Chapter 5.9, Problem 4AYU

(a)

To determine

To find: The scatter diagram treating time as independent variable and the number of cigarette as the dependent variable..

Explanation of Solution

Given:

The given table is shown in table 1

Table 1

Year	Export in Billion of Pieces
1995	231.1
1998	201.3
1999	151.4
2000	147.9
2001	133.9
2002	127.4
2003	121.5
2004	118.7

Calculation:

Consider the time as independent variable.

The scatter plot for the functions in table 1 is shown in Figure 1

  

Figure 1

(b)

To determine

To find: The exponential model from the given data.

Answer to Problem 4AYU

The exponential function is $y=231.1e^{-0.183x}$ .

Explanation of Solution

From the graph shown in Figure 1, the exponential function with the help of the graphing utility is,

  $y=231.1e^{-0.183x}$

(c)

To determine

To find: The function from of the exponential model in the form of $A\left(t\right)=A_0{e}^{kt}$ .

Answer to Problem 4AYU

The required model is $A\left(t\right)=231.1e^{-0.183x}$ .

Explanation of Solution

Given:

The exponential function is $y=231.1e^{-0.183x}$ .

Calculation:

Consider the given model is,

  $A\left(t\right)=A_0{e}^{kt}$

Compare the given function with the given model as,

  $A\left(t\right)=231.1e^{-0.183x}$

(d)

To determine

To find: The graph of the exponential function from part (b) or (c) on the scatter diagram.

Answer to Problem 4AYU

The required plot is shown in Figure 2

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=231.1e^{-0.183x}$

Calculation:

The required plot is shown in Figure 2

  

Figure 2

(e)

To determine

To find: The number of cigarettes that are exported from the united states in 2010.

Answer to Problem 4AYU

Therequired amount of cigarettes is $14.84\text{billions}$ /

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=231.1e^{-0.183x}$ .

Calculation:

Consider the given function then,

  $A\left(t\right)=231.1e^{-0.183x}$

Consider the time elapsed is 15 years then,

Then,

  $\begin{array}{c}A\left(15\right)=231.1e^{-0.183\left(15\right)}\\ =14.84\text{billions}\end{array}$

(f)

To determine

To find: The amount of radioactive material that will be left over 50 weeks.

Answer to Problem 4AYU

The required years is $8.36\text{years}$ .

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=231.1e^{-0.183x}$ .

Calculation:

Consider the given function is,

  $A\left(t\right)=231.1e^{-0.183x}$

Then,

  $\begin{array}{l}50=231.1e^{-0.183x}\\ x=8.36\text{years}\end{array}$