The population of Harristown, IL was 1327 people in 2000. By 2009, the populatio

Question

The population of Harristown, IL was 1327 people in 2000. By 2009, the population had decreased to 1208 people. Assume the population decreases exponentially. a. For the population of Harristown, find the following values. i. the 9-year growth factor: 1.01049 Preview ii. the 9-year percent change: 8.96 % Preview iii. the 1-year growth factor: .014 Preview iv. the 1-year percent change: 1.05 as % Preview b. Is the following true or false? The 9-year percent change is 9 times as large as the 1-year percent change. False c. Define a function P that gives the population of Harristown r years after 2000 P-G27-0.0104%) | Preview d. Define a function H that gives the population of Harristown in terms of d, the number of decades that have elapsed since 2000.

Explanation / Answer

2000 ---> 1327 ppl
2009 ---> 1208 ppl

A = 1327 e ^(bt)

Plug in t = 9 and A = 1208 :
1208 = 1327e^(9b)

Solving, we get
1208/1327 = e^(9b)

9b = ln(1208/1327)

b = -0.0104

So, we have
A = 1327e^(-0.0104t)

A = 1327(0.9896148952741206)^t

9 yr growth factor is :
0.9896148952741206^9

0.910324 ---> ANS
  
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9 yr percent change :
A = 1327(0.910324)^(t/9)

A = 1327(1 - 0.089676)^(t/9)

So, -8.9676% ---> ANS

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1 yr growth factor :
A = 1327(0.989615)^t

So, 0.989615 ---> ANS

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2 yr percent change :
A = 1327(1 - 0.010385)^t

So, -1.0385% ---> ANS

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FALSE

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A = 1327e^(-0.0104t)

This can be written as
A = 1327(1208/1327)^(t/9) ---> ANS

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