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3 years ago

If α, β are the zeroes of the polynomials f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) =

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

4 0

Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)explanation:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(xf(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1) + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

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Answer:

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Step-by-step explanation:

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Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

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KengaRu [80]

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This is equivalent to

$4\cdot\text{cis(}\frac{\pi}{4})\cdot5\text{cis(}\frac{\pi}{6})=(4\times5)cis(\frac{\pi}{4}+\frac{\pi}{6})$

Since

$\frac{\pi}{4}+\frac{\pi}{6}=\frac{6\pi+4\pi}{24}=\frac{10}{24}\pi=\frac{5}{12}\pi$

we have that

$4\cdot\text{cis(}\frac{\pi}{4})\cdot5\text{cis(}\frac{\pi}{6})=20cis(\frac{5\pi}{12})$

Therefore, the answer is the second option from top to bottom.

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