CREATE TABLE table_name_12 (firefox_ VARCHAR, _other_mozilla VARCHAR, chrome VARCHAR)

What percentage of browsers were using Firefox or other Mozilla browsers during the period in which 1.93% were using Chrome?
SELECT firefox_, _other_mozilla FROM table_name_12 WHERE chrome = "1.93%"
SELECT "firefox_", "_other_mozilla" FROM "table_name_12" WHERE "1.93%" = "chrome"