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Given the following series of dilutions, what is the total dilution factor.
But really, a single 1/10th dilution would suffice.
So our absolutely unavoidable error is plus or minus 500K/42million - about.1.
Another hint february 2013 calendar printable canada : You need to multiply your own use by 365 300 million.Voila, soup with clash of clans cheat tool v4.0 only 90,000 peas per liter.What counting method will you use.After that we could do standard 1/10th dilutions.In the end, I have 5 plates, labeled as shown below.That would be pretty difficult with a pipette.This is called multiplying by the reciprocal.Anyone can eat a soup that has 90 peas per liter.It tastes just like water.
Multiplying by the reciprocal of your sample size is the heart of counting bacteria by the serial diluation method.
I need a hint :Assume you are starting with 1,000,000,000 cells in one mL of solution another hint :After first dilution you are left with 10,000,000 cells in one mL of solution dilution factor 0 another hint :Dilution factor is the same each time another.
Take 1 mL of that and put it in 49 mLs of water.I need a hint : Remember you have basically counted 1/10,000th of the original bacteria.So our absolutely unavoidable error is plus or minus 500/182,000 - about.27.Simply provide concentration and dilution values to get serial dilution calculation.I need a hint : inverse of 1/20,000 is 20,000 another hint : 27 20,000 I think I have the answer:540,000 per mL I did a series of dilutions, with dilution factors.1,.1, and.01.And all that matters is that we get the ratio right hmm let s say we started with a hundredth of a pot of the old soup and added 99 hundredths of a pot of water.


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