75
An eighth remains.
8
You find the sample space by enumerating all of the possible outcomes. The sample space for three coins is [TTT, TTH, THT, THH, HTT, HTH, HHT, HHH].
The sample space when tossing a coin three times is [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]It does not matter if you toss one coin three times or three coins one time. The outcome is the same.
125
Approx 1/8 will remain.
Not sure what you mean by "had-lives". After 3 half lives, approx 1/8 would remain.
An eighth remains.
It will take two half-lives or about 60.34 years for three-fourths of a Cs-137 sample to decay.
The fraction that remains is 1/8.
1/8 of the original amount remains.
If the half-life of a nuclide is 44.5 days, then 178 days is four half-lives. After four half-lives, 0.0625 (0.54) of the original nuclide remains. If 50g is the original mass, then the mass after 4 half-lives is 3.125g. AT = A0 2(-T/H) A178 = (50) 2(-44.5/178) A178 = 3.125
If the half-life of bromine is 25 minutes, than in 75 minutes three half lives would have passed. If you started with a 4.0 mg sample, than after three half-lives only 0.50 mg would remain. Because you are dealing with a whole number of half lives, all you have to do is divide 4.0 by 2 three time. Alternatively, you could multiply 4.0 by one half raised to the third power. This would look like this when entered into a calculator: 4.0 X (0.5^3) = 0.5
If I take a radioactive sample of 400 moles of an unknown substance and let it decay to the point of three half-lives I would have 50 moles left of the sample. 1/2 of what is left will decay in the next half-life. At the end of that half-life I will have 25 moles left of the unknown substance or 4/25.
One sixteenth of a gram. 1st halflife- 1/2 gram 2nd, 1/4 3rd 1/8th 4th halflife, 1/16th
These Three Remain was created in 2005.
These Three Remain has 280 pages.