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NEW QUESTION: 1
A teacher is making five children stand in a row. Each child is assigned a number tag before being made
to stand in the row. The tags are not necessarily according to their positions.
Amy, Tara, Xenia, Yana, Pam are the children and they are given numbers 1, 2, 3, 4, 5.
The following conditions apply:
Exactly one number is given to a child.
Pam must be made to stand fourth and assigned number 1.
4 must be assigned to Yana.
Tara and Xenia must each be made to stand in one of the extreme positions.
Xenia cannot be given either number 2 or 3.
If Amy is made to stand second, which of the following must be true?
A. 2 is assigned to the child standing fifth.
B. Tara is standing first.
C. 3 is assigned to the child standing first.
D. 5 is assigned to the child standing second.
E. Yana is standing third.
Answer: E
NEW QUESTION: 2
Access control system can only set whether the user can open the door, can not be set when to open the door.
A. TRUE
B. FALSE
Answer: B
NEW QUESTION: 3


A. SELECT c.custidFROM Sales.Customers c LEFT OUTER JOIN Sales.Order o ON c.custid = o.custidWHERE orderid IS NULL
B. SELECT custid FROM Sales.CustomersINTERSECTSELECT custid FROM
Sales.Orders
C. SELECT c.custidFROM Sales.Customers c LEFT OUTER Sales.Order oON c.custid = o.custid
D. SELECT c.custidFROM Sales.Customers c INNER JOIN Sales.Order oON c.custid = o.custid
Answer: A
Explanation:
Explanation
Inner joins return rows only when there is at least one row from both tables that matches the join condition.
Inner joins eliminate the rows that do not match with a row from the other table. Outer joins, however, return all rows from at least one of the tables or views mentioned in the FROM clause, as long as those rows meet any WHERE or HAVING search conditions. All rows are retrieved from the left table referenced with a left outer join, and all rows from the right table referenced in a right outer join. All rows from both tables are returned in a full outer join.
References: https://technet.microsoft.com/en-us/library/ms187518(v=sql.105).aspx
NEW QUESTION: 4
Scenario










A. Option A
B. Option B
C. Option D
D. Option C
Answer: A
Explanation:
Using the "show ip int brief command" on R5 we can see the IP addresses assigned to this router. Then, using the
"show ip route" command on R1 we can see that to reach 10.5.5.5 and 10.5.5.55 the preferred path is via Serial 1/3,
which we see from the diagram is the link to R2.
