Advanced Math questions and answers Let P (x) = x4 x 1 Let kı = 5 mod 2 and k2 = 6 mod 2 Let A (x) = x3 k1x2 1 and B (x) = x2 k1x 1 Compute Clx) mod P (x) in GF (24) for (a) (b) (d) C (x) = A (x) B (x), C (x) = A (x)B (x), C (x) = () C (x) = B2 (x) Question Let P (x) = x4 x 1 Let kı = 5 mod 2 and k2 = 6 mod 2P*(x)(x*(x^21))=0 Step by step solution Step 1 Trying to factor as a Difference of Squares 11 Factoring x 21 Theory A difference of two perfect squares, A 2 B 2 can be factored into (AB) • (AB) Proof (AB) • (AB) = A 2 AB BA B 2 = A 2 AB AB B 2 = A 2 B 2 Note AB = BA is the commutative property of multiplicationBASIC STATISTICS 5 VarX= σ2 X = EX 2 − (EX)2 = EX2 − µ2 X (22) ⇒ EX2 = σ2 X − µ 2 X 24 Unbiased Statistics We say that a statistic T(X)is an unbiased statistic for the parameter θ of theunderlying probabilitydistributionifET(X)=θGiventhisdefinition,X¯ isanunbiasedstatistic for µ,and S2 is an unbiased statisticfor σ2 in a random sample 3
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Click here👆to get an answer to your question ️ If alpha, beta are roots of x^2 p ( x 1 ) c = 0 Show thati) ( alpha 1 ) ( beta 1 ) = 1 c &ii) alpha^22alpha1alpha^22alphac beta^22beta1beta^22betac = 1" # = 0 if trial i is a failure 1ftrialiisasuce X i The X i's are Bernoulli random variables with parameter p E(X i)=1*p0*(1!p)=p E(X)=E(X 1)E(X 2)KE(X n)=np Ex A group of N people throw their hats into the center of a room The hats are mixed, and= x(x1)(x2)1, and 0!
Prove that { 1 , 1 x , (1 x)^2 } is a basis for the vector space of polynomials of degree 2 or less Then express f(x) = 2 3x x^2 as a linear combinationThe probability of having x successes in n trials is (where x!The answer should be 22 As P (Q (x)) = P (x)R (x) P (x) = (x1) (x2) (x3) (Q (x) 1) (Q (x) 2) (Q (x) 3) = (x1) (x2) (x3)R (x) And R (x) is a 3 degree polynomial So, Q (x) should be a 2 degree polynomial We can write R (x) as (xa) (xb) (xb) because it is a 3 degree polynomial
Finally, using distributive property and foil, you would get x^2xx^23x2=(x^2x2)2x^22x2=x^2x23x^23x=0 3x(x1)x=0 and x=1 ( thats only if the big X is the same as the small xs) You are right, thank you SO MUCH!!12 p2 2px x2 is a perfect square It factors into (px)• (px) which is another way of writing (px)2 How to recognize a perfect square trinomial • It has three terms • Two of its terms are perfect squares themselves • The remaining term is twice the product of the square roots of the other two termsCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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Explanation According to Viete's Theorem if P (x) has a factor of (x −a) then a is a root of this polynomial, so in this case this polynomial has 2 roots x3 = − 1 and x4 = 2 To find the other roots you have to divide P (x) by (x − 2)(x 1) The result will be x2 5x 6 Then you can calculate the remaining rootsX (x1) (x2) (x3)=1 Simple and best practice solution for x (x1) (x2) (x3)=1 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let The Attempt at a Solution P (X 1 < X 2 < X 3) = triple integral of e (x1x2x3) dx 2 dx 1 dx 3 as x 2 goes from x 1 to x 3, x 1 goes from 0 to x 2 and x 3 goes from x 2 to infinity When I solve this integral I get 0 for an answer The back of the book suggests the answer is 1/6
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Expand (x1)^2 Rewrite as Expand using the FOIL Method Tap for more steps Apply the distributive property Apply the distributive property Apply the distributive property Simplify and combine like terms Tap for more steps Simplify each term Tap for more steps Multiply by Move to the left of Rewrite asGraph P(x)=(x1)(x1)(x2) Find the point at Tap for more steps Replace the variable with in the expression Simplify the result Tap for more steps Simplify each term Tap for more steps Raising to any positive power yields Raising to any positive power yields Multiply by Multiply byNew questions in Mathematics
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8 Comments 5 Solutions 2,391 Views Last Modified The solution to a problem is P (x) = (x1) (x2) (xn) (for some integer n) However, I want to expand this polynomial, so as to find the coefficient of the ith power It's quite clear that a_n=1, and a_0=n! p (x)=x²2√2x1 p (2√2)= (2√2)² (2√2*2√2)1 p (2√2)= (2√2)² (2√2)²1 =01 =1 Therefore, the answer is 1 Thank you Hope this will help you mitgliedd1 and 1355 more users found this answer helpful If x^2 x 1 is a factor of the polynomial 3x^2 8x^2 8x 3 5k, then the value of k is A 0 B 2/5 asked Apr 21 in Polynomials by Cammy ( 273k points) factorization of polynomials
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1x2at Sportwetten und Casino Today`s offer Complete offer Back Sport Sport if x= 1 p (1)= (1)21 p (1)=1 if x= 1 p (1)= (1)21 p (1)= 21 p (1)= 3 Pls marks this as the Brainliest answer!!!P(x n) the variance of X, denoted by σ² x is defined as σ² x = (x 1 μ x) 2 • p(x 1) (x 2 μ x) 2 • p(x 2) (x n μ x) 2 • p(xn) = E (x k μ x) 2 • p(x k) The positive square root of the variance is called the standard deviation of the random variable X and is denoted by σ xThe standard deviation makes use of the
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Number of successes in n trials We can write X as follows X=X 1 X 2 KX n where !(1)^n, but I've yet to figure a pattern for the othersX = x 1, x = x 2, , x = x k, where x 1, x 2, , x k are roots of Q(x) To find the vertical asymptotes of f(x) be sure that it is in lowest terms by canceling any common factors, and then find the roots of Q(x) 3 Oblique Asymptotes The rational function f(x) = P(x
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Then, f(x)g(x) = 4x 2 4x 1 = 1 Thus deg( f ⋅ g ) = 0 which is not greater than the degrees of f and g (which each had degree 1) Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f ( x ) = 0 to also be undefined so that it follows the rules of a norm in a Euclidean domainExpanding brackets Binomials The distributive property of multiplication over addition The distributive property of multiplication over subtractionDistri so to attain P (x) = 0, we go for random negative numbers, let us try x = 1 we get P (1) = 2, let us try x = 1/2, here P (1/2) = 0 is found, so this is our first value this way our first x = 1/2 is easily spotted then we need to find other 3 remaining values, we divide the given polynomial by (x05), the equation we get is 2x^3 14 x^2 24x 8,
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The probabilitythatX= 1, P(X=1)= P(1, 1)=1/36 The probability thatX =2, P(X=2)= P(1, 2), (2,1), (2, 2) = 3/36 Continuing we obtain P(X =1) = 1 36,P(X =2) = 3 36,P(X =3) = 5 36 P(X =4) = 7 36,P(X =5) = 9 36,P(X =6) = 11 36 We can then write the probability massfunction as pX(x)=P(X = x)= 2x − 1 36 forx=1, 2, 3, 4, 5, 6Tramwayniceix and 4 more users found this answer helpful Click here 👆 to get an answer to your question ️ Find p(x 1) if p(x) = x^2 – 3x – 1 everkayy everkayy Mathematics High School answered Find p(x 1) if p(x) = x^2 – 3x – 1 1 See answer everkayy is waiting for your help Add your answer and earn points
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77k views asked in Class IX Maths by ashu Premium (930 points) If p (x) = x2 – 2√2x 1, then p (2√2) is equal to (a) 0 (b) 1 (c) 4√2 (d) 8 √2 1 polynomialsIf p (x) = x^2 2 √ (2)x 1 , then p (2√ (2)) is equal to mathsDivide f2, the coefficient of the x term, by 2 to get \frac{f}{2}1 Then add the square of \frac{f}{2}1 to both sides of the equation This step makes the left hand side of the equation a perfect square
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Ex 42, 12 By using properties of determinants, show that 8(1&x&x2@x2&1&x@x&x2&1) = (1 – x3)2 Solving LHS 8(1&x&x2@x2&1&x@x&x2&1) Applying R1 → R1 R2 R3 = 8(𝟏𝐱𝟐𝐱&𝐱𝟏𝐱𝟐&𝐱𝟐𝐱𝟏@x2&1&x@x&x2&1) Taking (1 x x2) Common from 1st row = (𝟏𝐱𝐱𝟐) 8(1&1&1@x2&1&x@x&x2 Hence we can deduce that #x=1# is a zero of #P(x)# and #(x1)# a factor #x^36x^211x6 = (x1)(x^25x6)# Then to factor the remaining quadratic note that #5 = 23# and #6 = 2*3#, and hence #x^25x6 = (x2)(x3)# Either way we arrive at our result #x^36x^211x6 = (x1)(x2)(x3)#(1−x)2 X∞ k=0 (k1)2xk = S 2 = 1x (1−x)3 2 Geometric Distributions Suppose that we conduct a sequence of Bernoulli (p)trials, that is each trial has a success probability of 0 < p < 1 and a failure probability of 1−p The geometric distribution is given by
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Now, when we see the 2nd half of the question, where you have written "Find the Remainder when P(x) is divided by (x1)(x2)" I think that this is an incorrect statement as if you see carefully you might notice that here divident or P(x) ie 3x 8 (which we calculated above) is smaller than (x1)(x2) and such a division will not give aCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge andIf α is the positive root of the equation, p(x) = x 2 – x – 2 = 0, then lim x→α (√1cos(p(x)))/(x α 4) is equal to (1) 3/√2 (2) 3/2 (3) 1/√2 (4) 1/2
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Show 1 and 1 are only solutions to x^2 = 1 (mod p^2) for odd prime p Let p be an odd prime and n be an integer Show that 1 and 1 and the only solutions to x 2 ≡ 1 m o d p n Hint What does a ≡ b m o d m mean, then think a bit for k an integer= 1) E(X) = np = 3* 03 = 09 P(X=x)=!( )!! The underlying distribution makes no difference Therefore, the answer is 1 / 6 Had these not been IID then you'd need to do more work and the answer would depend on how they are distributed, but in this case, the variables are exchangeable Share answered Oct 30 ' at
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Solution Steps P ( x ) = ( x 4 ) ( x 2 ) ( x 1 ) P ( x) = ( x 4) ( x 2) ( x − 1) Apply the distributive property by multiplying each term of x4 by each term of x2 Apply the distributive property by multiplying each term of x 4 by each term of x 2Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!The points (x,y,z) of the sphere x 2 y 2 z 2 = 1, satisfying the condition x = 05, are a circle y 2 z 2 = 075 of radius on the plane x = 05 The inequality y ≤ 075 holds on an arc The length of the arc is 5/6 of the length of the circle, which is why the conditional probability is equal to 5/6
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X n x n − px (1p) nx VAR(X) = np(1p) = 3* 03 * 07 = 063 SD(X) = np(1p) Calculations shown for Binomial (n=3, p=03) = 0794 Note this is equivalent to counting success = 1 and failure = 0
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