Answer:
Part 1 : [tex]A=250 e^{-0.008223t}[/tex]
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,
[tex]A=A_0 e^{kt}[/tex]
If t = 0 min, A = 250 grams,
[tex]250=A_0 e^{0}[/tex]
[tex]\implies A_0 = 250[/tex]
If t = 250, A = 32 grams,
[tex]32 = A_0 e^{250k}[/tex]
[tex]32 = 250 e^{250k}[/tex]
[tex]0.128 = e^{250k}[/tex]
Taking ln both sides,
[tex]\ln(0.128) = 250k[/tex]
[tex]\implies k =\frac{\ln(0.128)}{250}=-0.008223[/tex]
Hence, the equation that shows this situation,
[tex]A=250 e^{-0.008223t}[/tex]
Part 2 : If A = 250/2 = 125,
[tex]125 = 250 e^{-0.008223t}[/tex]
[tex]0.5 = e^{-0.008223t}[/tex]
Taking ln both sides,
[tex]\ln(0.5) = -0.008223t[/tex]
[tex]\implies t =\frac{\ln(0.5)}{-0.008223}\approx 84[/tex]
Therefore, the half life of the substance would be 84 minutes.