Answer:
Let the required number be x. \[\begin{align} & \therefore \frac{-4}{15}\times x=\frac{-8}{9} \\ & x=\frac{-8}{9}\div \frac{-4}{15} \\ \end{align}\] \[x=\frac{8\times 15}{9\times 4}\] \[=\frac{10}{3}\] \[\therefore \]Other number is \[\frac{10}{3}\].
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