Question 852386
A general model can be {{{y=pe^(-kt)}}}, y=amount after time t, and p = initial amount, t is time in minutes.


{{{ln(y)=ln(p)+(-kt)*1}}}
{{{ln(y)=ln(p)-kt}}}
{{{ln(y)-ln(p)=-kt}}}
{{{ln(p)-ln(y)=kt}}}
{{{highlight_green(kt=ln(p)-ln(y))}}}
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Separately solve for k, and for t.
{{{k=(ln(p)-ln(y))/t}}}, and use this with the given half-life information to get the k value;
{{{t=(ln(p)-ln(y))/k}}}, and use this to .... No, you will not need this;
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The more specific model you have now is:
{{{y=256*e^(-kt)}}}, and you now have the appropriate information formed to get the value of k.





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You should have found {{{k=ln(2)/15=0.0462}}}.