Question

specific radioactive substance follows a continuous exponential decay model. It has a half-life of hours. At the start of the experiment, is present.

Answer 1

Answer:

$y = A_o (b)^t$

With $A_o = 82.6$ the initial amount and t the time on hours and t the time in hours. since the half life is 12 hours we can find the parameter of decay like this:

$41.3= 82.6(b)^{12}$

And solving for b we got:

$\frac{1}{2}= b^{12}$

And then we have:

$b= (\frac{1}{2})^{\frac{1}{12}}$

And the model would be given by:

$y(t) = 82.6 (\frac{1}{2})^{\frac{1}{12}}$

Step-by-step explanation:

Assuming this complete question: "A specific radioactive substance follows a continuous exponential decay model. It has a half-life of 12 hours. At the start of the experiment, 82.6g is present. "

For this case we can create a model like this one:

$y = A_o (b)^t$

With $A_o = 82.6$ the initial amount and t the time on hours and t the time in hours. since the half life is 12 hours we can find the parameter of decay like this:

$41.3= 82.6(b)^{12}$

And solving for b we got:

$\frac{1}{2}= b^{12}$

And then we have:

$b= (\frac{1}{2})^{\frac{1}{12}}$

And the model would be given by:

$y(t) = 82.6 (\frac{1}{2})^{\frac{1}{12}}$