How long did it take for the population to reach double that amount BookmarksWindowHelpA mccs.brightspace.comollegeCOVID-19 | My SMCCB Student Activity Notebook for Lesson 2 - Ma..https://mccs.brightspace.ccin millions)in millions)10,000 BCE118009789,000 BCE18501,2628,000 BCE19001,6507,000 BCE719502,5196,000 BCE1019602,9825,000 BCE1519703,6924,000 BCE2019804,4353,000 BCE2519854,8312,000 BCE3519905,2631,000 BCE5019955,674500 BCE10020006,070AD 120020056,454175079120106,86410tv@$%5

Answered: 1800 978 1850 1,262 1900 1,650 1950…

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter4: Graphing And Inverse Functions

Section: Chapter Questions

Problem 1GP

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Question

How long did it take for the population to reach double that amount

Bookmarks
Window
Help
A mccs.brightspace.com
ollege
COVID-19 | My SMCC
B Student Activity Notebook for Lesson 2 - Ma..
https://mccs.brightspace.cc
in millions)
in millions)
10,000 BCE
1
1800
978
9,000 BCE
1850
1,262
8,000 BCE
1900
1,650
7,000 BCE
7
1950
2,519
6,000 BCE
10
1960
2,982
5,000 BCE
15
1970
3,692
4,000 BCE
20
1980
4,435
3,000 BCE
25
1985
4,831
2,000 BCE
35
1990
5,263
1,000 BCE
50
1995
5,674
500 BCE
100
2000
6,070
AD 1
200
2005
6,454
1750
791
2010
6,864
10
tv
@
$
%
5

Expert Solution

Step 1

Taking the rate of increment as exponential first fits an exponential function to the given data.

Here Assume 1980 is t = 0 then 2010 will be t =30

The curve to be fitted is $y = a e^{b x}$

taking logarithm on both sides, we get
$\log_{10} (y) = \log_{10} (a) + b x \log_{10} (e)$

Y=A+Bx where $Y = \log_{10} (y), A = \log_{10} (a), B = b \log_{10} (e)$

which linear in Y,x
So the corresponding normal equations are
$\sum Y = n A + B \sum x \sum x Y = A \sum x + B \sum x^{2}$

The values are calculated using the following table

x	y	Y $= \log_{10} (y)$	x²	x⋅Y
0	4435	3.647	0	0
1	4831	3.684	1	3.684
2	5263	3.721	4	7.442
3	5674	3.754	9	11.262
4	6070	3.783	16	15.133
5	6454	3.81	25	19.049
6	6864	3.837	36	23.019
---	---	---	---	---
∑x=21	∑y=39591	∑Y=26.236	∑x²=91	∑x⋅Y=79.59

Substituting these values in the normal equations
7A+21B=26.236

21A+91B=79.59

Solving these two equations using the Elimination method,

we obtain A=3.654,B=0.032

$∴ a = a n t i \log_{10} (A) = a n t i \log_{10} (3.654) = 4502.98$

and $b = B \log_{10} (e) = 0.0320.434 = 0.073$

Now substituting these values in the equation is y=aebx, we get

$y = 4502.98 e^{0.073 x}$