYou could try to be just a little more vague when asking your question though  ;)
742071
You put tallies in the table.(Small lines in set of five).eg If digit 3 comes 8 times in a distribution, you put four tallies and cross them with fifth tally diagonally and than after a space add three more tallies,because its easy to count set of five in large data.if 8 comes 4 times than you put four tallies.(////)
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
1002465
there are different reasons. 3 is a very powerful number in a lot of cultures. also, in fairy tales they like to isolate people, things, etc. having thins come in 3s allow them to create a patten, and then have the third person or thing break this pattern.
73838
Its' probably a rounding error. It would depend on which masses for the elements you use, how precise you want to be.
179072
the answers is more complicated that 1, or 40, its asking for four different numbers.  He can put stones on easch side of the scale, so addition of weights as well as subtration can be used.  For example the weights x = 1, y = 3, z = 9, w = 27 you can get many weights.\n1 = [x], 2 = [y-x], 3 = [y], 4 = [x+y],  5 = [z-y-x], 6 = [z-y], 7 = [x+z-y], etc. 34 = [x+z+w-y], 35 = [y+w-x], 36 = [y+w], 37 = [x+z+w], 38 = [y+z+w-x], 39 = [y+z+w] 40 = [x+y+z+w]\n\ni have to go watch hustle and flow, but thats the idea
864833
There is no one answer to this question as it depends on the data being represented.\n\nfor example demonstrating the ditribution of a population such as percentages of ethnic minorities it might be more suitable to use a piechart as each wedge is clearly showing the amount of the whole,\n\nhowever comparing the reults of a survey it might be more suitable to use a bar chart as you may wish to be show a comparison between the results instead of as a proportion
335069
On one axis type, the other size.\nOn one axis color, the other size.\nOn one axis toxicity, on the other, size.\n\nAny of those are interchangable.
379102
The range is the lowest number to the highest number. You don't need to subract them (well, depends on the case I guess).
402289
Seperate balls in 3 packs of 3.\n\nFirst weigh pack 1 against pack 2. If both are the same, then your heavy ball is in pack 3. If one weighs more, you ball is in that pack.\n\nSecond you weigh 2 balls in the heavy pack. If both balls are the same, it's the 3rd ball. If one weighs more, that's your ball.\n\nToodles.
826531
tiny amounts of chemical can stick to differnt things, like stir bar, stir rod, thermometer and if you remove these items before the experiment is over then you remove some chemicals.  Other kinetics events that your not considering. as in pressure, humididtym and so on.  If you used a Ph probe then maybe it wasn't calibrated right or isn't properly maintained
808727
I hope that you are not going to be using that specific order though!  Looks pretty comprehensive otherwise, and the only remaining answer for your two pics/graphs is: you need enough to display all your results.
270417
Succinctly, for the normal ditribution, the three averages are equal, that is the mean, median and mode are all the same. Not so in other cases.
630210
Weighted moving averages assign a value to some of the numbers in a series to give them more of an effect on the resulting average.  It is the same effect as if you had extended the series and put some of the numbers in more times than they already were.\n\nI sometimes use weighted moving averages while examining forex charts.  The specific type of weighted average I like to use is an Exponential Moving Average (EMA).\n\nFor example, if I were looking at the hourly chart of the EUR/USD I could add the 20 bar Simple Moving Average (SMA) to show me what the average price is over the last 20 hours.  The SMA is not weighted, so all the bars will have the same effect on the average, which means if the price takes off it will not change quickly to reflect that.  If I use the weighted EMA, however, it will give more value to the most recent bars causing it to change more quickly as price changes.
789193
a 2x3 ANOVA has two factors one with 2 levels and one has 3 levels a 1x6 would mean that one factor only has one level which doesn't make sense because then it isnt a variable