1). Piggy backing is a technique for
a) Flow control b) sequence c) Acknowledgement d) retransmission
ans: c piggy backing
2). The layer in the OST model handles terminal emulation
a) session b) application c) presentation d) transport
ans: b application
3) ans: a odd numbers of errors
5) c 20
6) a 120
7) b synchronies the access
9) b the operating system
10) a 177333
11) d used as a network layer protocol in network and windows system
12) b has to be unique in the sub network
13)Q. There is an employer table with key fields as employer no. Data in every
n'th row are needed for a simple following queries will get required results.
A) select A employee no. From employee A , where exists from employee B
where A employee no. >= B employee having (count(*) mod n)=0
b) select employee no. From employee A, employee B where
A employee no. >= B employ no. Grouply employee no. Having (count(*) mod n=0 )
c) both a& b
d)none of the above
14)Q. Type duplicates of a row in a table customer with non uniform key field
customer no. You can see
a) delete from customer where customer no. Exists
( select distinct customer no. From customer having count )
b) delete customer a where customer no. In
(select customer b where customer no. Equal to b customer no. ) and a rowid >
b rowid
c) delete customer a where customer no. In
( select customer no. From customer a, customer b )
d) none of the above
Lucent Technologies Computer Awareness Interview Questions - Section 1
1). Piggy backing is a technique for
a) Flow control b) sequence c) Acknowledgement d) retransmission
ans: c piggy backing
2). The layer in the OST model handles terminal emulation
a) session b) application c) presentation d) transport
ans: b application
3) ans: a odd numbers of errors
5) c 20
6) a 120
7) b synchronies the access
9) b the operating system
10) a 177333
11) d used as a network layer protocol in network and windows system
12) b has to be unique in the sub network
13)Q. There is an employer table with key fields as employer no. Data in every
n'th row are needed for a simple following queries will get required results.
A) select A employee no. From employee A , where exists from employee B
where A employee no. >= B employee having (count(*) mod n)=0
b) select employee no. From employee A, employee B where
A employee no. >= B employ no. Grouply employee no. Having (count(*) mod n=0 )
c) both a& b
d)none of the above
14)Q. Type duplicates of a row in a table customer with non uniform key field
customer no. You can see
a) delete from customer where customer no. Exists
( select distinct customer no. From customer having count )
b) delete customer a where customer no. In
(select customer b where customer no. Equal to b customer no. ) and a rowid >
b rowid
c) delete customer a where customer no. In
( select customer no. From customer a, customer b )
d) none of the above
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