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The volume of a certain lake V=1.0 107m3, input water quantity of tributary Qin=0.5 108m3/a, input a certain pollutant concentration of Cin=3mg/L. The initial concentration

The volume of a certain lake V=1.0 107m3, input water quantity of tributary Qin=0.5 108m3/a, input a certain pollutant concentration of Cin=3mg/L. The initial concentration of this pollutant in the lake at 0:00 on January 1, 2020, C0=1.5mg/L, and the attenuation of the pollutant in the lake follows a first-order reaction kinetics law. Sampling and monitoring of lake outflow were conducted, and the results are shown in the table on the right.
(1) Please construct a zero dimensional water quality model for the lake for this pollutant. Using optimization methods and monitoring data from 1, 2, 4, 5, 7, 8, 10, and 11 months, determine the value of the pollutant reaction rate constant k (1/a). (Reminder: According to preliminary estimates, the value of k is between 0 and 0.25/a.)
(2) Using observation data from March, June, September, and December, select error based evaluation indicators for model validation. Compare the error levels between the parameter estimation stage and the model validation stage, and provide validation conclusions.
(3) Predict the concentration of pollutants in the lake at 0:00 on July 1, 2023.
(4) We found that there were issues with the observation data for August and November. After removing these two data points, we conducted parameter estimation and validation again. Has the conclusion changed? Discuss it.
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\begin{tabular}{|l|l|c|} \hline monitoring date & time & pollutantconcentration(mg/L) \\ \hline 20200115 & 10:00am & 1.845 \\ \hline 20200215 & 10:00am & 2.167 \\ \hline 20200315 & 10:00am & 2.504 \\ \hline 20200415 & 10:00am & 2.598 \\ \hline 20200515 & 10:00am & 2.731 \\ \hline 20200615 & 10:00am & 2.720 \\ \hline 20200715 & 10:00am & 2.750 \\ \hline 20200815 & 10:00am & 3.016 \\ \hline 20200915 & 10:00am & 2.887 \\ \hline 20201015 & 10:00am & 2.819 \\ \hline 20201115 & 10:00am & 3.024 \\ \hline 20201215 & 10:00am & 2.904 \\ \hline \end{tabular}

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