What does this $(($RANDOM \% 10800)) cron entry mean?

Question

bash

There is a cron job that runs on one machine (centos) but fails on another (debian).
What does this entry mean

sleep $(($RANDOM \% 10800))

.
Pause before executing for a random amount of time , then escape and % 10800 ?

Answer 1

sleep $(($RANDOM % 10800))means an arbitrary pause within 3 hours (10800 seconds).
but the slash before the percentage is most likely a mistake.