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Lesechka [4]
3 years ago
7

On a certain airline, customers are assigned a row number when they purchase their ticket, but the four seats within the row are

first come, first served during boarding. If Karen and Georgia end up with random seats in the same row on a sold-out flight, what is the probability that they sit next to each other?
Mathematics
1 answer:
maksim [4K]3 years ago
6 0

Answer:

The probability that they sit next to each other is 50%.

Step-by-step explanation:

Consider the provided information.

It is given that there are four seats within the row are first come, first served during boarding.

There are 4 seats and 2 customers (Karen and Georgia)

The total number of ways in which Karen and Georgia can sit is: ^4C_2

Now if they will sit together, then consider  Karen and Georgia as a single unit.

Thus, the number of ways in which they can sit together is: ^3C_1

The required probability is:

P=\frac{^3C_1}{^4C_2} \\\\P=\frac{3}{6}\\\\P=\frac{1}{2}

Hence, the probability that they sit next to each other is 50%.

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Answer:

\hat p=0.645\\\\ME=0.233\\\\x=645 \ individuals

Step-by-step explanation:

-Given the boundaries as 0.412 and 0.878

-\hat p is the point estimate for the population proportion and is calculated as follows:

\hat p=\frac{Upper \ Bound+Lower \ Bound}{2}\\\\=\frac{0.878+0.412}{2}\\\\=0.645\\\\

#The margin of error, ME can be calculated for the confidence intervals using the formula:

ME=\frac{Upper \ Bound-Lower \ Bound}{2}\\\\=\frac{0.878-0.412}{2}\\\\=0.233

#The number of individuals in the sample is the product of the point estimate and population size:

\hat p=\frac{x}{n}\\\\x=\hat pn\\\\=0.645\times 1000\\\\=645

Hence, there are 645 individuals in the sample.

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Answer: 170

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