Question

The measures in the table describe the weights of animals that visited a vet on one day, in pounds. Mean Median Mode Mean Absolute Deviation 12.9 12.0 12.0 2.4 Select from the drop-down menus to correctly complete each statement. On average, the weight of a pet visiting this vet on this day is about pounds away from pounds. If the MAD of weights for another day was 1.5, then that day's weights would be than the weights of pets seen on this day.

Answer 1

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

If the MAD of weights for another day was 1.5, then that day's weights would be less variable than the weights of pets seen on this day.

Step-by-step explanation:

Given: The measures in the table describe the weights of animals that visited a vet on one day, in pounds.

Mean = 12.9

Median= 12.0

Mode = 12.0

Mean Absolute Deviation = 2.4

We know that the mean absolute deviation (MAD) of a data set is the mean distance between each and every data value and the mean.It tells about the variation in a data set.

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

Also, If the MAD of weights for another day was 1.5, and since 1.5< 2.4.

Then that day's weights would be less variable than the weights of pets seen on this day.

Answer 2

Answer:

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

If the MAD of weights for another day was 1.5, then that day's weights would be less variable than the weights of pets seen on this day.

I took the test and this was correct