Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - 7.6 Modeling Data: Exponential, Logarithmic, and Quadratic Functions - Exercise Set 7.6 - Page 476: 32

Answer

See below:

Work Step by Step

(a) Consider the given graph: Exponential function was used to model the population data because from the graph it can be observed that the data points increase very rapidly. Hence, we use exponential function of the form \[f\left( x \right)={{b}^{x}},b>1\]. An exponential function has to be used to model the population data given in graph. (b) Now, simplify the model as $y=3.476{{\left( 1.023 \right)}^{x}}$ This can be expressed on the graphic calculator as shown below: \[{{Y}_{\text{1}}}\text{=3}\text{.476*}\left( \text{1}\text{.023} \right)\text{ }\!\!\hat{\ }\!\!\text{ }x\] Thus, the model is expressed on the graphic calculator as shown above. Hence, the model is expressed on the graphic calculator as explained above. (c) Consider the model shown in the above figure. $y=3.476{{\left( 1.023 \right)}^{x}}$ Substitute the value of x as \[101\] for the year\[2000\], because x represents the year after 1899. It is substituted as shown below: $\begin{align} & y=3.476{{\left( 1.023 \right)}^{101}} \\ & =34.55563 \\ & =34.56 \end{align}$ The population shown by the bar graph in the year \[2000\] is 44.6 million. This shows that the calculated value using the model underestimate the value shown in the bar graph. Now, subtract the calculated value from the value obtained from the bar graph for the year \[2000\]. \[44.6-43.4=1.2\] Thus, the calculated value underestimates the value from the bar graph by 1.5 million for the year \[2000\]. Hence, the calculated number is \[34.56\,\text{million}\] and this value underestimates the value from the bar graph by \[0.24\,\text{million}\] for the year\[2000\]. (d) Consider the model shown in the above figure. $y=3.476{{\left( 1.023 \right)}^{x}}$ Substitute the value of x as \[131\] for the year \[2030\], because x represents the year after 1899. It is substituted as shown below: $\begin{align} & y=3.476{{\left( 1.023 \right)}^{131}} \\ & =68.35775 \\ & =68.35\text{ million} \end{align}$ The population shown by the bar graph in the year \[2030\] is \[68.35\,\text{million}\]. This shows that the calculated value using the model underestimates the value shown in the bar graph. Now, subtract the calculated value from the value obtained from the bar graph for the year \[2030\]. $70.3-68.35=1.95\,\text{million}$ Thus, the calculated value overestimates the value from the bar graph by \[1.95\,\text{million}\] for the year \[2030\]. Hence, the calculated number is \[68.35\,\text{million}\] and it underestimates the value from the bar graph by \[1.95\,\text{million}\] for the year \[2030\].
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