Using a standard plate count for microorganisms, 0.1ml of a 10^ -2 dilution of water yielded 280 colonies...?
606825
Here is the actual formula:\n\nColony forming unit / mL = cfu/mL = # of colony * (1/dilution factor) * correction factor\n\ncorrection factor is the reciprocal of the volume you used in inocculation.\n\nOverall for your sample -->\n\ncfu/mL= 280*(1/10^-2)*(1/10^-1) = 28 * 10^4
222016
I am in a Microbiology college course. The closest I can come to your question from my text is about 50 NM (nanometer).
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
758825
If you are trying to make a 1:80 dilution of some solution, by adding it to 3 L of water,  you need the 3 L of water to be 79 x the solution amount, S.\n\nThis will satisfy the need for S/(S+3) = 1/80, since 3=79S, and S+3=80S.\n\nTherefore, S=3/79, which rounds to 0.037975 L, or 37.975 mL.\n\nIf you are simply trying to combine some amount of solution so that 3 L is 80 times as much as the amount of the solution, then you need 3/80 or 0.0375L or 37.5 mL of the solution and the mixing ratio will be 1/80 or 1:80.  \n\nIt wasn't clear whether you wanted the mixing ratio or the dilution factor (also a ratio), so I've given you both.
869195
I will assume because it is not clear that the number of bacteria is given by the formula:\n\nf(t) = 720*(3/2)^t\nWhen t = 0, this formula will give\nf(0) = 720 * (3/2)^0 = 720 * 1 = 720, which is consistent with the data.  I will also assume that t is given in hours. \n\nAt 11 am t = 3 hours \n\nPutting the time in the formula:\n\nf(3) = 720 * 1.5^3 = 2430\n\nIf the formula is different from what I understood, just put t=3 in the right formula.
87641
There are hundreds if not thousands of insects per each human so if you count them, fish, etc.  You have thousands to one.  If you want to count bacteria as animals then it is billions to one.
116087
keeping with the same unit fractor harry is correct.\n \n .2 ml
460952
Two hundred quattuordecillions\n\nCan be written using scientific notation as 2.0 x 10^47
428544
Mmmmmmmm....microbes.
1179743
if you need to see bacterial morphology, it's probably 100x objective (total magnification = 1000x).  This would be under oil immersion.  This is the only magnification i know of that comes standard on a compound microscope that would allow you to see bacterial morphology clearly....anything lower (ie-400x) and you won't be able to see much more than some tiny speckles on a slide.
883446
Hi, it depends on particular organisms. Check the link below for basic mathematical formulas you may be able to use.\n\nA study is done in which the number of bacteria in a liquid culture is monitored over a period of several hours. Ten observations were made:\n\n\nTwo questions need to be answered:\n\nIs this data consistent with an exponential growth model: N = N0ekt, where \nN0 = the number of bacteria present initially (at t = 0)\nk = a growth rate constant\nN = the number of bacteria present at a given time t,\nand e is the fundamental mathematical constant forming the base for natural logarithms. \nIf the answer to the first question is "yes", then what are the values of N0, k, and t2 (the so-called doubling time) for this bacterial system?\nThe experimental measurements listed above indicate the number of bacteria are growing with time. If we plot the observed values of N (vertically) against the elapsed time, t, in hours, we get a graph (available on web page below).\n\nYou can draw a smooth curve through the experimental points as shown in this figure, and the points all seem to lie relatively close to this curve. Further, this curve looks quite a bit like the graph of an exponential function, but from what we see, it might just as well be part of a parabola or of one of a number of other curves that result when various algebraic forms are graphed. This graph suggests that the answer to the first question may be "yes", but we would usually want stronger evidence than this to draw a definite conclusion. The trouble here is that it is difficult to distinguish precisely and reliably between the curves are the graphs of quite different mathematical functions.\n\nThere is, of course, one curve that is easily recognized by everyone - the straight line! In this situation, the logarithmic function can be used to rewrite the exponential growth formula into a form that will allow us to create a plot that does give a straight line when data is the result of an exponential growth (or decay) process. Start with the exponential growth formula:\n\nN = N0 ekt\n\nSince this involves an exponential with base e, it is "natural" to use natural logarithms - those based on the number e. Using the more common notation ‘ln’ for ‘loge’, we simply take the logarithm of both sides of the previous equation:\n\nln N = ln(N0 ekt)\n\n= ln N0 + ln(ekt)\n\n= ln N0 + kt ln e\n\n= ln N0 + kt\n\nHere, we have used two of the three basic properties of logarithms to manipulate the right-hand side, namely that\n\nlog (AB) = log A + log B \n\nlog (Am) = m log A \n\nFurther, in the last step we took advantage of the fact that by choosing to work with natural logarithms, ln e = logee = 1. Thus, the exponential growth formula can be written in the logarithmic form\n\nln N = kt + ln N0\n\nIf the exponential growth formula is correct and we were to plot ln N along the vertical axis of our graph (rather than N itself), and t along the horizontal axis, then the equation of the plot we are preparing will have the generic form\n\nvertical quantity = constant x horizontal quantity + another constant \n\nBut this is just the equation of a straight line. The plot that results using the data given above is available on web site below.\n\nAlthough the values written on the vertical axis are values of N, this graph is ruled vertically so that the vertical distances are actually proportional to ln N. Since one scale is logarithmic (the vertical one) whereas the other graph scale, the horizontal one, is the ordinary uniform spacing type, this is called a ‘semilogarithmic plot’.\n\nAt any rate, the figure indicates that the points are scattered about a straight line, the scatter due to small random errors in measurement. Since the plot of ln N vs t gives a straight line, we have confirmed that\n\nln N = kt + ln N0 (equation *) \n\nis a valid formula for this data, and therefore, so is its equivalent, N = N0 ekt, valid for this data.
1096663
A=1000 +.05*t\nB=1500+d*t\nday^-1 cancels with the unit of time days\n3(1000+.05*t)=1500+d*t\n3000+.15*t=1500+d*t\n3000-1500=d*t-.15*t\n1500=d*t-.15*t\n1500=(d-.15)t\n1500/(d-.15)=t\nt is how long it well take.\nIf d is equal to or less then .15 culture B well never have 3 times as many cells as A.
263670
Hmmm... judging by your interest in cleaning products and dishwashing detergents, I'm guessing you're in that industry or just a student looking for some answers for a project.\n\n-Insufficient data - \nAre you referring to dishwater in a more sterile kitchen or a dirty restaurant?  Is the piping clean or are we talking rusty piping?  Filtered water or non filtered water? Hmmm... so, at STP you want to figure out this. \n\nHere's what you would need.  First figure out the type of microbes that would typically exist in your sink.  Or if you don't have one, get some kind of bacteria.  Take the whole thing, perform your experiment.  Take some water samples.  Finally, send the water samples to a lab immediately for a count.  You could get a concentration of microbes per given volume.  My advice -- it's all I can do.
900972
You could use a turbidometric method that could detect the diffraction of light through the liquid and correlate it with the amount of microbial growth in it.  \n\nYou could also use a calibrated loop to plate the liquid and then count colonies.  I know they do this with urine cultures in hospital labs at times.  The loop picks up a certain amount of liquid like 10 microliters or so and then you can correlate the growth with this 10 microliter figure.
779549
OK , i'll give it try.  we are assuming that the water is distilled because that affects the specific gravity.  grams per cc of h20.  4 oz. of water =1 cup......so 1oz of water =.0296 liters.......4 x that  = .1184 liters.....devide that by 1000 to get the mass or gram  value of 1.184 cubic centimeters = 1 cup of water...give me 10 pts for effort.!!  I think i messed up on the conversion but believe the answer is correct!
193064
the micrometer (µm). 1 µm = 10^-6 meter
383531
========\n\nApparently purity, calculation or measurement errors have been there.\n\nYou repeated the thrice from the same solution to get an average.\n\nTry another solution.