Respuesta :
Answer:
a) [tex] A(t) = 250 (1-0.17)^t = 250(0.83)^t [/tex]
b) [tex] A(t=14) = 250 (0.83)^{14}= 18.408 mg[/tex]
c) For this case we want to find when the quantity is below 25 mg, so we can do this:
[tex] 25 = 250 (0.83)^t [/tex]
We can divide both sided by 250 and we got:
[tex] 0.1 = 0.83^t [/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(0.1) = t ln (0.83)[/tex]
And if we solve for t we got:
[tex] t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years[/tex]
So the answer for this case would be 12.358 years after.
Step-by-step explanation:
Part a
For this case we know that the decay rate per year is about 17% or 0.17 in fraction and the initial amount is 250 mg for 2009, we can define the model like this:
[tex] A(t) = A_o (1-r)^t[/tex]
Where A represent the amount of the substance in mg, r the decay rate = 0.17 and t the number of years after 2009.
Our model for this case would be:
[tex] A(t) = 250 (1-0.17)^t = 250(0.83)^t [/tex]
Part b
For this case we have that t= 2023-2009=14 years, so then we can find the amount of substance like this:
[tex] A(t=14) = 250 (0.83)^{14}= 18.408 mg[/tex]
Part c
For this case we want to find when the quantity is below 25 mg, so we can do this:
[tex] 25 = 250 (0.83)^t [/tex]
We can divide both sided by 250 and we got:
[tex] 0.1 = 0.83^t [/tex]
Now we can apply natural log on both sides and we got:
[tex] ln(0.1) = t ln (0.83)[/tex]
And if we solve for t we got:
[tex] t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years[/tex]
So the answer for this case would be 12.358 years after.
