Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - Quiz: 2

Answer

There were around $6.76$ million enrollments in 2006 and around $7.235$ million enrollments in 2010.. 2010 is the midpoint of 2008 and 2012, the last two entries in the table. Thus, the midpoint of the last two entries is: $=\left(\frac{2008+2012}{2}, \frac{6.97+7.50}{2}\right) \\=\left(\frac{4020}{2}, \frac{14.47}{2}\right) \\=(2010, 7.235)$ There were around $7.235$ million enrollments in 2010.

Work Step by Step

The coordinates of the midpoint the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula: $$=\left(\dfrac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$ 2006 is the midpoint of 2004 and 2008, the first two entries in the table. Thus, the midpoint of the first two entries is: $=\left(\frac{2004+2008}{2}, \frac{6.55+6.97}{2}\right) \\=\left(\frac{4012}{2}, \frac{13.52}{2}\right) \\=(2006, 6.76)$ There were around $6.76$ million enrollments in 2006. 2010 is the midpoint of 2008 and 2012, the last two entries in the table. Thus, the midpoint of the last two entries is: $=\left(\frac{2008+2012}{2}, \frac{6.97+7.50}{2}\right) \\=\left(\frac{4020}{2}, \frac{14.47}{2}\right) \\=(2010, 7.235)$ There were around $7.235$ million enrollments in 2010.
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