You need to 'echo' the value in $flavor.
<?php $flavor= "choco"; echo "<p>Your favorite flavor of ice cream is choco"; echo "vanilla"; echo ".</p>"; echo "<p>Hal's favorite flavor is cookie dough, also!</p>"; ?>
Jonathan GrieveTreehouse Moderator 87,698 Points
It looks like what you need to do is echo a variable value directly into a string.
Since you're using double quotes
"" for your string, you can do this by simply typing the variable name in those, to replace the hard coded names in there and you should be good to go! :-)