An eighth remains.
75
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
1/8 of the original amount remains.
After three half-lives, only 1/8 (or 12.5%) of the original radioactive sample remains. This is because each half-life reduces the amount of radioactive material by half, so after three half-lives, you would have (1/2) * (1/2) * (1/2) = 1/8 of the original sample remaining.
It is 1/8 .
Not sure what you mean by "had-lives". After 3 half lives, approx 1/8 would remain.
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.
Approx 1/8 will remain.
There are three models of a fraction namely area model, liner model and set model.
Not sure what you mean by "had-lives". After 3 half lives, approx 1/8 would remain.
Three half lives have elapsed. This can be determined by calculating how many times the original sample size must be halved to get to one eighth: (1/2) * (1/2) * (1/2) = 1/8.
It will take two half-lives or about 60.34 years for three-fourths of a Cs-137 sample to decay.
91.16% of the daughter product has formed after 3.5 half lives.
The idea is to convert the percent to a fraction (divide it by 100), and then solve the equation: (1/2)^x = (that fraction) (Note: Using "^" for "power".) In the general case, solving this equation requires logarithms. But in this specific case, you can just try out different whole numbers for "x". # 1/2 lives vs fraction 1 0.5 2 0.25 3 0.125 4 0.0625 5 0.03175 6 0.015875 The answer falls between 5 and 6 half lives, closer to 5.