Simplify: $(\dfrac{1}{4})^{-\frac{1}{2}}·\dfrac{(\sqrt{4ab^{-1}})^3}{(0.1)^{-2}(a^{3}b^{-4})^{\frac{1}{2}}}$ .

And find the value when $a=2019,b=625$ .

The answer is 4.

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