CREATE TABLE table_name_31 (internet_explorer VARCHAR, safari VARCHAR, chrome VARCHAR)

What internet explorer has 7.89% as the safari, and 8.22% as the chrome?
SELECT internet_explorer FROM table_name_31 WHERE safari = "7.89%" AND chrome = "8.22%"
SELECT "internet_explorer" FROM "table_name_31" WHERE "7.89%" = "safari" AND "8.22%" = "chrome"