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Rom4ik [11]
4 years ago
8

An airline has 83% of its flights depart on schedule. It has 68% of its flight depart and arrive on schedule. Find the probabili

ty that a flight that departs on schedule also arrives on schedule. Round the answer to two decimal places.
Mathematics
2 answers:
arsen [322]4 years ago
7 0
The probability that a flight that departs on schedule also arrives on schedule
83%×68%=56%
SVEN [57.7K]4 years ago
7 0

Answer:

Hence, the probability is:

0.82

Step-by-step explanation:

Let A denote the event that the flight departs on time.

Let B denote the event that the flight arrives on schedule.

Also, A∩B denotes the event that the flight that departs on time also arrives on schedule.

Let P denotes the probability of an event.

We are asked to find:

                              P(B|A)

We know that:

P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}

Hence, we have:

P(A)=0.83 x.

P(A∩B)=0.68 x.

where x are the total number of flights.

Hence,

P(B|A)=\dfrac{0.68x}{0.83x}\\\\P(B|A)=0.82

Hence,  the probability that a flight that departs on schedule also arrives on schedule is:

                                     0.82

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0.12, 19/50, 5/12. Convert the fractions into decimals by dividing the numerator by the denominator to find the answer to questions like this more easily.
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3 years ago
What is the next number in the sequence? 3, 6, 12, , ... place an x by the pattern of the sequence. add 2 subtract 2 times 2?
ch4aika [34]
3,6,12...

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8 0
3 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
When finding x-intercepts, y is equal to zero. So, we can set the whole function equal to_______.
finlep [7]

Answer:

ZERO

ZERO

Step-by-step explanation:

When finding x-intercepts, y is equal to zero. So, we can set the whole function equal to ZERO.

Then we set each binomial equal to ZERO.

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3 years ago
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suter [353]
The answer is 90%. Because 10% love math you subtract 100
4 0
3 years ago
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