Chapter 6.7, Problem 35E

a.

To determine

To show : That an exponential model fits the data and then find an exponential model for the data.

Answer to Problem 35E

The exponential function that models the data is $y=\left(0.514\right){\left(1.462\right)}^x$ .

Explanation of Solution

Given information :

The given data is $\begin{array}{cccccc}\text{Age, }x& 1& 2& 3& 4& 5\\ \text{Weight, }y& 0.751& 1.079& 1.702& 2.198& 3.438\end{array}$ .

Calculation :

Create a table of data pairs $\left(x,\text{ ln}y\right)$ .

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ \ln y& -0.285& 0.076& 0.532& 0.788& 1.235\end{array}$

Plot the above transformed points as shown below:

  

Since, the point lie close to a line, so an exponential model should be a good fit for the original data.

Find an exponential model $y=a{b}^x$ by choosing any two points on the line, such as $\left(1,-0.285\right)$ and $\left(5,\text{ 1}\text{.235}\right)$ . Use these points to write an equation of the line. Then solve for y.

Therefore,

  $\begin{array}{l}\ln y+0.285=\frac{1.235+0.285}{5-1}\left(x-1\right)\text{ (Equation of the line)}\\ \ln y+0.285=0.38\left(x-1\right)\text{ (Simplify) }\\ \ln y=0.38x-0.665\\ y=e^{0.38x-0.665}\text{ (Exponentiate each side using base e)}\\ y=e^{0.38x}\left(e^{-0.665}\right)\text{ (Use peoperty of exponents)}\\ y=\left(0.514\right){\left(1.462\right)}^x\text{ (Simplify)}\end{array}$

Hence,

The exponential function that models the data is $y=\left(0.514\right){\left(1.462\right)}^x$ .

b.

To determine

To explain : By what percentage does the weight of the Atlantic cod increases each year in this period of time.

Answer to Problem 35E

The weight of the Atlantic cod increases each year in this period of time is 46.2%.

Explanation of Solution

Given information :

The given data is $\begin{array}{cccccc}\text{Age, }x& 1& 2& 3& 4& 5\\ \text{Weight, }y& 0.751& 1.079& 1.702& 2.198& 3.438\end{array}$ .

Calculation :

From part (a), the exponential model that represents the data is

  $y=\left(0.514\right){\left(1.462\right)}^x$

Let $r$ be the rate of weight increases each year in this period of time. Therefore,

  $\begin{array}{l}1+r=1.462\\ r=1.462-1\\ r=0.462\end{array}$

Therefore, percentage increase is 46.2%.

Hence,

The weight of the Atlantic cod increases each year in this period of time is 46.2%.