In a football tournament, the Bees scored 9 less than three times as many points as the Hornets. The Wasps scored 28 more points

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.

1. The number 10 indicates the time taken to create the centerpiece, while 30 signifies the minutes needed to craft the paper cranes when he arrives home.

Here are 3 questions with their respective answers.

1) Find




Answer: 4.

Explanation:

This expression indicates the limit as the function f(x) approaches 2 from the right side.

You should apply the function (the line) from the right side of 2 and get as close to x = 2 as you can.

That line has an open circle at y = 4, which is the limit we are looking for.

2) Analyze the graph to see if the limit exists.

Answer:





To find each limit, utilize the function approaching from the direction of x.

It's important to note that since the two limits differ, it is concluded that the limit of the function as x approaches 2 does not exist.

3)
Answer: -1




To determine the limit as the function approaches 3 from the left, follow the line ending with an open circle at (3, -1).

Hence, the limit is -1.