How to pass arguments to a script?

To pass arguments to a Bash script, you can include them after the script name when you run it. Here's how to do it:

Create a script (e.g., script.sh):

#!/bin/bash
echo "First argument: $1"
echo "Second argument: $2"

Make the script executable:
```
chmod +x script.sh
```
Run the script with arguments:
```
./script.sh "Hello" "World"
```

In this example:

$1 refers to the first argument ("Hello").
$2 refers to the second argument ("World").

You can pass as many arguments as you need, and they can be accessed using $3, $4, etc. The special variable $@ can be used to refer to all arguments.

If you have any further questions or need clarification, feel free to ask!