Chapter 1.1, Problem 129E

a.

To determine

To graph: The data given and the graph the model.

Explanation of Solution

Given:

Following data is given in the form of a table

Year	Bachelor’s degrees, B (in thousands)
2005	826
2006	855
2007	875
2008	895
2009	916
2010	943
2011	982
2012	1026
2013	1053
2014	1068
2015	1082
2016	1099

The data can be approximated by the linear model $B=26.42t+690.9,5\le t\le 16$ , where t represents the year, with t = 5 corresponding to 2005.

Graph:

By using the above data, points and the model can be plotted as

  

Interpretation:

The graph shows that the points plotted are very close to the graph of the linear model.

b.

To determine

To find: the bachelor’s degrees earned by women for each year from 2005 through 2016 by the linear model.

Explanation of Solution

Given:

The linear model $B=26.42t+690.9,5\le t\le 16$ , where t represents the year, with t = 5 corresponding to 2005.

Calculation:

Table below shows the data for each year from 2005 to 2016.

Year	Bachelor’s degrees, B (in thousands)
2005	823
2006	849.42
2007	875.84
2008	902.26
2009	928.68
2010	955.1
2011	981.52
2012	1007.94
2013	1034.36
2014	1060.78
2015	1087.2
2016	1113.62

c.

To determine

To compare: the given data with the data obtained from linear model.

Explanation of Solution

Given:

Following data is given in the form of a table

Year	Bachelor’s degrees, B (in thousands)
2005	826
2006	855
2007	875
2008	895
2009	916
2010	943
2011	982
2012	1026
2013	1053
2014	1068
2015	1082
2016	1099

The data can be approximated by the linear model $B=26.42t+690.9,5\le t\le 16$ , where t represents the year, with t = 5 corresponding to 2005.

Calculation:

Year	Bachelor’s degrees, B (in thousands) (data given)	Bachelor’s degrees, B (in thousands) (data from the linear model )	Difference between the two values
2005	826	823	3
2006	855	849.42	5.58
2007	875	875.84	-0.84
2008	895	902.26	-7.62
2009	916	928.68	-12.68
2010	943	955.1	-12.1
2011	982	981.52	0.48
2012	1026	1007.94	18.06
2013	1053	1034.36	18.64
2014	1068	1060.78	7.22
2015	1082	1087.2	-5.2
2016	1099	1113.62	-14.62

As can be seen from the table, the difference between the given data and the data obtained from the linear model is very less thus; the model is best suited for the data.

d.

To determine

To find: the slope and the y -intercept of the model.

Explanation of Solution

Given:

The linear model: $B=26.42t+690.9,5\le t\le 16$ , where t represents the year, with t = 5 corresponding to 2005.

Concept used; The equation of a straight line is given by $y=mx+C$ , where m is the slope of the line and C is the y -intercept.

Calculation:

We have the equation of linear model as $B=26.42t+690.9,5\le t\le 16$ , compare the equation with the standard form of straight line $y=mx+C$ , we get

  $m=26.42$ and $C=690.9$ .

Thus the slope of the model is, $m=26.42$ and the y -intercept of the model is, $C=690.9$ .

Slope tells how much the value of y changes with the change in x .

The y -intercept of the linear model shows the value at t = 0, as we know t = 5 corresponds to 2005, t = 0 will correspond to 2000. Thus the y -intercept gives the number of bachelor’s degrees earned by women in the year 2000.

e.

To determine

To find: the number of bachelor’s degrees earned by women in the year 2022

Explanation of Solution

Given:

The linear model $B=26.42t+690.9,5\le t\le 16$ , where t represents the year, with t = 5 corresponding to 2005.

Calculation:

t = 5 corresponds to 2005, therefore 2022 is given by t = 22,

Putting the value of t as 22 in the linear model we get,

  $\begin{array}{l}B=26.42(22)+690.9\\ B=581.24+690.9\\ B=1272.14\end{array}$

Therefore, the number of bachelor’s degrees earned by women in the year 2022 is 1272.14 thousands