Chapter 5.9, Problem 3AYU

(a)

To determine

To find: The scatter diagram treating time as independent variable.

Answer to Problem 3AYU

The required plot is shown in Figure 1

Explanation of Solution

Given:

The given table is shown in table 1

Table 1

Time hours x	Population y
0	100
1	88.3
2	75.9
3	69.4
4	59.1
5	51.8
6	45.5

Calculation:

Consider the time as independent variable.

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

  

Figure 1

v

(b)

To determine

To find: The exponential model from the given data.

Answer to Problem 3AYU

The exponential function is $y=100e^{-0.13x}$ .

Explanation of Solution

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

  $y=100e^{-0.13x}$

(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 3AYU

The required model is $A\left(t\right)=100e^{-0.13x}$ .

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=100e^{-0.13x}$ .

Calculation:

Consider the given model is,

  $A\left(t\right)=A_0{e}^{kt}$

Compare the given function with the given model as,

  $A\left(t\right)=100e^{-0.13x}$

(d)

To determine

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

Answer to Problem 3AYU

The required plot is shown in Figure 2

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=100e^{-0.13x}$ .

Calculation:

The required plot is shown in Figure 2

  

Figure 2

(e)

To determine

To find: The half-life of the radio active material.

Answer to Problem 3AYU

The required time is of $5.33\text{weeks}$ .

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=100e^{-0.13x}$ .

Calculation:

Consider the given function then,

  $\begin{array}{l}A\left(t\right)=100e^{-0.13x}\\ 50=100e^{-0.13x}\\ x=5.33\text{weeks}\end{array}$

(f)

To determine

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

Answer to Problem 3AYU

The amount of material is $0.15\text{grams}$ .

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=100e^{-0.13x}$ .

Calculation:

Consider the given function is,

  $A\left(t\right)=100e^{-0.13x}$

Then,

  $\begin{array}{c}A\left(t\right)=100e^{-0.13\left(50\right)}\\ =0.15\text{grams}\end{array}$

(g)

To determine

To find: The time at which there will only be 20 grams of the radioactive material.

Answer to Problem 3AYU

The required duration is of $12.38\text{weeks}$ .

Explanation of Solution

Given:

The exponential function is $A\left(t\right)=100e^{-0.13x}$ .

Calculation:

Consider the given function is,

  $A\left(t\right)=100e^{-0.13x}$

Then,

  $\begin{array}{l}20=100e^{-0.13x}\\ x=12.38\text{weeks}\end{array}$