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SVEN [57.7K]
3 years ago
15

If the radius of a circle is 3 meters, how long is the arc subtended by an angle measuring 30°?

Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
Circumference of entire circle = 2 * PI * radius
2 * PI * 3 = <span> <span> <span> 18.8</span></span></span>5 meters
a 30 degree arc is (30/360) or 1/12 of a circle.
So the arc equals (18.85 meters / 12) or <span> <span> <span> 1.57</span></span></span> meters.


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Answer:


Step-by-step explanation:

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The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
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Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

Z = \frac{X - \mu}{s}

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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