What percentage of users were using Firefox according to the source that reported 21.70% used Internet Explorer?
CREATE TABLE table_name_22 (firefox VARCHAR, internet_explorer VARCHAR)
SELECT firefox FROM table_name_22 WHERE internet_explorer = "21.70%"
### Context: CREATE TABLE table_name_22 (firefox VARCHAR, internet_explorer VARCHAR) ### Question: What percentage of users were using Firefox according to the source that reported 21.70% used Internet Explorer? ### Answer: SELECT firefox FROM table_name_22 WHERE internet_explorer = "21.70%"