I need help please im just sick and my head hurts. please show work if you help. ignore the a

Mathematics

1 answer:

KIM [24]2 years ago

6 0

ans: 48

exp: <aed= 89

<aec= 29+12 --> 41

<ced= aed- aec= 89-41

--> 48

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Pls ,pls,plssss help my

Mumz [18]

plsss go ahead and tell your problem

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3 years ago

A small airline overbooks flights on the assumption that several passengers will not show up. Suppose that the probability that

olchik [2.2K]

We have 22 tickets sold, and 20 seats. This means that at least 2 passengers must not show up (otherwise, at least 21 passengers will be present, and there wouldn't be space for them).

Considering each passenger as independent, you can think of this experiment. Suppose you toss a coin for each passenger. If the coin lands on heads, the passenger shows up. If it lands on tails, the passenger doesn't show up.

But the coin is unfair: it has a 0.91 probability of landing on heads, and thus 0.09 probability of landing on tails.

This implies that the probability of having exactly $k$ tails is

$\binom{22}{k} \cdot 0.09^k \cdot 0.91^{22-k}$

We already concluded that at least two passengers must not show up. So, if our coins lands on tails less than twice, we've lost. So, the losing probability is

$\displaystyle\binom{22}{0}\cdot 0.09^0 \cdot 0.91^{22} + \binom{22}{1}\cdot 0.09^1 \cdot 0.91^{21} \approx 0.39$

Finally, remember the rule to negate events:

$P(E) = 1-P(\lnot E)$

So, if we lose with probability 0.39, we win with probability

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3 years ago

What is 19/4 as a mixed number?

Bumek [7]

We have to divide using long division.
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4| 19
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The divisor becomes the denominator, the remainder becomes the numerator, and the quotient becomes the whole number.

So we have:

$\sf4\dfrac{3}{4}$