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2 years ago

In AJKL, the measure of ZL=90°, JL = 55, LK = 48, and KJ = 73. What

1 answer:

3 0

Together I think it’s the same way that you can get the one to the one you have on your computer problems you have no proof that I have it on my computer problems I have lost all the same information I I don’t have any proof that you are in my account so so much you have to pay for it and I’m sorry you don’t know where I can get you I know I

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The problem of matching aircraft to passenger demand on each flight leg is called the flight assignment problem in the airline i

Darina [25.2K]

Answer:

Step-by-step explanation:

Given that the demand for the 6 p.m. flight from Toledo Express Airport to Chicago's O'Hare Airport on Cheapfare Airlines is normally distributed with a mean of 132 passengers and a standard deviation of 42

Let X be the no of passengers who report

X is N(132, 42)

Or Z is $\frac{x-132}{42}$

a) Suppose a Boeing 757 with a capacity of 183 passengers is assigned to this flight.

the probability that the demand will exceed the capacity of this airplane

= $P(X>183) = P(Z>1.21) =$

$=0.5-0.3869\\=0.1131$

b) the probability that the demand for this flight will be at least 80 passengers but no more than 200 passengers

= $P(80\leq x\leq 200)\\= P(-1.23\leq z\leq 1.62)\\$

=0.4474+0.3907

=0.8381

c) the probability that the demand for this flight will be less than 100 passengers

$=P(x$

d) If Cheapfare Airlines wants to limit the probability that this flight is overbooked to 3%, how much capacity should the airplane that is used for this flight have? passengers

= $P(Z>c)=0.03\\c=1.88\\X=132+1.88(42)\\=210.96$

e) 79th percentile of this distribution

= $132+0.81(42)\\= 166.02$

7 0

3 years ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$

$P(at least 1 dull blade)=1-0.2931$

P(at least 1 dull blade)=0.7068

5 0

3 years ago

A square is cut from a rectangle. The side length of the square is half of the unknown width w. The area of the shaded region is

Virty [35]

Answer:

See below

Step-by-step explanation:

The side of the square is w/2, then its area is:

A = (w/2)² = w²/4

The shaded area is the difference of areas of the rectangle and the square:

14w - w²/4 = 84

This can be simplified:

56w - w² = 336
w² - 56w + 336 = 0

8 0

2 years ago

Robin kept a log of the number of hours she spent birdwatching during September and the cumulative number of birds she observed.

BlackZzzverrR [31]

Answer:

I dont understand there is no equation to solve or any math at all

4 0

3 years ago

40° (3x + 10)º If anyone can help me on this it will be helpful

Igoryamba

Answer:

3x+10 = 2x+40

3x - 2x = +40 - 10

x = 30

Step-by-step explanation:

hope helps have a good day

7 0

3 years ago