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Ipatiy [6.2K]
4 years ago
10

Factor the expression x^4 -1 using the polynomial identify x^2-y^2)=(x-y)(x+y)

Mathematics
1 answer:
Orlov [11]4 years ago
3 0

Answer:

(x - 1)(x + 1)(x² + 1)

Step-by-step explanation:

Step 1: Factor Difference of Squares

(x² - 1)(x² + 1)

Step 2: Factor Difference of Squares for non-complex binomial root

(x - 1)(x + 1)(x² + 1)

And we have our answer!

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Which is expression is equivalent to 4b
GalinKa [24]

Answer: b(4) or 4(b)

Step-by-step explanation:

5 0
4 years ago
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Exercise 5.2. Suppose that X has moment generating function
soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Step-by-step explanation:

Given:

moment generating function  of X as:

MX(t) = \frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}

a)  Now

Mean, E(X) = M_{X}'(t=0)

Thus,

M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}

or

M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}

also,

E(X^{2})=M_{X}''(t=0)

Thus,

M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}

or

M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}

Therefore,

Mean, E(X) = M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}

or

Mean, E(X) = - 0.5

and

E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}

or

E(X^{2}) = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

M_{X}(t)=E(e^{tX})

or

⇒ =\sum e^{tx}p(x)

⇒ =\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}

Therefore,

at x = 0, P(x) = \frac{1}{2}

at x= - 4 ,P(x) = \frac{1}{3}

at x = 5, P(x) = \frac{1}{6}

Thus,

E(X) =\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})

or

E(X) = - 0.5

also,E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})

E(X^{2})  = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

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HELPP ASAP
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Answer:

(4, -5)

(3, -2)

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The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

                                       = 100 × 0.03

                                       = 3

Therefore the we find the probability that there are 3 or less errors on the page as

     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

   Using the formula for Poisson distribution for P(x = X ) = \frac{e^{-\lambda}\lambda^X}{X!}

Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

 The probability that there are 3 or less errors in 100 pages is 0.648.                                

7 0
3 years ago
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An oblique prism has a square base and height of 3 centimeters.
ExtremeBDS [4]
I could be very much wrong but maybe 45 cm3
3 0
3 years ago
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