The output for the first print statement `print(1,2,3,4,sep="*")` will be `1*2*3*4`. This is because the `sep="*"` argument changes the default separator between values from a space to an asterisk.

The output for the second print statement `print(1,2,3,4,sep="#",end="@")` will be `1#2#3#4@`. This is because the `sep="#"` argument changes the default separator between values from a space to a hash symbol, and the `end="@"` argument changes the default ending of the print statement from a newline to an at symbol.

Question

The output for the first print statement `print(1,2,3,4,sep="*")` will be `1*2*3*4`. This is because the `sep="*"` argument changes the default separator between values from a space to an asterisk.

The output for the second print statement `print(1,2,3,4,sep="#",end="@")` will be `1#2#3#4@`. This is because the `sep="#"` argument changes the default separator between values from a space to a hash symbol, and the `end="@"` argument changes the default ending of the print statement from a newline to an at symbol.

Knowee AI · Accepted Answer

The output for the first print statement `print(1,2,3,4,sep="*")` will be `1*2*3*4`. This is because the `sep="*"` argument changes the default separator between values from a space to an asterisk.

The output for the second print statement `print(1,2,3,4,sep="#",end="@")` will be `1#2#3#4@`. This is because the `sep="#"` argument changes the default separator between values from a space to a hash symbol, and the `end="@"` argument changes the default ending of the print statement from a newline to an at symbol.

What will be the output for the following print statements?print(1,2,3,4,sep="*")print(1,2,3,4,sep="#",end="@")

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What will be the output for the following print statements?

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