what does this code mean

I have the following code running as a crontab on a jboss server, could anyone explain each line, thanks.

#!/bin/bash

for i in {1..20}
do
        #curl $(curl -k -i -m 10 https://xxx.xxx.com/jboss-status -k -i 2> /dev/null | grep "Location:" |sed 's/Location: //') > /var/www/html/f5-status 2> /dev/null
        curl -kLm 10 https://localhost/jboss-status > /var/www/html/f5-status 2> /dev/null
        sleep 15
done

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Jason YuAsked:
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Bill PrewCommented:
Basically the script loops 20 times, and inside the loop retrieves the contents of the web page "https://localhost/jboss-status" via the curl utility.  The result is output to the file "/var/www/html/f5-status", and errors are disposed of.  A ten second timeout is used on the curl command, an insecure connection is allowed, and if the page has move and redirects to the new location that will be followed.  A 15 second delay is added to each pass of the loop before getting the page content again.

The # indicates a comment in bash, so the first curl line has been commented out.

~bp
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Gerwin Jansen, EE MVETopic Advisor Commented:
The first line means that the script is being executed by the /bin/bash shell. Minor note on the output, it is overwritten each time because of the single > If you would retain output of each check add another Iike this :

... >> /var/www ...
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