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3 years ago

What are the roots of the polynomial equation x^3-7x=6x-12?

1 answer:

4 0

Hello,

Answer B

P(x)=x^3-13x+12
P(1)=1-13+12=0 ==>1 is root
P(3)=3^3-13*3+12=0 ==>3 is root
P(-4)=(-4)^3-13*(-4)+12=0 is root

P(x)=(x-1)(x-3)(x+4)

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Answer:

$\frac{5}{13}$

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3 years ago

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Answer:

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(b) The probability that exactly 2 are defective given that at least 2 purchased PS4s are defective is 0.3846.

Step-by-step explanation:

Let X = number of defective PS4s.

It is provided that 4 PS4s of 8 are defective.

The probability of selecting a defective PS4 is:

$P(X)=p=\frac{4}{8}=0.50$

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The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.50.

The probability function of a Binomial distribution is:

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

(a)

Compute the probability of exactly 2 defective PS4s among them as follows:

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Thus, the probability of exactly 2 defective PS4s among them is 0.3125.

(b)

Compute the probability that exactly 2 are defective given that at least 2 purchased PS4s are defective as follows:

$P(X=2|X\geq 2)=\frac{P(X=2\cap X\geq2)}{P(X\geq2)} =\frac{P(X=2)}{P(X\geq 2)}$

The value of P (X = 2) is 0.3125.

The value of P (X ≥ 2) is:

$P(X\geq 2)=1-P(X$

Then the value of P (X = 2 | X ≥ 2) is:

$P(X=2|X\geq 2)=\frac{P(X=2)}{P(X\geq 2)}=\frac{0.3125}{0.8125} =0.3846$

Thus, the probability that exactly 2 are defective given that at least 2 purchased PS4s are defective is 0.3846.

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4 years ago

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2 years ago

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