An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he
1 answer:
Answer:
2330.51 miles
Step-by-step explanation:
Given that the speed of the airplane = 610 mph.
The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.
So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.
As, distance = time x speed, so
The distance covered at 270 degrees, miles.
The distance covered at 310 degrees, miles.
Total distance covered, d, is the magnitude of the sum of vectors and
as shown in the figure.
The angle between the vectors and
,
degree.
Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},
miles
Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.
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Answer:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
And we have :
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).
Step-by-step explanation:
For this case we know that we have a sample of n = 500 students and we have a percentage of expected return for their sophomore years given 66% and on fraction would be 0.66 and we are interested on the distribution for the population proportion p.
We want to know if we can apply the normal approximation, so we need to check 3 conditions:
1) Randomization: We assume that we have a random sample of students
2) 10% condition, for this case we assume that the sample size is lower than 10% of the real population size
3) np = 500*0.66= 330 >10
n(1-p) = 500*(1-0.66) =170>10
So then we can use the normal approximation for the distribution of p, since the conditions are satisfied
The population proportion have the following distribution :
And we have :
And we can use the empirical rule to describe the distribution of percentages.
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
Using the 68-95-99.7% rule we expect 68% of the values between 0.639 (63.9%) and 0.681 (68.1%), 95% of the values between 0.618(61.8%) and 0.702(70.2%) and 99.7% of the values between 0.596(59.6%) and 0.724(72.4%).