Answer:
Part 1 : ![A=250 e^{-0.008223t}](https://tex.z-dn.net/?f=A%3D250%20e%5E%7B-0.008223t%7D)
Part 2 : Half life is 84 minutes ( approx )
Step-by-step explanation:
Part 1 : Suppose the function that shows the amount( in grams ) of the substance after t minutes,
![A=A_0 e^{kt}](https://tex.z-dn.net/?f=A%3DA_0%20e%5E%7Bkt%7D)
If t = 0 min, A = 250 grams,
![250=A_0 e^{0}](https://tex.z-dn.net/?f=250%3DA_0%20e%5E%7B0%7D)
![\implies A_0 = 250](https://tex.z-dn.net/?f=%5Cimplies%20A_0%20%3D%20250)
If t = 250, A = 32 grams,
![32 = A_0 e^{250k}](https://tex.z-dn.net/?f=32%20%3D%20A_0%20e%5E%7B250k%7D)
![32 = 250 e^{250k}](https://tex.z-dn.net/?f=32%20%3D%20250%20e%5E%7B250k%7D)
![0.128 = e^{250k}](https://tex.z-dn.net/?f=0.128%20%3D%20e%5E%7B250k%7D)
Taking ln both sides,
![\ln(0.128) = 250k](https://tex.z-dn.net/?f=%5Cln%280.128%29%20%3D%20250k)
![\implies k =\frac{\ln(0.128)}{250}=-0.008223](https://tex.z-dn.net/?f=%5Cimplies%20k%20%3D%5Cfrac%7B%5Cln%280.128%29%7D%7B250%7D%3D-0.008223)
Hence, the equation that shows this situation,
![A=250 e^{-0.008223t}](https://tex.z-dn.net/?f=A%3D250%20e%5E%7B-0.008223t%7D)
Part 2 : If A = 250/2 = 125,
![125 = 250 e^{-0.008223t}](https://tex.z-dn.net/?f=125%20%3D%20250%20e%5E%7B-0.008223t%7D)
![0.5 = e^{-0.008223t}](https://tex.z-dn.net/?f=0.5%20%3D%20e%5E%7B-0.008223t%7D)
Taking ln both sides,
![\ln(0.5) = -0.008223t](https://tex.z-dn.net/?f=%5Cln%280.5%29%20%3D%20-0.008223t)
![\implies t =\frac{\ln(0.5)}{-0.008223}\approx 84](https://tex.z-dn.net/?f=%5Cimplies%20t%20%3D%5Cfrac%7B%5Cln%280.5%29%7D%7B-0.008223%7D%5Capprox%2084)
Therefore, the half life of the substance would be 84 minutes.