CREATE TABLE table_name_98 (average VARCHAR(255), evening_gown VARCHAR(255), preliminaries VARCHAR(255), interview VARCHAR(255))
How many average scores have preliminary scores over 9.25, interview scores more than 9.44, and evening gown scores of 9.77?
    
SQL: SELECT COUNT(average) FROM table_name_98 WHERE preliminaries > 9.25 AND interview > 9.44 AND evening_gown = 9.77