Building and solving an equation to model the situation, it is found that it takes 4 min for the prairie dog to reach 13 feet underground.
- Initially, the dog is 5 feet underground.
- Each minute, the dog goes 2 more feet underground.
Hence, the underground height of the dog after t seconds is given by:

The time it takes for the dog to reach 13 feet underground is <u>t for which h(t) = 13</u>, hence:





It takes 4 min for the prairie dog to reach 13 feet underground.
A similar problem is given at brainly.com/question/25290003
Answer:

Here is for you.
Step-by-step explanation:
Answer:
C. The difference of the medians is 4 times the interquartile range.
Step-by-step explanation:
The diagram show two box plots.
<u>Mahoney's box plot:</u>
Median 
Interquartile range 
<u>Martin's box plot:</u>
Median 
Interquartile range 
The interquartile ranges are the same.
The difference of the medians 
Hence, the difference of the medians is 4 times the interquartile range.