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astra-53 [7]
3 years ago
11

Please help me with this

Mathematics
1 answer:
sesenic [268]3 years ago
5 0
Ok so the average is 50

average of x number of things is
(add value of all things)/x=average

4 test scores needed
so divide by 4
3 test scores ar known
average is 50
x= missing score

(43+48+42+x)/4=50
times 4 both sides
133+x=200
minus 133 both sides
x=67

a score of 67 is needed
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A person traveling from Seattle to Sydney has three airlines to choose from. 40% of travelers choose airline A, and this airline
dalvyx [7]

Answer:

46.67% probability that they flew with airline B.

Step-by-step explanation:

We have these following probabilities:

A 40% probability that a traveler chooses airline A.

A 35% probability that a traveled chooses airline B.

A 25% probability that a traveler chooses airline C.

If a passenger chooses airline A, a 10% probability that he arrives late.

If a passenger chooses airline B, a 15% probability that he arrives late.

If a passenger chooses airline C, a 8% probability that he arrives late.

If a randomly selected traveler is on a flight from Seattle which arrives late to Sydney, what is the probability that they flew with airline B?

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

What is the probability that the traveler flew with airline B, given that he was late?

P(B) is the probability that he flew with airline B.

So P(B) = 0.35

P(A/B) is the probability of being late when traveling with airline B. So P(A/B) = 0.15.

P(A) is the probability of being late. This is the sum of 10% of 40%(airline A), 15% of 35%(airline B) and 8% of 25%(airline C).

So

P(A) = 0.1*0.4 + 0.15*0.35 + 0.08*0.25 = 0.1125

Probability

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.35*0.15}{0.1125} = 0.4667

46.67% probability that they flew with airline B.

5 0
3 years ago
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