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For the following difference equation: y(t) = 0.15 + 0.57 y(t-1) + epsilon(t), where y(t) represents the value of y at time t, and epsilon

For the following difference equation: y(t) = 0.15 + 0.57 y(t-1) + epsilon(t), where y(t) represents the value of y at time t, and epsilon (t) represents the error term at time t, the solution is y(t) = k + c, where k is:

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