Question

The populations P (in thousands) of a certain town in North Carolina, from 2006 through 2012 can be modeled by P= 5.9e, wher

(a) Find the value of k for the model. Round your result to four decimal places.
k=
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(b) Use your model to predict the population in 2018. (Round your answer to the nearest person.)
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PRECALCRMRP7 3.5.038.
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Answer 1

a) k = 0.0381 (4 d.p.)

b) P = 11,714 (nearest person)

Explanation:

The populations P (in thousands) of a certain town in North Carolina, from 2006 through 2012 can be modeled by
`P= 5.9e^(kt)` where t is the year, with t = 6 corresponding to 2006.

Part (a)

If the population was 8000 in 2008, then P = 8 when t = 8.

Substitute these values into the formula and solve for k.

`\begin{aligned}P&=5.9e^(kt)\\\\8&=5.9e^(8k)\\\\(8)/(5.9)&=e^(8k)\\\\\ln \left((8)/(5.9)\right)&=\ln \left(e^(8k)\right)\\\\\ln \left((8)/(5.9)\right)&=8k\ln \left(e\right)\\\\\ln \left((8)/(5.9)\right)&=8k\\\\k&=(1)/(8)\ln \left((8)/(5.9)\right)\\\\k&=0.0380611488...\\\\k&=0.0381\; \sf (4\;d.p.)\end{aligned}`

Therefore, the value of k is 0.0381 (rounded to four decimal places).

Part (b)

To use the model to predict the population in 2018, substitute the found value of k = 0.0381 and t = 18 into the equation and solve for P:

`\begin{aligned}P&=5.9e^(kt)\\\\P&=5.9e^(0.0381 \cdot 18)\\\\P&=5.9e^(0.6858)\\\\P&=11.713620979...\end{aligned}`

As P is measured in thousands, to find the actual population, multiply the P-value by 1,000.

Therefore, the predicted population in 2018 is 11,714 (rounded to the nearest person).