<span>D. Let x represent the number of trays of dog bone treats made and y represent number of trays of oatmeal dog treats made.</span>
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.
Divide 4 by 6. Easier if you write it as a fraction: 4/6 This can be reduced to 2/3. This means that each person gets 2/3 of a pie equally.
Answer:
They are dependent because we have to select from people who are given cards.
Step By Step Explanation:
So we'll take away people not given cards first den find the probability of selecting people with cards over the total number of people present .
Probability we'll be equal to = number of people with card(C) two persons/total number of people
Where C represent combination
