Question 1154570
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that over each hour, 25% of the amount present at the beginning of the hour is eliminated.
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{{{A(t)=A[o]e^(kt)}}}
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{{{0.75=1.00e^(1*k)}}}

{{{e^(k)=0.75}}}

{{{ln(e^k)=ln(0.75)}}}

{{{k=ln(0.75)}}}
{{{highlight(k=-0.2877)}}}



{{{highlight(A(t)=A[o]e^(-0.2877t))}}}

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You could round to  {{{k=-0.29}}}  if you want.