Refer to Schema A. Which query correctly shows only the total quantity for order ID 5001? A ) SELECT SUM(T.qty) FROM tblORDER T WHERE T.orderID = 5001; B ) SELECT SUM(T.qty) FROM tblORDER T GROUP BY T.orderID HAVING T.orderID = 5001; C ) SELECT T.qty FROM tblORDER WHERE T.orderID = 5001; D ) SELECT SUM(T.qty) FROM tblORDER WHERE T.orderID = 5001;  

For a column named Product Name how must it be specified in CREATE TABLE, ALTER TABLE or any query?

Refer to Schema A. What is the result of the following? DECLARE @CustID INT; SET @CustID = 2; SELECT COUNT(*)   FROM tblORDER O JOIN tblPRODUCT P ON O.prodID = P.prodID WHERE O.custID = @CustID AND P.category = ‘Tech’;  

Refer to Scheme A Consider the following query:SELECT P.pName AS [Name], P.unitPrice * 1.1 AS [NewPrice]FROM tblPRODUCT PWHERE [NewPrice] > 100; What is the result?

Refer to Schema A. Consider this program: DECLARE @Limit INT; SET @Limit = 1; SELECT COUNT(*) FROM tblORDER O WHERE O.qty > @Limit; What is the result?

Extra Credit (2 points) I understand that immediately after completion of each part of the exam – I must submit the pictures of my: Cheat Sheet (even if blank) Scratch paper (even if blank) I understand the failure to do so will result in loss of extra credit points; and also potentially a zero on the exam.

Refer to Schema B. What is the result of the following? WITH LoanCount AS (   SELECT L.memID ,COUNT(*) AS [Num]   FROM tblLOAN L   GROUP BY L.memID ) SELECT M.lastName ,LC.[Num] FROM tblMEMBER M JOIN LoanCount LC ON M.memID = LC.memID WHERE LC.[Num] >= 2;

Refer to Schema B. What is the result of: SELECT COUNT(*) – COUNT(DISTINCT L.bookID) FROM tblLOAN L WHERE L.durationDays IS NULL;

Refer to Schema B. If a LEFT JOIN is performed from tblMEMBER to tblLOAN, which member’s lastName will show NULL for the loanID column?

Ingrowing nail