List all week 37 results when week 33 is 14.3%.
CREATE TABLE table_23680576_3 (week_37 VARCHAR, week_33 VARCHAR)
SELECT week_37 FROM table_23680576_3 WHERE week_33 = "14.3%"
<s>[INST] Write mssql query to answer the following question given the database schema. Schema: CREATE TABLE table_23680576_3 (week_37 VARCHAR, week_33 VARCHAR) Question: List all week 37 results when week 33 is 14.3%. [/INST] SELECT week_37 FROM table_23680576_3 WHERE week_33 = "14.3%" </s>