Q&A about the site for students, teachers, and linguists wanting to discuss the finer points of the Italian language
-- Enter Query Title SELECT TagName AS Tags, AVG(Reputation) AS [Average Answerer's Reputation], COUNT(*) AS [Number of Answerers] FROM ( SELECT DISTINCT pa.OwnerUserId, pt.TagId FROM Users u JOIN Posts pa ON pa.OwnerUserId = U.Id JOIN Posts pq ON pq.Id = pa.ParentId JOIN PostTags pt ON pt.PostId = pq.Id ) ut JOIN Tags t ON t.Id = ut.TagId JOIN Users u ON u.Id = ut.OwnerUserId -- WHERE u.Reputation GROUP BY t.Id, t.TagName HAVING COUNT(*) > 1000 ORDER BY 2 DESC -- Enter Query Description