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Is there a command to start a program after a delay?

Something like shutdown 0 for firefox where firefox launch after a delay.

I was following this tutorial where it says adding sleep 150 adds a delay of 150 seconds before it starts the program. But it doesn't seem to be working. The program never launches on startup after I made this edit.

Sandeep C
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    What exactly do you want to achieve? I read it as _"start firefox 150 seconds after system startup/boot/login"_ but according to your question's title `sleep 150; firefox` in a command line would do. – PerlDuck Mar 29 '18 at 13:25
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    Do you want this at boot/reboot of the computer, or delayed after you run a command (manually)? Or some other way? Please explain. – sudodus Mar 29 '18 at 13:32
  • @sudodus I think I figured out: according to the tutorial he wants a new entry in _Startup applications_ that starts firefox with a delay of 150 seconds. – PerlDuck Mar 29 '18 at 13:33
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    @PerlDuck, Yes, this is what I think too, but let us wait for an answer from Sandeep C. - Firefox and other GUI programs should be started that way. Programs that need not communicate with the desktop (that read from files and write to files) can be started via `crontab` '@reboot'. – sudodus Mar 29 '18 at 13:37
  • Yes, @PerlDuck identified my requirement correctly. For example, I edited the command to `sleep 120; dropbox start -i` after reading the tutorial. But, Dropbox never launches on startup now. So what is the correct method of implementing this? Does `bash -c "sleep 10m && dropbox start -i"` work? – Sandeep C Mar 30 '18 at 09:56

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