Table schema:
CREATE TABLE table_name_98 (average VARCHAR, evening_gown VARCHAR, preliminaries VARCHAR, interview VARCHAR)

Question: 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