Chapter 3.4, Problem 30AYU

(a)

To determine

To graph:

The scatter plot and define the relation appears to exist between temperature and chirps.

Explanation of Solution

Given Information:

temperature, x	Chirps pers second, C
$88.6$	$20.0$
$93.3$	$19.8$
$80.6$	$17.1$
$69.7$	$14.7$
$69.4$	$15.4$
$79.6$	$15.0$
$80.6$	$16.0$
$76.3$	$14.4$
$75.2$	$15.5$

The data list the temperature and the number of chirps per second for the striped ground cricket.

Graph:

  

Interpretation:

By using the graphing utility, the scatter plot as above.

The above scatter plot shows the data of the list of temperature and the number of chirps per second for the striped ground cricket.

The relation is quadratic function appears to exist between temperature and chirps per second.

(b)

To determine

To calculate:

The linear or quadratic model based on (a).

Answer to Problem 30AYU

The model is $C\left(x\right)=0.009x^2-1.3x+59.5$ .

Explanation of Solution

Given information:

temperature, x	Chirps pers second, C
$88.6$	$20.0$
$93.3$	$19.8$
$80.6$	$17.1$
$69.7$	$14.7$
$69.4$	$15.4$
$79.6$	$15.0$
$80.6$	$16.0$
$76.3$	$14.4$
$75.2$	$15.5$

The data list the temperature and the number of chirps per second for the striped ground cricket.

Calculation:

  $C\left(x\right)=0.009x^2-1.3x+59.5$

  

Hence, the model is $C\left(x\right)=0.009x^2-1.3x+59.5$ .

(c)

To determine

To calculate:

The chirps per second if the temperature is ${80}^{\circ }\text{F}$ .

Answer to Problem 30AYU

The chirps per second if the temperature ${80}^{\circ }\text{F}$ is $C\left(80\right)=16.3$ .

Explanation of Solution

Given information:

temperature, x	Chirps pers second, C
$88.6$	$20.0$
$93.3$	$19.8$
$80.6$	$17.1$
$69.7$	$14.7$
$69.4$	$15.4$
$79.6$	$15.0$
$80.6$	$16.0$
$76.3$	$14.4$
$75.2$	$15.5$

The data list the temperature and the number of chirps per second for the striped ground cricket.

Calculation:

  $\begin{array}{l}C\left(x\right)=0.0095x^2-1.3x+59.5\\ C\left(80\right)=0.0095{\left(80\right)}^2-1.3\left(80\right)+59.5\\ C\left(80\right)=16.3\end{array}$