ssukarieh1
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Please help with this question 
thanks
HELP WITH QUESTION PLEASEE (1 Viewer)
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Please help with this question thanks//Hopefully it might be start cannot be bothered doing it all
Suppose we could model the amount of the chemical by
k * e ^ (-alpha*t) where alpha > 0 and t is the time in weeks.
Therefore k is the initial amount of the chemical.
The percentage of the chemical is given by
k * e ^ (-alpha*t) /k = e^-( alpha *t)
So the percentage of the chemical left is e^(-alpha*t)
For suspect A there is time of 1 week.
e^-alpha*(x) = 16% (1)
e^-alpha*(x+1) = 15% (2)
Where 0<x<1
(1) * (2) = e^-alpha*(x) * e^-alpha*(x+1)
=> e^-alpha(2x+1) = 2.4%
Suppose we could model the amount of the chemical by
k * e ^ (-alpha*t) where alpha > 0 and t is the time in weeks.
Therefore k is the initial amount of the chemical.
The percentage of the chemical is given by
k * e ^ (-alpha*t) /k = e^-( alpha *t)
So the percentage of the chemical left is e^(-alpha*t)
For suspect A there is time of 1 week.
e^-alpha*(x) = 16% (1)
e^-alpha*(x+1) = 15% (2)
Where 0<x<1
(1) * (2) = e^-alpha*(x) * e^-alpha*(x+1)
=> e^-alpha(2x+1) = 2.4%
Let )
be the concentration of the chemical at time 
weeks after the discovery of the dumping. Let 
be the concentration at the time of the dumping, which occurred at some time 
. We know that the concentration of the chemical decays according to the equation
for some constant
and the solution of this differential equation is
for some constant
. We know that the concentration has fallen from 
at time 
to 
at time 
and to 
at time 
. Using the latter two pieces of information, we get
)
and
)
Dividing equation (1) by equation (2) yields
 \end{align*})
So, we can find
, where 
:
} \\ \frac{C_0}{0.16C_0} &= e^{-kT} \\ 0.16 &= e^{kT} \\ \ln{0.16} &= kT \\ \ln\left(\frac{4}{25}\right) &= T\ln\left(\frac{16}{15}\right) \\ T &= \frac{\ln{4} - \ln{25}}{\ln{16}-\ln{15}} \\ &= \frac{2\ln{2} - 2\ln{5}}{4\ln{2}-\ln{15}} \\ &= -28.395... \\ &\approx\ \text{28 weeks and 3 days prior to $t = 0$} \end{align*})
So, based on decay rates, the contamination / dumping occurred 28 weeks and 3 days prior to it being discovered.
Suspect A was imprisoned for 26 weeks ending 1 week prior to the discovery, and so was imprisoned 27 weeks prior to the discovery of the contamination, and so was free at the time the dumping occurred.
Suspect B was imprisoned for 26 weeks ending 13 weeks prior to the discovery, and so was imprisoned at 39 weeks prior to the discovery and was still imprisoned when the dumping occurred.
The contamination decay rates are consistent with Suspect A being free at the time the crime was committed but Suspect B being incarcerated at that time and so unable to commit the crime. Based on these results, Suspect B is innocent and Suspect A remains under suspicion and is potentially the offender.
for some constant
for some constant
and
Dividing equation (1) by equation (2) yields
So, we can find
So, based on decay rates, the contamination / dumping occurred 28 weeks and 3 days prior to it being discovered.
Suspect A was imprisoned for 26 weeks ending 1 week prior to the discovery, and so was imprisoned 27 weeks prior to the discovery of the contamination, and so was free at the time the dumping occurred.
Suspect B was imprisoned for 26 weeks ending 13 weeks prior to the discovery, and so was imprisoned at 39 weeks prior to the discovery and was still imprisoned when the dumping occurred.
The contamination decay rates are consistent with Suspect A being free at the time the crime was committed but Suspect B being incarcerated at that time and so unable to commit the crime. Based on these results, Suspect B is innocent and Suspect A remains under suspicion and is potentially the offender.
The calculation quoted below is correct for determining the time when the concentration falls to 
.
However, it has been pointed out below that the question actually asked for the concentration to fall to 0.05% of its initial value, and thus to
.
The correct calculation and answer is given in a post further below.
My mistake is preserved as a reminder that everyone can and does mistakes, that you should check independently, and that it is helpful to others to point out when mistakes are made. Thank you to @notme123 for pointing out this mistake.
However, it has been pointed out below that the question actually asked for the concentration to fall to 0.05% of its initial value, and thus to
The correct calculation and answer is given in a post further below.
My mistake is preserved as a reminder that everyone can and does mistakes, that you should check independently, and that it is helpful to others to point out when mistakes are made. Thank you to @notme123 for pointing out this mistake.
The time when the concentration falls to can be calculated easily:
} \\ \frac{0.05C_0}{0.16C_0} &= e^{-kt} \\ \frac{16}{5} &= e^{kt} \\ \ln{\left(\frac{16}{5}\right)} &= t\ln\left(\frac{16}{15}\right) \\ t &= \frac{4\ln{2} - \ln{5}}{4\ln{2}-\ln{15}} \end{align*})
This is the time taken for the ground to become safe for replanting crops from the time the dumping occurred. Compensation happens from the time of the crime, so the period of compensation is
Last edited:
You are absolutely correct, I have made a mistake because the question did say 0.05%, which is a concentration of
I shall make a post correcting my answer. Thanks for pointing this out, @notme123.
Correcting the answer from above after @notme123 noted that I had made a mistake:
The time when the concentration falls to 0.05% of its initial value, that is, to can be calculated easily:
} \\ \frac{0.0005C_0}{0.16C_0} &= e^{-kt} \\ \frac{1600}{5} &= e^{kt} \\ \ln{320} &= t\ln\left(\frac{16}{15}\right) \qquad \qquad \text{as it was shown above that $k=\ln\left(\frac{16}{15}\right)$} \\ t &= \frac{\ln{320}}{\ln{16}-\ln{15}} \\ &= \frac{\ln{(64 \times 5)}}{\ln{16}-\ln{15}} \\ &= \frac{6\ln{2} + \ln{5}}{4\ln{2}-\ln{15}} \end{align*})
This is the time taken for the ground to become safe for replanting crops from the time the dumping occurred. Compensation happens from the time of the crime, so the period of compensation is
The time when the concentration falls to 0.05% of its initial value, that is, to