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

`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:

`(1)/(2)= b^(12)`

And then we have:

`b= ((1)/(2))^{(1)/(12)}`

And the model would be given by:

`y(t) = 82.6 ((1)/(2))^{(1)/(12)}`

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:

`(1)/(2)= b^(12)`

And then we have:

`b= ((1)/(2))^{(1)/(12)}`

And the model would be given by:

`y(t) = 82.6 ((1)/(2))^{(1)/(12)}`