$m^{2}$ =4(2m+5)
-2(14-12x)=-4(10-6x)
$\frac{11y-1}{5}$ = $y^{2}$ +1
$\sqrt{\x}$ = -2
∫ (3 - t)^(3/2) - (3 + t)^(3/2) + 1 dt
Where the limits of integration are: 3 < t < -3
Given the stationary AR(2) process:
X_t=5/6 X_(t-1)-1/6 X_(t-2)+e_t
Find ρ0 , ρ1 and ρ2. ...
2(k-5)+3k=k+6
9xu+6xv-3xw+6uy+4vy-2wy
8.3+10.35/0.007
$y^{2}$ =10y-16
