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2 years ago

6

Past records indicate that 15 percent of the flights for a certain airline are delayed. Suppose flights are randomly selected on

e at a time from all flights. Assume each selection is independent of another. Which of the following is closest to the probability that it will take 5 selections to find one flight that is delayed?
a. 0.0783
b. 0.0921
c. 0.4780
d. 0.5220
e. 0.5563

1 answer:

Zarrin [17]2 years ago

3 0

Answer:

a. 0.0783

Step-by-step explanation:

For each flight, we have these following probabilities:

A 15% probability that a flight is delayed

An 85% probability that a flight is on time.

Which of the following is closest to the probability that it will take 5 selections to find one flight that is delayed?

This is the probability that the first four flights are on time and the fifth is late. So

$P = 0.85*0.85*0.85*0.85*0.15 = 0.0783$

So the correct answer is

a. 0.0783

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Step-by-step explanation:

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Since these are isosceles triangles, the base angles are the same:

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Therefore the congruency statement can be written two different ways.

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Choosing between B) and C):

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2 years ago

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Answer:

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2 years ago

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2 years ago

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Step-by-step explanation:

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In which $\sigma$ is the standard deviation of the population and n is the size of the sample. In this problem, $\sigma = 0.25$

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This is n when M = 0.1. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

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$M = z*\frac{\sigma}{\sqrt{n}}$

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

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This is n when M = 0.05. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

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Rounding up, 97

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2 years ago

What is the answer please

Marina86 [1]

Hey there!

Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.

I hope this helps!

8 0

2 years ago