Glencoe Algebra 2 Student Edition C2014
Glencoe Algebra 2 Student Edition C2014
1st Edition
ISBN: 9780076639908
Author: McGraw-Hill Glencoe
Publisher: MCG
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Chapter SH, Problem 2.15EP

a.

To determine

To represent: a scatter plot, graph a line fit, and describe the correlation of the given condition.

a.

Expert Solution
Check Mark

Answer to Problem 2.15EP

“Strong Positive Correlation.”

Explanation of Solution

Given:

The table shows the number shows the LCD televisions sold in the recent year:

    Year Numbers of TVs sold
    2006

      733

    2007

      838

    2008

      943

    2009

      1048

    2015

      ?

Calculation:

Consider the data that represents the number of LCD televisions sold in various years at an electronic store. Assume that x represents the number of years since 2000.

To draw a scatter plot in Excel, first select data to plot and go to “Insert” menu, then go to “charts” and click on “Scatter plot”. Label both axis and the scatter plot of the given data is as follows:

  Glencoe Algebra 2 Student Edition C2014, Chapter SH, Problem 2.15EP , additional homework tip  1

The points (6,733) and(9,1048) will represent the data well. Draw a line through these two points. The line of fir is as follows:

  Glencoe Algebra 2 Student Edition C2014, Chapter SH, Problem 2.15EP , additional homework tip  2

The given data shows a “Strong Positive Correlation.”

b.

To determine

To write: a prediction equation by using two ordered pairs.

b.

Expert Solution
Check Mark

Answer to Problem 2.15EP

  y=105x+103 .

Explanation of Solution

Given:

The given ordered pairs (6,733) and(9,1048)

Calculation:

To find an equation of a line through. The slope of the line will be,

  m=y2y1x2x1=104873396=3153=105

The equation of the line will be,

  yy1=m(xx1)y733=105(x6)y=105x630+733y=105x+103

Therefore, the equation of the line is y=105x+103 ,

c.

To determine

To predict: the missing value of the given table.

c.

Expert Solution
Check Mark

Answer to Problem 2.15EP

  1678 .

Explanation of Solution

Given:

The table shows the number shows the LCD televisions sold in the recent year:

    Year Numbers of TVs sold
    2006

      733

    2007

      838

    2008

      943

    2009

      1048

    2015

      ?

Calculation:

To compute the number of LCDs sold in 2015, substitute x=15 in equation y=105x+103 ,

  y=105×15+103y=1575+103y=1678

Therefore, the predicted number of LCDs sold in 2015 is equal to 1678

Chapter SH Solutions

Glencoe Algebra 2 Student Edition C2014

Ch. SH - Prob. 1.11EPCh. SH - Prob. 1.12EPCh. SH - Prob. 1.13EPCh. SH - Prob. 1.14EPCh. SH - Prob. 1.15EPCh. SH - Prob. 1.16EPCh. SH - Prob. 1.17EPCh. SH - Prob. 1.18EPCh. SH - Prob. 1.19EPCh. SH - Prob. 1.20EPCh. SH - Prob. 1.21EPCh. SH - Prob. 1.22EPCh. SH - Prob. 1.23EPCh. SH - Prob. 1.24EPCh. SH - Prob. 1.25EPCh. SH - Prob. 2.1EPCh. SH - Prob. 2.2EPCh. SH - Prob. 2.3EPCh. SH - Prob. 2.4EPCh. SH - Prob. 2.5EPCh. SH - Prob. 2.6EPCh. SH - Prob. 2.7EPCh. SH - Prob. 2.8EPCh. SH - Prob. 2.9EPCh. SH - Prob. 2.10EPCh. SH - Prob. 2.11EPCh. SH - Prob. 2.12EPCh. SH - Prob. 2.13EPCh. SH - Prob. 2.14EPCh. SH - Prob. 2.15EPCh. SH - Prob. 2.16EPCh. SH - Prob. 2.17EPCh. SH - Prob. 2.18EPCh. SH - Prob. 2.19EPCh. SH - Prob. 2.20EPCh. SH - Prob. 2.21EPCh. SH - Prob. 2.22EPCh. SH - Prob. 2.23EPCh. SH - Prob. 2.24EPCh. SH - Prob. 2.25EPCh. SH - Prob. 3.1EPCh. SH - Prob. 3.2EPCh. SH - Prob. 3.3EPCh. SH - Prob. 3.4EPCh. SH - Prob. 3.5EPCh. SH - Prob. 3.6EPCh. SH - Prob. 3.7EPCh. SH - Prob. 3.8EPCh. 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SH - Prob. 5.6EPCh. SH - Prob. 5.7EPCh. SH - Prob. 5.8EPCh. SH - Prob. 5.9EPCh. SH - Prob. 5.10EPCh. SH - Prob. 5.11EPCh. SH - Prob. 5.12EPCh. SH - Prob. 5.13EPCh. SH - Prob. 5.14EPCh. SH - Prob. 5.15EPCh. SH - Prob. 5.16EPCh. SH - Prob. 5.17EPCh. SH - Prob. 5.18EPCh. SH - Prob. 5.19EPCh. SH - Prob. 5.20EPCh. SH - Prob. 5.21EPCh. SH - Prob. 5.22EPCh. SH - Prob. 5.23EPCh. SH - Prob. 5.24EPCh. SH - Prob. 5.25EPCh. SH - Prob. 6.1EPCh. SH - Prob. 6.2EPCh. SH - Prob. 6.3EPCh. SH - Prob. 6.4EPCh. SH - Prob. 6.5EPCh. SH - Prob. 6.6EPCh. SH - Prob. 6.7EPCh. SH - Prob. 6.8EPCh. SH - Prob. 6.9EPCh. SH - Prob. 6.10EPCh. SH - Prob. 6.11EPCh. SH - Prob. 6.12EPCh. SH - Prob. 6.13EPCh. SH - Prob. 6.14EPCh. SH - Prob. 6.15EPCh. SH - Prob. 6.16EPCh. SH - Prob. 6.17EPCh. SH - Prob. 6.18EPCh. SH - Prob. 6.19EPCh. SH - Prob. 6.20EPCh. SH - Prob. 6.21EPCh. SH - Prob. 6.22EPCh. SH - Prob. 6.23EPCh. SH - Prob. 6.24EPCh. SH - Prob. 6.25EPCh. SH - Prob. 7.1EPCh. SH - Prob. 7.2EPCh. SH - Prob. 7.3EPCh. 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SH - Prob. 9.8EPCh. SH - Prob. 9.9EPCh. SH - Prob. 9.10EPCh. SH - Prob. 9.11EPCh. SH - Prob. 9.12EPCh. SH - Prob. 9.13EPCh. SH - Prob. 9.14EPCh. SH - Prob. 9.15EPCh. SH - Prob. 9.16EPCh. SH - Prob. 9.17EPCh. SH - Prob. 9.18EPCh. SH - Prob. 9.19EPCh. SH - Prob. 9.20EPCh. SH - Prob. 9.21EPCh. SH - Prob. 10.1EPCh. SH - Prob. 10.2EPCh. SH - Prob. 10.3EPCh. SH - Prob. 10.4EPCh. SH - Prob. 10.5EPCh. SH - Prob. 10.6EPCh. SH - Prob. 10.7EPCh. SH - Prob. 10.8EPCh. SH - Prob. 10.9EPCh. SH - Prob. 10.10EPCh. SH - Prob. 10.11EPCh. SH - Prob. 10.12EPCh. SH - Prob. 10.13EPCh. SH - Prob. 10.14EPCh. SH - Prob. 10.15EPCh. SH - Prob. 10.16EPCh. SH - Prob. 10.17EPCh. SH - Prob. 10.18EPCh. SH - Prob. 10.19EPCh. SH - Prob. 10.20EPCh. SH - Prob. 10.21EPCh. SH - Prob. 10.22EPCh. SH - Prob. 10.23EPCh. SH - Prob. 10.24EPCh. SH - Prob. 10.25EPCh. SH - Prob. 10.26EPCh. SH - Prob. 11.1EPCh. SH - Prob. 11.2EPCh. SH - Prob. 11.3EPCh. SH - Prob. 11.4EPCh. SH - Prob. 11.5EPCh. SH - Prob. 11.6EPCh. 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