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:
Step-by-step explanation:
For resolution, the Euclidean division is necessary:
- The polynomial is expressed as 16-2x³-6x²+x+9=-2x³-6x²+x+9
- We will divide this by: 4+3x-2=3x+2 and analyze the remainder
A visual representation of the process is provided
- The remainder is 47/9, hence the value to subtract is 47/9
To determine the time interval δt, we must subtract the starting time from the ending time. In this scenario, the first value in the coordinates signifies time:
δt=50 - 0
δt= 50s
The time interval is 50s.
Answer:
1. x=±4
2. t=±9
3. r=±10
4. x=±12
5. s=±5
Step-by-step explanation:
1. x^2 = 16
By extracting the square root on both sides
x=±4
2. t^2=81
Again, take the square root of each side
t=±9
3. r^2-100=0
r=±10
4. x²-144=0
We rewrite as x²=144
Applying square roots
x=±12
5. 2s²=50
s=±5 ..
