Chapter A.4, Problem 77E

a.

To determine

Create a table showing the numbers of households banking online and the numbers of households paying bills online for the given years.

Answer to Problem 77E

  $\begin{array}{ccc}Year\left(t\right)& \begin{array}{l}Banking\\ \left(millions\right)\end{array}& \begin{array}{l}PayingBills\\ \left(millions\right)\end{array}\\ 5& 46.9& 17\\ 6& 57.6& 25.6\\ 7& 63.5& 27.3\\ 8& 67.3& 28.8\\ 9& 69.9& 30.8\\ 10& 71.9& 33.7\end{array}$

Explanation of Solution

Given information:

The table shows the numbers of U.S. households (in millions) banking online and paying bills online from $2005$ through $2010$ .

  

Mathematical models for the data are

Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

And

Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

Where $t$ represents the year, with $t=5$ corresponding to $2005$ .

Using the models, create a table showing the numbers of households banking online and the numbers of households paying bills online for the given years.

Calculation:

Consider $t=5$ for $2005$ , $t=6for2006$ , $t=7for2007$ , $t=8for2008$ , $t=9for2009$ , $t=10for2010$

Now calculate Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$ and Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$ for $t=5$ for $2005$ , $t=6for2006$ , $t=7for2007$ , $t=8for2008$ , $t=9for2009$ , $t=10for2010$

For $t=5$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 5+121.8}{-0.40\times 5+1.0}\\ =46.9\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(5\right)}^2-6.54\left(5\right)+24.6}{0.015{\left(5\right)}^2-0.28\left(5\right)+1.0}\\ =17\end{array}$

For $t=6$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 6+121.8}{-0.40\times 6+1.0}\\ =57.6\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(6\right)}^2-6.54\left(6\right)+24.6}{0.015{\left(6\right)}^2-0.28\left(6\right)+1.0}\\ =25.6\end{array}$

For $t=7$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 7+121.8}{-0.40\times 7+1.0}\\ =63.7\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(7\right)}^2-6.54\left(7\right)+24.6}{0.015{\left(7\right)}^2-0.28\left(7\right)+1.0}\\ =27.3\end{array}$

For $t=8$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 8+121.8}{-0.40\times 8+1.0}\\ =67.3\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(8\right)}^2-6.54\left(8\right)+24.6}{0.015{\left(8\right)}^2-0.28\left(8\right)+1.0}\\ =28.8\end{array}$

For $t=9$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 9+121.8}{-0.40\times 9+1.0}\\ =69.9\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(9\right)}^2-6.54\left(9\right)+24.6}{0.015{\left(9\right)}^2-0.28\left(9\right)+1.0}\\ =30.8\end{array}$

For $t=10$ ,

Banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

  $\begin{array}{l}=\frac{-33.74\times 10+121.8}{-0.40\times 10+1.0}\\ =71.9\end{array}$

Paying bills $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

  $\begin{array}{l}=\frac{-33.74{\left(10\right)}^2-6.54\left(10\right)+24.6}{0.015{\left(10\right)}^2-0.28\left(10\right)+1.0}\\ =33.7\end{array}$

Now table showing the numbers of households banking online and the numbers of households paying bills online for the given years.

  $\begin{array}{ccc}Year\left(t\right)& \begin{array}{l}Banking\\ \left(millions\right)\end{array}& \begin{array}{l}PayingBills\\ \left(millions\right)\end{array}\\ 5& 46.9& 17\\ 6& 57.6& 25.6\\ 7& 63.5& 27.3\\ 8& 67.3& 28.8\\ 9& 69.9& 30.8\\ 10& 71.9& 33.7\end{array}$

Hence, the table is created.

b.

To determine

Compare the values given by the models with the actual data.

Answer to Problem 77E

The estimate is close to the actual number of households.

Explanation of Solution

Given information:

The table shows the numbers of U.S. households (in millions) banking online and paying bills online from $2005$ through $2010$ .

  

Mathematical models for the data are

Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

And

Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

Where $t$ represents the year, with $t=5$ corresponding to $2005$ .

Compare the values given by the models with the actual data.

Calculation:

Consider the given table,and created table now on comparing both,

  

Now table showing the numbers of households banking online and the numbers of households paying bills online for the given years.

  $\begin{array}{ccc}Year\left(t\right)& \begin{array}{l}Banking\\ \left(millions\right)\end{array}& \begin{array}{l}PayingBills\\ \left(millions\right)\end{array}\\ 5& 46.9& 17\\ 6& 57.6& 25.6\\ 7& 63.5& 27.3\\ 8& 67.3& 28.8\\ 9& 69.9& 30.8\\ 10& 71.9& 33.7\end{array}$

Hence, the estimate is close to the actual number of households.

c.

To determine

Determine a model for the ratio of the number of households paying bills online to the number of households banking online.

Answer to Problem 77E

  $\boxed{\frac{-0.1228{\left(t\right)}^3+2.923{\left(t\right)}^2-16.38t+24.6}{-0.5016{\left(t\right)}^3+11.2742t^2-67.844\left(t\right)+121.8}}$

Explanation of Solution

Given information:

The table shows the numbers of U.S. households (in millions) banking online and paying bills online from $2005$ through $2010$ .

  

Mathematical models for the data are

Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

And

Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

Where $t$ represents the year, with $t=5$ corresponding to $2005$ .

Determine a model for the ratio of the number of households paying bills online to the number of households banking online.

Calculation:

To find the ratio divide Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$ to Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$ we get,

  $\begin{array}{l}\frac{\text{Number paying bills online}}{\text{Number banking online}}\\ =\frac{-33.74{\left(t\right)}^2-6.54\left(t\right)+24.6}{0.015{\left(t\right)}^2-0.28\left(t\right)+1.0}÷\frac{-33.74t+121.8}{-0.40t+1.0}\\ =\frac{-33.74{\left(t\right)}^2-6.54\left(t\right)+24.6}{0.015{\left(t\right)}^2-0.28\left(t\right)+1.0}\times \frac{-0.40t+1.0}{-33.74t+121.8}\\ =\frac{\left(-33.74{\left(t\right)}^2-6.54\left(t\right)+24.6\right)\left(-0.40t+1.0\right)}{\left(0.015{\left(t\right)}^2-0.28\left(t\right)+1.0\right)\left(-33.74t+121.8\right)}\\ =\frac{-0.1228{\left(t\right)}^3+2.923{\left(t\right)}^2-16.38t+24.6}{-0.5016{\left(t\right)}^3+11.2742t^2-67.844\left(t\right)+121.8}\end{array}$

Hence, the ratio is $\boxed{\frac{-0.1228{\left(t\right)}^3+2.923{\left(t\right)}^2-16.38t+24.6}{-0.5016{\left(t\right)}^3+11.2742t^2-67.844\left(t\right)+121.8}}$ .

d.

To determine

To find the ratios for the given years.

Answer to Problem 77E

  $\begin{array}{ccccccc}Year& 2005& 2006& 2007& 2008& 2009& 2010\\ Ratio& 0.3625& 0.449& 0.4292& 0.4282& 0.4403& 0.4685\end{array}$

Explanation of Solution

Given information:

The table shows the numbers of U.S. households (in millions) banking online and paying bills online from $2005$ through $2010$ .

  

Mathematical models for the data are

Number banking online $=\frac{-33.74t+121.8}{-0.40t+1.0}$

And

Number paying bills online $=\frac{-33.74t^2-6.54t+24.6}{0.015t^2-0.28t+1.0}$

Where $t$ represents the year, with $t=5$ corresponding to $2005$ .

Use the model from part (c) to find the ratios for the given years. Interpret your results.

Calculation:

Consider the ratio model,

  $\frac{\text{Number paying bills online}}{\text{Number banking online}}$ = $\boxed{\frac{-0.1228{\left(t\right)}^3+2.923{\left(t\right)}^2-16.38t+24.6}{-0.5016{\left(t\right)}^3+11.2742t^2-67.844\left(t\right)+121.8}}$

Now find the ratio for the given years,

For $t=5$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(5\right)}^3+2.923{\left(5\right)}^2-16.38\left(5\right)+24.6}{-0.5016{\left(5\right)}^3+11.2742{\left(5\right)}^2-67.844\left(5\right)+121.8}\\ 0.3625\end{array}$

For $t=6$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(6\right)}^3+2.923{\left(6\right)}^2-16.38\left(6\right)+24.6}{-0.5016{\left(6\right)}^3+11.2742{\left(6\right)}^2-67.844\left(6\right)+121.8}\\ 0.4449\end{array}$

For $t=7$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(7\right)}^3+2.923{\left(7\right)}^2-16.38\left(7\right)+24.6}{-0.5016{\left(7\right)}^3+11.2742{\left(7\right)}^2-67.844\left(7\right)+121.8}\\ 0.4292\end{array}$

For $t=8$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(8\right)}^3+2.923{\left(8\right)}^2-16.38\left(8\right)+24.6}{-0.5016{\left(8\right)}^3+11.2742{\left(8\right)}^2-67.844\left(8\right)+121.8}\\ 0.4282\end{array}$

For $t=9$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(9\right)}^3+2.923{\left(9\right)}^2-16.38\left(9\right)+24.6}{-0.5016{\left(9\right)}^3+11.2742{\left(9\right)}^2-67.844\left(9\right)+121.8}\\ 0.4403\end{array}$

For $t=10$ ,

  $\begin{array}{l}ratio=\frac{-0.1228{\left(10\right)}^3+2.923{\left(10\right)}^2-16.38\left(10\right)+24.6}{-0.5016{\left(10\right)}^3+11.2742{\left(10\right)}^2-67.844\left(10\right)+121.8}\\ 0.4685\end{array}$

  $\begin{array}{ccccccc}Year& 2005& 2006& 2007& 2008& 2009& 2010\\ Ratio& 0.3625& 0.449& 0.4292& 0.4282& 0.4403& 0.4685\end{array}$

Hence, the table shows the ratio.