Question

the populations ( p ) (in thousands) of a certain city from 2000 through 2007 can be modeled by ( p = 1676.3e^{kt} ), where ( t ) represents the year, with ( t = 0 ) corresponding to 2000. in 2006, the population of the city was about 1,960,000. (a) find the value of ( k ). (round your answer to five decimal places.) ( k = ) nice work. is the population increasing or decreasing? explain. because ( k ) is positive, the population is decreasing. because ( k ) is positive, the population is increasing. because ( k ) is negative, the population is decreasing. because ( k ) is negative, the population is increasing. great job! (b) use the model to find the populations of the city in 2010 and 2015. (round your answers to three decimal places.) 2010 ( p = ) thousand 2015 ( p = ) thousand (c) according to the model, during what year will the population reach 2.2 million? the population will reach 2.2 million in great job.

Explanation:

Step1: Find \( t \) for 2010

Since \( t = 0 \) corresponds to 2000, for 2010, \( t=2010 - 2000=10 \).
The population model is \( P = 1676.3e^{0.02606t} \).
Substitute \( t = 10 \) into the formula:
\( P=1676.3e^{0.02606\times10}=1676.3e^{0.2606} \)
We know that \( e^{0.2606}\approx1.297 \) (using a calculator for the exponential function).
\( P = 1676.3\times1.297\approx2174.1 \) (in thousands).

Step2: Find \( t \) for 2015

For 2015, \( t=2015 - 2000 = 15 \).
Substitute \( t = 15 \) into the formula \( P = 1676.3e^{0.02606t} \).
\( P=1676.3e^{0.02606\times15}=1676.3e^{0.3909} \)
Since \( e^{0.3909}\approx1.478 \) (using a calculator for the exponential function).
\( P=1676.3\times1.478\approx2477.6 \) (in thousands).

Answer:

2010: \( P\approx2174.1 \) thousand
2015: \( P\approx2477.6 \) thousand