We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population decreased from 15800 to 6540 between 3:15 and 4:00, we can solve for this constant of proportionality. Treat the minutes as decimal values for the hour:

\(\displaystyle 6540=15800e^{k(4-3.25)}\)

\(\displaystyle \frac{6540}{15800}=e^{0.75k}\)

\(\displaystyle 0.75k=ln(\frac{6540}{15800})\)

\(\displaystyle k=\frac{ln(\frac{6540}{15800})}{0.75}=-1.176\)

Using this, we can calculate the expected value from 4:00 to 8:30:

\(\displaystyle P=6540e^{(8.5-4)(-1.176)} \approx 33\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 1800 to 2500 between 2012 and 2015, we can solve for this constant of proportionality:

\(\displaystyle 2500=1800e^{k(2015-2012)}\)

\(\displaystyle \frac{25}{18}=e^{3k}\)

\(\displaystyle 3k=ln(\frac{25}{18})\)

\(\displaystyle k=\frac{ln(\frac{25}{18})}{3}=0.1095\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=2500e^{(2018-2015)(0.1095)} \approx 3472\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 4900 to 7193 between 3:54 and 4:21, we can solve for this constant of proportionality. Treat the minutes as decimals of an hour by dividing by 60:

\(\displaystyle 7193=4900e^{k(4.35-3.9)}\)

\(\displaystyle \frac{7193}{4900}=e^{0.45k}\)

\(\displaystyle 0.45k=ln(\frac{7193}{4900})\)

\(\displaystyle k=\frac{ln(\frac{7193}{4900})}{0.45}=0.853\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=7193e^{(5.20-4.35)(0.853)} \approx 14852\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 178 to 413 between 3:15 and 3:45, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing them by 60:

\(\displaystyle 413=178e^{k(3.75-3.25)}\)

\(\displaystyle \frac{413}{178}=e^{0.5k}\)

\(\displaystyle 0.5k=ln(\frac{413}{178})\)

\(\displaystyle k=\frac{ln(\frac{413}{178})}{0.5}=1.683\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=413e^{(5.333-3.75)(1.683)} \approx 5929\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 12 to 43 between Monday and Wednesday, we can solve for this constant of proportionality. Assign the days numbers based on the occurence in the week:

\(\displaystyle 43=12e^{k(4-2)}\)

\(\displaystyle \frac{43}{12}=e^{2k}\)

\(\displaystyle 2k=ln(\frac{43}{12})\)

\(\displaystyle k=\frac{ln(\frac{43}{12})}{2}=0.638\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=43e^{(7-4)(0.638)} \approx 292\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 193678 to 254183 between 2014 and 2015, we can solve for this constant of proportionality:

\(\displaystyle 254183=193678e^{k(2015-2014)}\)

\(\displaystyle \frac{254183}{193678}=e^{k}\)

\(\displaystyle k=ln(\frac{254183}{193678})=0.272\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=254183e^{(2021-2015)(0.272)} \approx 1299915\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 19000 to 58000 between 2013 and 2014, we can solve for this constant of proportionality:

\(\displaystyle 58000=19000e^{k(2014-2013)}\)

\(\displaystyle \frac{58}{19}=e^{k}\)

\(\displaystyle k=ln(\frac{58}{19})=1.116\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=58000e^{(2017-2014)(1.116)} \approx 1649855\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 17902 to 39551 between March and May, we can solve for this constant of proportionality. Assign the months their number in the calendar to determine time elapsed:

\(\displaystyle 39551=17902e^{k(5-3)}\)

\(\displaystyle \frac{39551}{17902}=e^{2k}\)

\(\displaystyle 2k=ln(\frac{39551}{17902})\)

\(\displaystyle k=\frac{ln(\frac{39551}{17902})}{2}=0.396\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=39551e^{(7-5)(0.396)} \approx 87321\)

We're told that the rate of decrease of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population decreased from 45000 to 1800 between 1900 and 1972, we can solve for this constant of proportionality:

\(\displaystyle 1800=45000e^{k(1972-1900)}\)

\(\displaystyle \frac{1800}{45000}=e^{72k}\)

\(\displaystyle 72k=ln(\frac{1800}{45000})\)

\(\displaystyle k=\frac{ln(\frac{1800}{45000})}{72}=-0.0447\)

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

\(\displaystyle p(t)=p_0e^{kt}\)

Where \(\displaystyle p_0\) is an initial population value, \(\displaystyle t\)  represents a measure of elapsed time relative to this population value, and \(\displaystyle k\) is the constant of proportionality.

Since the population increased from 100 units to 1500 between 2008 and 2015, we can solve for this constant of proportionality:

\(\displaystyle 1500=100e^{k(2015-2008)}\)

\(\displaystyle 15=e^{7k}\)

\(\displaystyle 7k=ln(15)\)

\(\displaystyle k=\frac{ln(15)}{7}=0.387\)

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

\(\displaystyle P=1500e^{(2055-2015)(0.387)} \approx 7924462944\)

Better unplug that production facility.