CO₂ level CO₂ level (in ppm) (in ppm) 1980 338.2 1996 362.7 1982 341.6 1998 366.7 1984 344.9 2000 369.6 1986 346.7 2002 373 1988 351.3 2004 377.2 1990 354.5 2006 381.5 1992 356.8 2008 385.1 1994 358.4 Year с 380 370 360 350 340 Year 1000 1005 1000 1005 2006 2005 3000 t 2010 EXAMPLE 2 The table lists the average carbon dioxide level in the atmosphere, measured in parts per million at observatory from 1980 to 2008. Use the data in the table to find a model for the carbon dioxide level. SOLUTION We use the data in the table to make a scatter plot, where t represents time (in years) and C represer the CO₂ level (in parts per million, ppm). Notice that the data points appear to lie close to a straight line, so it's natural to choose a linear model in this case. But there are many possible lines that approximate these data points, so which one should we use? From the graph appears that one possibility is the line that passes through the first and last data points. The slope of this line, roun to four decimal places, is 2008 and its equation is or C 380 - 338.2 1980 C = The equation above gives one possible linear model for the carbon dioxide level; it is graphed below. 370 28 (t-1980)
CO₂ level CO₂ level (in ppm) (in ppm) 1980 338.2 1996 362.7 1982 341.6 1998 366.7 1984 344.9 2000 369.6 1986 346.7 2002 373 1988 351.3 2004 377.2 1990 354.5 2006 381.5 1992 356.8 2008 385.1 1994 358.4 Year с 380 370 360 350 340 Year 1000 1005 1000 1005 2006 2005 3000 t 2010 EXAMPLE 2 The table lists the average carbon dioxide level in the atmosphere, measured in parts per million at observatory from 1980 to 2008. Use the data in the table to find a model for the carbon dioxide level. SOLUTION We use the data in the table to make a scatter plot, where t represents time (in years) and C represer the CO₂ level (in parts per million, ppm). Notice that the data points appear to lie close to a straight line, so it's natural to choose a linear model in this case. But there are many possible lines that approximate these data points, so which one should we use? From the graph appears that one possibility is the line that passes through the first and last data points. The slope of this line, roun to four decimal places, is 2008 and its equation is or C 380 - 338.2 1980 C = The equation above gives one possible linear model for the carbon dioxide level; it is graphed below. 370 28 (t-1980)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Transcribed Image Text:CO₂ level
(in ppm)
1980 338.2 1996
362.7
1982 341.6 1998 366.7
1984 344.9 2000 369.6
1986 346.7 2002 373
1988 351.3 2004 377.2
1990 354.5 2006 381.5
1992 356.8 2008 385.1
1994 358.4
CO₂ level
(in ppm)
Year
C
380
370
360
L
350
Video Example)
Year
340
t
1980 1985 1990 1995 2000 2005 2010
EXAMPLE 2 The table lists the average carbon dioxide level in the atmosphere, measured in parts per million at an
observatory from 1980 to 2008. Use the data in the table to find a model for the carbon dioxide level.
SOLUTION We use the data in the table to make a scatter plot, where t represents time (in years) and C represents
the CO₂ level (in parts per million, ppm).
Notice that the data points appear to lie close to a straight line, so it's natural to choose a linear model in this case.
But there are many possible lines that approximate these data points, so which one should we use? From the graph, it
appears that one possibility is the line that passes through the first and last data points. The slope of this line, rounded
to four decimal places, is
and its equation is
or
C-
с
380
370
360
C =
The equation above gives one possible linear model for the carbon dioxide level; it is graphed below.
350
- 338.2
2008 - 1980
340
=
28
WebAssign Plot
(t - 1980)
![Notice that our model gives values higher than most of the actual CO₂ levels. A better linear model is obtained by a
procedure from statistics called linear regression. If we use a graphing calculator, we enter the data from the table into
the data editor and choose the linear regression command. (With Maple we use the fit[leastsquare] command in the
stats package; with Mathematica we use the Fit command.) The machine gives the slope (round to two decimal
places) and y-intercept (round to the nearest integer) of the regression line as
m =
b =
So our least squares model for the CO₂ level is
C =
In the figure below, we graph the regression line as well as the data points. Comparing with the previous graph, we
see that it gives a better fit than our previous linear model.
с
380
370
360
350
340
1980 1985 1990 1995 2000 2005 2010
t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2950fafb-198b-44f1-a559-ce3dae66bd5f%2Fa83726ad-b754-48e9-b4af-b65009e33204%2Fdf76rw9_processed.png&w=3840&q=75)
Transcribed Image Text:Notice that our model gives values higher than most of the actual CO₂ levels. A better linear model is obtained by a
procedure from statistics called linear regression. If we use a graphing calculator, we enter the data from the table into
the data editor and choose the linear regression command. (With Maple we use the fit[leastsquare] command in the
stats package; with Mathematica we use the Fit command.) The machine gives the slope (round to two decimal
places) and y-intercept (round to the nearest integer) of the regression line as
m =
b =
So our least squares model for the CO₂ level is
C =
In the figure below, we graph the regression line as well as the data points. Comparing with the previous graph, we
see that it gives a better fit than our previous linear model.
с
380
370
360
350
340
1980 1985 1990 1995 2000 2005 2010
t
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