Answer:
Approximately 24–36 pounds
Step-by-step explanation:
Given:
Weight of standard American Eskimo dog has mean= 30 pounds
A standard deviation= 2 pounds
For 99.7% of the dogs according to normal distribution would lie between 3 standard deviations on either side of the mean.
i.e. lower bound= Mean- 3 (standard deviation)
= 30-3(2)
=30-6
=24
Upper bound = Mean +3 (standard deviation)
= 30+3(2)
=30+6
=36
Hence the range of weights : 24-36 pounds
correct answer Approximately 24–36 pounds!
Answer:
24 centimeters
Step-by-step explanation:
You can just count the squares around the rectangle since each of those squares is a centimeter. Hence the name centimeter grid.
<h3>
Answer: (C) 8/30</h3>
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Explanation:
There are 5+3 = 8 people in choir and running club. I added the values in the overlapped region of the circles "choir" and "running club".
This is out of 2+5+6+7+3+4+3 = 30 people total.
Divide the two results to form the fraction 8/30
Ideally we should reduce to get 4/15, but your teacher has chosen not to reduce.
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
