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P(J>2) = P(J=0)+P(J=1)     [split it up]

P(X=t)= (V^t)/t!*e^V       where V=4 in this case  [use the formula]

P(J>2) = 4^0/0!*e^4 + 4^1/1!*e^4

          =1/e^4 + 4/e^4  =  5e^-4...

First solve for the exact point on the line by substituting 5 into the original equation. You should get y=+-4. 
Now implicitly differentiate the equation: 4x-6y(dy/dx)=0. Rearranging this will yiel...

We know in a quadratic x^2 +bx + c = 0, -b/a = α + β and c/a = αβ. 

Therefore, α + β = -(-2k) = 2k, and αβ = k - 1. (Both are divided by the coefficient in front of x which is 1 so can be ignored.<...

P- (3,0) y=ln(x/3)     u=x/3    y=ln(u) ​​​​​​            du = 1/3  dy = 1/u = 3            dx       du dy= du x dy dx dx  du   = 1/3 x 3 = 1 gradient at normal = -1 equ...

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