Chapter 9.6, Problem 24E

To determine

To find: Which city has greater average daily temperature Lexington, Kentucky or Louisville, Kentucky.

Answer to Problem 24E

Louisville, Kentucky.

Explanation of Solution

Given information:

The average daily temperature (in Fahrenheit) of Louisville, Kentucky is modeled by $y=-22\cos \left(\frac{\pi }{6}t\right)+57$

where $y$ is the temperature and $t$ is the number of months since January.

Below graph shows the average daily temperature of Lexington, Kentucky.

  

Calculation:

The average daily temperature (in Fahrenheit) of Louisville, Kentucky is modeled by $y=-22\cos \left(\frac{\pi }{6}t\right)+57$

where $y$ is the temperature and $t$ is the number of months since January.

Plotting the graph of $y=-22\cos \left(\frac{\pi }{6}t\right)+57$ using a graphing utility as below-

  

Below graph shows the average daily temperature of Lexington, Kentucky.

  

Comparing above two graph it is clear that Louisville, Kentucky has greater average daily temperature as graph of Louisville, Kentucky is above the graph of Lexington, Kentucky.