Answer:
Step-by-step explanation:
hey i got the answer straight from my teach but ony a and c if you have the answer to b let me know : ( A) To determine the mean absolute deviation, first calculate the mean. Find the sum of the values: 8 plus 6 plus 9 plus 6 plus 14 plus 9 plus 5 plus 7 equals 64. Then divide the sum by the number of values: 64 divided by 8 equals 8. The mean is 8.
Next, find the distance of each value from the mean :
|8-8|=0 |6-8|=2 |9-8|=1 |6-8|=2 |14-8|=6 |9-8|=1 |5-8|=3 |7-8|=1
0 plus 2 plus 1 plus 2 plus 6 plus 1 plus 3 plus 1 equals 16. Then divide the sum of absolute deviations by the number of data values: 16 divided by 8 equals 2. The MAD of the data is 2.
C. Since the MAD for the vegetable plant data is so much higher than the MAD for the flower data, this indicates that the heights of the vegetable plants vary more than the heights of the flowers ALSO do you go to k12
Answer:
who think
Step-by-step explanation:
Answer:
1. Multiply like the - isn't there
2.add it
3. Idk I just out something XD
Step-by-step explanation:
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
It decreases
The further two objects are, the gravitational attraction decreases and the closer two objects are, the gravitational attraction increases
