Question 328263
Radioactive decay problems are modeled using an exponential function,
{{{N(t)=N0*e^(kt)}}}
where N0 is the initial amount of material, k is a constant for each material. 
Half life means that 1/2 of the initial amount remains. 
We use this to calculate k.
{{{(1/2)N0=N0*e^(k(432))}}}
{{{e^(432k)=(1/2)}}}
{{{432k=ln(1/2)}}}
{{{k=ln(1/2)/432=-1.605x10^(-3)}}}
.
.
.
So now find N(329).
{{{kt=(329)(-1.605x10^(-3))=-0.527883}}}
{{{N(329)=45*e^(-0.527883)}}}
{{{N(329)=45*(0.58985)}}}
{{{N(329)=26.543}}}grams