calculating percentage error in experiment- help!?
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
338000
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
986574
if you are using a TI-83, use the Z-Interval test funcition.\n\nManual working:\n\nSampling error = 2.06(SD/sqrt n)\n\n\nlower boundary =  20 - 2.06(3/sqrt 25) = 18.764\nupper boundary =  20 + 2.06(3/sqrt 25) = 21.236\n\n96% confidence limits is\n18.764 < mean < 21.236
100648
It depends on your teacher / professor.  But typically,\nA+ 98 - 100%\nA  93 - 97%\nA- 90 - 92%\nB+ 88 - 89%\nB  83 - 87%\nB- 80 - 82%\nC+ 78 - 79%\nC  73 - 77%\nC- 70 - 72%\nD+ 68 - 69%\nD  63 - 67%\nD- 60 - 62%\nF  0 - 59%\n\nHope this helps.  Good luck and have a nice day!
982793
0.0631, 3/5, .63\n\nCan't read your whole question
896594
The width of the confidence interval is 40, so the error about the mean is 20. You have a sample of 25 batteries, so the number of degrees of freedom is 24. The critical t-score for 24 degrees of freedom and a confidence level of .95 is approximately 2.063899.\n\nWe can use the t-score to approximate the sample standard deviation:\n\nt = error / (s/sqrt(n))\nSolving for s we get \ns = error*sqrt(n)/t, \ns = 20*sqrt(25)/2.063899\ns = 48.45199
195594
heres 1 from my university. dont know how much itll help.\n\nhttp://www.upscale.utoronto.ca/PVB/Harrison/ErrorAnalysis/index.html
683869
0.01 is already decimal..\nif you want 0.01%, it should be 0.01/100=0.0001
752897
Here is a link to a document that clearly explains the case:\nhttp://www.uvm.edu/~jleonard/AGRI85/spring2004/Standard_Error_Bars_in_Excel.html
793664
What's your problem - your in college - do your own homework (this son of a statistician is unimpressed).  If you don't know how to construct a confidence interval look at http://facweb.bhc.edu/EastCampus/leeb/ma108/06_01w/tsld032.htm\n\nGood Luck!!!
871703
Sorry, if the result can be greater than 1 then you ain't estimating a proportion.  For proportions each unit in the sample gives you either 0 or 1 as a result.\n\nYou need to swith to the general case where you calculate the standard deviation from the sum of squared differences from the mean, etc. PQ/n or whatever cannot be applied.
311582
the correct answer should be # 2 because the main point of the experiment is that "the same steps don't always produce the same results"
140069
If all questions are worth equal value, each is worth 2%. 1 wrong is 98%, 2 wrong is 96%, and so on. 9 wrong would be 82%.\n\nYou can use the formula y=100-2x, where y is the percentage right, and x is the number you got wrong. Or for any test, y=100-(100/z)x, where y is the percentage right, x is the number you got wrong, and z is the number of questions on the test.
429498
To calculate an answer to this question you need to specify a few other parameters. For example the size of the tree is important (the larger the tree the smaller the percentage), the location of the pot is important too, is it one metre away from the trunk of the tree? Over what period of time are you interested?  If you factor in things like prevailing weather, wind, species of tree etc. then it becomes very difficult to calculate a percentage.\n\nMy best guess is that less than a percent of the trees flowers would end up in the pot.  You could of course experiment with a pot but I don't fancy your chances of counting the flowers on the tree if you pick a big tree.\n\nWhat inspired you to ask this question? Is it an assignment?\n\nRegards \n\n\n\n\nr
167644
Divide it using decimals then multply the decimal by 100 and that will be your percent\n\nIs it just me or did someone else use this same exact answer below.  seriously people use your own imaginations and brains.
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
608759
Missed/Percentage Correct\n\n0-100%\n1-99%\n2-98%\n3-98%\n4-97%\n5-96%\n6-95%\n7-94%\n8-93%\n9-93%\n10-92%\n11-91%\n12-90%\n13-89%\n14-88%\n15-88%\n16-87%\n17-86%\n18-85%\n19-84%\n20-83%\n21-83%\n22-82%\n23-81%\n24-80%\n25-79%\n26-78%\n27-78%\n28-77%\n29-76%\n30-75%\n31-74%\n32-73%\n33-73%\n34-72%\n35-71%\n36-70%\n37-69%\n38-68%\n39-68%\n40-67%\n41-66%\n42-65%\n43-64%\n44-63%\n45-63%\n46-62%\n47-61%\n48-60%\n49-59%\n50-58%\n51-58%\n52-57%\n53-56%\n54-55%\n55-54%\n56-53%\n57-53%\n58-52%\n59-51%\n60-50%\n61-49%\n62-48%\n63-48%\n64-47%\n65-46%\n66-45%\n67-44%\n68-43%\n69-43%\n70-42%\n71-41%\n72-40%\n73-39%\n74-38%\n75-38%\n76-37%\n77-36%\n78-35%\n79-34%\n80-33%\n81-33%\n82-32%\n83-31%\n84-30%\n85-29%\n86-28%\n87-28%\n88-27%\n89-26%\n90-25%\n91-24%\n92-23%\n93-23%\n94-22%\n95-21%\n96-20%\n97-19%\n98-18%\n99-18%\n100-17%\n101-16%\n102-15%\n103-14%\n104-13%\n105-13%\n106-12%\n107-11%\n108-10%\n109-9%\n110-8%\n111-8%\n112-7%\n113-6%\n114-5%\n115-4%\n116-3%\n117-3%\n118-2%\n119-1%\n120-0%