The measures in the table describe the weights of animals that visited a vet on one day, in pounds. Mean Median Mode Mean Absolu

Answer: On average, a pet's weight during this vet visit is approximately 2.4 pounds away from 12.9 pounds.

If the MAD for another day's weights was 1.5, then that day’s weights would be less variable compared to the weights of pets seen today.

Step-by-step explanation:

Given: The data in the table outlines the weights of animals visiting a vet one day, in pounds.

Mean = 12.9

Median= 12.0

Mode = 12.0

Mean Absolute Deviation = 2.4

It’s understood that the mean absolute deviation (MAD) of a dataset indicates the average distance between each data point and the mean. It reflects the variation present in the dataset.

Therefore, the average weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.

Moreover, if another day had a MAD of 1.5, and since 1.5 < 2.4,[TAG_42]]

it implies that the weights on that day would be less variable compared to those of pets encountered on this day.

Answer:

On average, a pet visiting this vet today weighs around 2.4 pounds away from 12.9 pounds.

If the MAD of another day’s weights was found to be 1.5, that day's weights would be less variable compared to the weights of pets that were seen today.

I took the test and this was correct