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Anit [1.1K]
4 years ago
7

Divide the following polynomials, then place the answer in the proper location on the grid. Write answer in descending powers of

y. (2y3 + 3y2 - 4y - 5) (y + 1)
Mathematics
1 answer:
Varvara68 [4.7K]4 years ago
7 0
Let's rewrite the given expression correctly:
 (2y3 + 3y2 - 4y - 5) / (y + 1)
 Dividing the expression between y + 1 we have:
 2y2 + y - 5
 We can check the result:
 (2y2 + y - 5) * (y + 1)
 = (2y3 + y2 - 5y + 2y2 + y - 5)
 = (2y3 + 3y2 - 4y - 5)
 Division OK.
 Answer:
 
The result of the division is:
 
2y2 + y - 5
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Read 2 more answers
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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