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arlik [135]
2 years ago
9

Work out 5/25+2/50+7/100

Mathematics
1 answer:
bixtya [17]2 years ago
8 0

Answer:

The answer is 31/100

Step-by-step explanation:

5/25 + 2/50 + 7/100 = 31/100

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Answer:

Weight of copper: 62,000 lbs. (31 tons). Weight of framework: 250,000 lbs. (125 tons). Weight of concrete foundation: 54,000,000 lbs. Thickness of copper sheeting: 3/32 of an inch, the thickness of two pennies placed together.  the answer in 125

Step-by-step explanation:

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186:403 in simpleast form<br><br>​
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6:13

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United Flight 15 from New York's JFK airport to San Francisco uses a Boeing 757-200 with 182 seats. Because some people with res
Tcecarenko [31]

Answer:

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

Step-by-step explanation:

For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

A probability is said to be unusually low if it is lower than 5%.

For this problem, we have that:

There are 200 reservations, so n = 200.

A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So \pi = 0.0995.

Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.

X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find P(X < 18), with \pi = 0.0995, n = 200. We can use a binomial probability calculator, and we find that:

P(X < 18) = 0.2927.

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

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3 years ago
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I think it’s d because that’s where the z would come from with the AEC right angle.
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