Q. 30 - Precalculus Enhanced with Graphing Utilities

Question

Which Model? A cricket makes a chirping noise by sliding its wings together rapidly. Perhaps you have noticed that the number of chirps seems to increase with the temperature. The following data list the temperature (in degrees Fahrenheit) and the number of chirps per second for the striped ground cricket. 

(a) Using a graphing utility, draw a scatter diagram of the data treating temperature as the independent variable. What type of relationship appears to exist between temperature and chirps per second?

 (b) Based on your response to part (a), find either a linear or quadratic model that best describes the relationship between temperature and chirps per second.

 (c) Use your model to predict the chirps per second if the temperature is ${80}^{\circ }F.$ 

Step-by-Step Solution

Answer

(a) The scatter diagram is,

(b) $T\left(a\right)=0.00998a^2-1.3a+59.5.$

(c) If the temperature is ${80}^{\circ }F$ the chirps are approximately $16.2$ per second.

Part (a) Step 1. Given information

The given table is, 

We need to draw a scatter diagram of the data treating temperature as the independent variable and determine the type of relationship that appears to exist between temperature and chirps per second.

Step 2. Graphing

The scatter diagram of the data is, 

Part (b) Step 1. Given information

The given table is,

Based on the result to part (a) we need to determine either a linear or quadratic model to describe the relationship between temperature and chirps per second. 

Step 2. Simplify

The quadratic model is of the form,

$y=a{x}^2+bx+c.$

Using the calculator and the values given in the table we determine the value of constant $a,b, and c.$

So, $a=0.00948, b=-1.3, c=59.5.$

Therefore, $y=0.00948x^2-1.3x+59.5.$

So, $T\left(a\right)=0.00998a^2-1.3a+59.5.$

Part (c) Step 1. Given information

The given table is,

We need to use our model to predict the chirps per second if the temperature is ${80}^{\circ }F.$

Step 2. Simplify

$T\left(a\right)=0.00998a^2-1.3a+59.5.$

If $a=80.$

$T\left(80\right)=0.00998{\left(80\right)}^2-1.3\left(80\right)+59.5.$

          $\approx 16.2.$