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Bad White [126]
3 years ago
7

x = (a+b)2, y = a2+2ab+b2 and z = a2+b2-2ab (a) determine the sum of the numerical co-efficients of z terms. (b) determine y+z a

nd y-z (c) If a = 3 and b = - 2, Prove that x = y​
Mathematics
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

The given equations are;

x = (a + b)²

y = a² + 2·a·b + b²

z = a² + b² - 2·a·b

(a) The numerical coefficients of z terms are

1, 1, -2

The sum of the numerical coefficients = 1 + 1 - 2 = 0

(b) y + z is found by substituting the values of 'y' and 'z', in the expression y + z, as follows;

y + z = a² + 2·a·b + b² + a² + b² - 2·a·b = 2·a² + 2·b²

y + z = 2·a² + 2·b²

y - z = a² + 2·a·b + b² - (a² + b² - 2·a·b) = 4·a·b

(c) Given that a = 3, and b = -2, we have;

x = (a + b)² = a² + a·b + a·b + b² = a² + 2·a·b + b² = y

Therefore, x = y, for all values of 'a', and 'b'

Step-by-step explanation:

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Let T1 be a poisson distribution with mean λ = 2, then

P(T1=0) = e^{-2}

P(T1=1) = 2 * e^{-2}

P(T1=2) = 2 * e^{-2}

P(T1=3) = \frac{4}{3} \, e^{-2}

P(T1=4) = \frac{2}{3}\, e^{-2}

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Lets do the same with a Poisson distribution T2 with mean λ = 3

P(T2=0) = e^{-3}

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P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2}  + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2}  + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2}  + 0.7*\frac{9}{2} e^{-3}) = 0.0471

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