Chapter 4.2, Problem 35PS

Solution

(A)

To find: Identify the decay factor and percentage increase

The decay factor is $0.9977$ and percentage increase is $0.2239\%$ .

Given information:

The given equation is :

  $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ ,where $w$ is the age in weeks of the chicken and $w\ge 22$ .

Calculation:

In order to determine the decay factor rewrite the given equation as:

  $\begin{array}{l}E=179.2{\left({\left(0.89\right)}^{\frac{1}{52}}\right)}^W\\ E=179.2{\left(0.9977\right)}^W\end{array}$

Apply the property of exponent $a^{mn}={\left(a^m\right)}^n$ .

The decay factor will be:

  $1-0.9977=0.002239$

The percentage decrease will be:

  $\begin{array}{l}\%d=0.002239\cdot 100\\ =0.2239\%\end{array}$

The decay factor is $0.9977$ and percentage increase is $0.2239\%$ .

(b)

To find: Draw the graph of the model.

Given information:

The given equation is :

  $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ ,where $w$ is the age in weeks of the chicken and $w\ge 22$ .

Calculation:

In order to draw the graph substitute different values of $w\ge 22$ .

Let substitute $w=30$ in $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ gives the value of $E$ as:

  $\begin{array}{l}E=179.2{\left(0.89\right)}^{\frac{30}{52}}\\ E=167.54\end{array}$

So the point will be $\left(30,40.83\right)$ .

Now substitute $w=40$ in $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ gives the value of $E$ as:

  $\begin{array}{l}E=179.2\left(0.89\right)\frac{40}{52}\\ E=163.83\end{array}$

Similarly different values of $w$ will give different values of $E$ .

Draw these points on the graph .

The graph is given below:

  

(C)

To find: Estimate the egg production of the chicken that is 2.5 year old

The egg production of a chicken is 134 eggs.

Given information:

The given equation is :

  $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ ,where $w$ is the age in weeks of the chicken and $w\ge 22$ .

Calculation:

In order to find the egg production of chicken in 2.5 years convert 2.5 into weeks.

  $52\cdot 2.5=130weeks$

Now substitute $w=130$ in $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ gives:

  $\begin{array}{l}E=179.2{\left(0.89\right)}^{\frac{130}{52}}\\ E=133.91\\ E\approx 134\end{array}$

The egg production of a chicken is 134 eggs.

(d )

To find: Rewrite the equation so that time can measured in years.

The given equation $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ can be rewritten as $E=179.2{\left(0.89\right)}^t$ .t is number of years

Given information:

The given equation is :

  $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ ,where $w$ is the age in weeks of the chicken and $w\ge 22$ .

Calculation:

In the equation $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$

  $w$ is the number of weeks and in a year has $52$ weeks .

So the time $t$ represents the number of years which is :

  $t=\frac{w}{52}$

So the equation becomes:

  $E=179.2{\left(0.89\right)}^t$

The given equation $E=179.2{\left(0.89\right)}^{\frac{W}{52}}$ can be rewritten as $E=179.2{\left(0.89\right)}^t$ .t is number of years.