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timofeeve [1]
3 years ago
6

in 2012, the population of a city was 63,000. by 2017, the population was reduced to approximately 54,100. identify any equation

s that are appropriate models for the population of the city, and explain why the others are not
Mathematics
1 answer:
Akimi4 [234]3 years ago
4 0

Answer:

<em></em>f(x) = 63000(0.97)^x<em></em>

<em></em>

Step-by-step explanation:

Given

f(0) = 63000 ---x = 0, in 2012

f(5) = 54100 -- x = 5, in 2017

Required

Select all possible equations

Because there is a reduction in the population, as time increases; the rate must be less than 1.

An exponential function is represented as:

f(x) = ab^x

Where

b = rate

rate > 1 in options (a) and (b) i.e. 1.03

This implies that (a) and (b) cannot be true

For option (c), we have:

f(x) = 63000(0.97)^x

Set x = 0

f(0) = 63000(0.97)^0 = 63000*1=63000\\

Set x = 5

f(5) = 63000(0.97)^5 = 63000*0.8587=54098.1 \approx 54100

<em>This is true because the calculated values of f(0) and f(5) correspond to the given values</em>

For option (d), we have:

f(x) = 52477(0.97)^x

Set x = 0

f(0) = 52477(0.97)^0 - 52477* 1 = 52477

<em>This is false because the calculated value of f(0) does not correspond to the given value</em>

For option (e), we have:

f(x) = 63000(0.97)^\frac{1}{5x}

Set x = 0

f(0) = 63000(0.97)^\frac{1}{5*0} = 63000(0.97)^\frac{1}{0} =undefined

<em>This is false because the f(x) is not undefined at x = 0</em>

For option (f), we have:

f(x) = 52477(0.97)^{5x

Set x = 0

f(0) = 52477(0.97)^{5*0} = 52477(0.97)^0 =52477*1= 52477

<em>This is false because the calculated value of f(0) does not correspond to the given value</em>

<em>From the computations above, only (c) </em>f(x) = 63000(0.97)^x<em> is true</em>

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