A set of data following a normal distribution has a mean of 192.3 and a standard deviation of 11.8. Find the probability that a
2 answers:
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First lets convert everything to feet, and feet/min
1/2 mile = 5280/2 = 2640 feet
90 mi/hr = (90)(5280)/(60) = 7920 ft/min
Next lets find total distance plane is away from runway.
Set up right triangle where total distance is hypotenuse.
Next we will use a little trig. Let theta be the angle of elevation from runway to the plane. The rate of descent is the vertical component of the planes speed. Find vertical component using sine.
Finally substitute ratio from right triangle for sin theta.
The rate of descent is about 598 feet per minute.
1/2 mile = 5280/2 = 2640 feet
90 mi/hr = (90)(5280)/(60) = 7920 ft/min
Next lets find total distance plane is away from runway.
Set up right triangle where total distance is hypotenuse.
Next we will use a little trig. Let theta be the angle of elevation from runway to the plane. The rate of descent is the vertical component of the planes speed. Find vertical component using sine.
Finally substitute ratio from right triangle for sin theta.
The rate of descent is about 598 feet per minute.