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! 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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P(x)=x^3-2x^2-8x-1 g(x)=x+1- 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 ' atP(X =0) = 1 16,P(X =1) = 4 16,P(X =2) = 6 16,P(X =3) = 4 16,P(X =4) = 1 16 (1) Notice that the denominators of the five fractions are the same and the numerators of the five fractions are 1, 4, 6, 4, 1 The numbers in the numeratorsis a setof binomial coefficients 1 16 = 4 0 1 16, 4 16 = 4 1 1 16, 6 16 = 4 2 1 16, 4 16 = 4 3 1 16, 1 16 = 4
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyThe probability of having x successes in n trials is (where x!Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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 = 0Click here👆to get an answer to your question ️ If p(x) = x^2 2 √(2)x 1 , then p(2√(2)) is equal to Join / Login > 10th > Maths > Polynomials > Relationship between ZeroesNew questions in Mathematics
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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 helpfulFinally, 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!!Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
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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 polynomials 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 Explanation (x −1)(x 2) aaaaaaaaaaaaax − 1 aaaaaaaaaaa × x 2 −−−− − aaaaaaaaaaaaax2 − x aaaaaaaaaaaaaaaa2x − 2 aaaaaaaaaaaaaa¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2 x − 2
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(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 byDivide 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 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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If 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 For example, $(1x)^2 = 1 2x 1 x^2$, and so the coefficients $1, 2, 1$ translate into the column vector $ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ Because $\dim \mathbf{P}_2 = 3$, this set is a basis if and only if these three vectors are linearly independent
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