The population function is given by:

$P(t) = 1200 \times 2^{0.1009t}$

where $t$ is the number of years since 2011, and $P(t)$ gives the population in thousands.

(a) Finding the population in 2019:

To find the population in 2019, we first determine the value of $t$ for 2019. Since $t$ is the number of years since 2011:

$t = 2019 - 2011 = 8$

Now, substitute $t = 8$ into the population function:

$P(8) = 1200 \times 2^{0.1009 \times 8}$

First, calculate the exponent $0.1009 \times 8 = 0.8072$. Then calculate $2^{0.8072}$.

$P(8) = 1200 \times 2^{0.8072}$

Finally, multiply by 1200 to get the population in 2019.

(b) Predicting the population in 2028:

Similarly, for 2028:

$t = 2028 - 2011 = 17$

Substitute $t = 17$ into the population function:

$P(17) = 1200 \times 2^{0.1009 \times 17}$

First, calculate $0.1009 \times 17 = 1.7153$. Then calculate $2^{1.7153}$.

$P(17) = 1200 \times 2^{1.7153}$

Now, multiply by 1200 to get the predicted population in 2028.

I'll compute these values now.### Results: (a) The population in 2019 is approximately 2,099.77 thousand (or 2.1 million).

(b) The predicted population in 2028 is approximately 3,940.38 thousand (or 3.94 million).

Would you like further details or have any questions?

Here are five related questions to expand on this topic:

1. How does the growth rate affect long-term population predictions?
2. What would the population have been in 2011, based on this model?
3. How can exponential growth models be adjusted for real-world factors like migration or natural disasters?
4. Can you calculate the year when the population will reach 5 million?
5. How does the base of the exponential function $2$ affect the speed of population growth?

Tip: Exponential functions grow faster over time, so even small increases in the exponent can lead to significant changes in predictions.