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We prove that for $c<1$, the expected number of clauses satisfiable is $\\cn-\\Theta(1/n)$; for large $c$, it is $((3/4)c + \\Theta(\\sqrt{c}))n$; for $c = 1+\\eps$, it is at least $(1+\\eps-O(\\eps^3))n$ and at most $(1+\\eps-\\Omega(\\eps^3/\\ln \\eps))n$; and in the ``scaling window'' $c= 1+\\Theta(n^{-1/3})$, it is $cn-\\Theta(1)$. 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