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GuDViN [60]
3 years ago
6

A designer wants to create a whisper chamber in the shape of an ellipse. He has a warehouse space with a longest length of 50 ya

rds, which he decides will be the major axis of his elliptical chamber. He determines the best spots for his guests to stand to experience his whisper chamber will be 10 yards from the center of the warehouse space, which will act as the foci. How far out from the center, along the minor axis should he build out his whisper chamber?
53.9 yd

45.8 yd

26.9 yd

22.9 yd
Mathematics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

22.9 yards

Step-by-step explanation:

Since b² = a² - c² where a = vertex of major axis, 2a = 50 yards the length of the major axis. So , a = 50/2 = 25 yards. c = focus of chamber = 10 yards from center and b = vertex of minor axis.

So, b = ±√(a² - c²)

= ±√(25² - 10²)

= ±√(625 - 100)

= ±√525

= ±22.91 yards

≅ ± 22.9 yards

Since b = length of minor axis from center of chamber = 22.91 yards. So, he should build the whisper chamber 22.9 yards out from the center of the chamber.

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Step-by-step explanation:

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The probability that the aircraft is overloaded is 97.98%, which means the pilot should take the action.

In a Normal distribution with mean ц and standard deviation σ, the z-score of a measure x is given by:

Z = X-ц / σ

· It measures how many standard deviations the measure is from the mean.

· After finding Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

· By the Central Limit Theorem, the sampling distribution of sample means of the size n has standard deviation σ

σ = σ / \sqrt{n}    

σ is standard deviation

n is the sample size.

Given that the mean and the standard deviation of the population is 176.1 lb and 35.4 respectively.

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For a sample of 43 passengers, we have

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The probability that the aircraft is loaded is

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The probability that the aircraft is overloaded is 97.98%

Know more about Normal probability Distribution: -brainly.com/question/9333901

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