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OpenStudy (anonymous):
Please help with this growth/decay problem
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OpenStudy (anonymous):
How did that person get .85 for a?
OpenStudy (anonymous):
N=No e^kt they have used 85 % of it so whats left is No-0.15No=No e^1343k No(1-0.15)=No e^1343k 0.85= e^1343k
OpenStudy (anonymous):
Where does it say that they used 85?
OpenStudy (anonymous):
what was left is 15% according to the prob
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OpenStudy (anonymous):
oh... okay I kind of understand. So if 20% of the sample decays, you would use .80 instead of .85?
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