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 increased from 112 to 357 between 1:00 and 2:00, we can solve for this constant of proportionality:

\(\displaystyle 357=112e^{k(2-1)}\)

\(\displaystyle \frac{357}{112}=e^{k}\)

\(\displaystyle k=ln(\frac{357}{112})\)

\(\displaystyle k=1.159\)

Using this, we can calculate the expected value from 2:00 to 3:00:

\(\displaystyle P=357e^{(3-2)(1.159)} \approx 1138\)

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 increased from 29 to 167 between 3:00 and 6:00, we can solve for this constant of proportionality:

\(\displaystyle 167=29e^{k(6-3)}\)

\(\displaystyle \frac{167}{29}=e^{3k}\)

\(\displaystyle 3k=ln(\frac{167}{29})\)

\(\displaystyle k=\frac{ln(\frac{167}{29})}{3}=0.584\)

Using this, we can calculate the expected value from 6:00 to 8:30:

\(\displaystyle P=167e^{(8.5-6)(0.584)} \approx 719\)

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 1038 to 817 between 2010 and 2015, we can solve for this constant of proportionality:

\(\displaystyle 817=1038e^{k(2015-2010)}\)

\(\displaystyle \frac{817}{1038}=e^{5k}\)

\(\displaystyle 5k=ln(\frac{817}{1038})\)

\(\displaystyle k=\frac{ln(\frac{817}{1038})}{5}=-0.048\)

Using this, we can calculate the expected value from 2015 to 2017:

\(\displaystyle P=817e^{(2017-2015)(-0.048)} \approx 742\)

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 5430  to 3740 between 2008 and 2015, we can solve for this constant of proportionality:

\(\displaystyle 3740=5430e^{k(2015-2008)}\)

\(\displaystyle \frac{3740}{5430}=e^{7k}\)

\(\displaystyle 7k=ln(\frac{3740}{5430})\)

\(\displaystyle k=\frac{ln(\frac{3740}{5430})}{7}=-0.0533\)

Using this, we can calculate the expected value from :

\(\displaystyle P=3740e^{(2020-2015)(-0.0533)} \approx 2865\)

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 increased from 1130 to 1460 between 3:15 and 4:30, we can solve for this constant of proportionality. Convert the minutes to decimal values of an hour:

\(\displaystyle 1460=1130e^{k(4.5-3.25)}\)

\(\displaystyle \frac{1460}{1130}=e^{1.25k}\)

\(\displaystyle 1.25k=ln(\frac{1460}{1130})\)

\(\displaystyle k=\frac{ln(\frac{1460}{1130})}{1.25}=0.205\)

Using this, we can calculate the expected value from 4:30 to 6:45:

\(\displaystyle P=1460e^{(6.75-4.5)(0.205)} \approx 2316\)

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 increased from 10800 to 25200 between April and June, we can solve for this constant of proportionality. Use the months' number in the calendar:

\(\displaystyle 25200=10800e^{k(6-4)}\)

\(\displaystyle \frac{25200}{10800}=e^{2k}\)

\(\displaystyle 2k=ln(\frac{25200}{10800})\)

\(\displaystyle k=\frac{ln(\frac{25200}{10800})}{2}=0.424\)

Using this, we can calculate the expected value in November from June:

\(\displaystyle P=25200e^{(11-6)(0.424)} \approx 209945\)

Good luck, humanity.

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:

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