Respuesta :
Answer:
(a). 1,617,599
(b). In year 2018.
Step-by-step explanation:
We have been given that population P (in thousands) of Phoenix, Arizona from 1980 through 2009 can be modeled by [tex]p=788e^{0.0248t}[/tex], where, [tex]t = 0[/tex] corresponds to 1980.
(a). To find the population of Phoenix in 2009, we will substitute [tex]t=2009-1980=29[/tex] in our given equation as:
[tex]p=788e^{0.0248*29}[/tex]
[tex]p=788e^{0.7192}[/tex]
[tex]p=788(2.0527903213187)[/tex]
[tex]p=1617.5987731991356[/tex]
Since our given population is in thousands, so we will multiply 1617.5987731991356 by 1,000 as:
[tex]1617.5987731991356\times 1,000=1,617,598.7731991356\approx 1,617,599[/tex]
Therefore, the population of Phoenix would be 1,617,599 in 2009.
(b). To find the time, when Phoenix have a population of 2,000,000, we can set an equation as:
[tex]2,000=788e^{0.0248t}[/tex]
We will substitute 2,000 as our given equation represents population in thousands.
Let us solve for t.
[tex]\frac{2,000}{788}=\frac{788e^{0.0248t}}{788}[/tex]
[tex]2.538071065989=e^{0.0248t}[/tex]
Now, we will use natural log to solve for t as:
[tex]\text{ln}(2.538071065989)=\text{ln}(e^{0.0248t})[/tex]
[tex]\text{ln}(2.538071065989)=0.0248t*\text{ln}(e)[/tex]
[tex]0.9314043696842032297=0.0248t*1[/tex]
[tex]\frac{0.9314043696842032297}{0.0248}=\frac{0.0248t}{0.0248}[/tex]
[tex]37.5566278098469=t[/tex]
[tex]t\approx 38[/tex]
Now, we need to find 38 years after 1980 that is [tex]1980+38=2018[/tex].
Therefore, in 2018 Phoenix will have a population of 2,000,000.
