Question: create table movies ( movie_id integer, name varchar(1000), score integer ); create table cast ( movie_id integer, cast_id integer, cast_name varchar(1000) ); Using the tables

create table movies ( movie_id integer, name varchar(1000), score integer );

create table cast ( movie_id integer, cast_id integer, cast_name varchar(1000) );

Using the tables created above, find the movies with score > 80 and has no cast members with name David. Output format: movie_id, score.

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