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Answers by adarsh_1997

0 votes
31
Er diagram
0 votes
32
MADE EASY TEST-SERIES
Consider R(A,B,C,D,E,F,G) be a relational schema with the following functional dependencies: AC->G, D->EG, BC->D, CG->BD, ACD->B, CE->AG The number of different minimal cover possible are___________?
Consider R(A,B,C,D,E,F,G) be a relational schema with the following functional dependencies:AC->G, D->EG, BC->D, CG->BD, ACD->B, CE->AGThe number of different minimal cov...
1 votes
33
test algorithms
i marked option a) answer is b)
i marked option a) answer is b)
0 votes
34
#virtualgate
they have given b as the answer...i think answer is d....plz check!
they have given b as the answer...i think answer is d....plz check!
1 votes
35
MADE EASY DB
please explain this how to solve this kind of questions
please explain this how to solve this kind of questions
1 votes
36
Acebook
Consider the following code: co-begin-end X=2; co begin X=X+3; X=X+1; co end The final possible values of 'X' are: a)3,5 b)3,4,5 c)3,5,6 d)5,3,2
Consider the following code:co-begin-end X=2; co begin X=X+3; X=X+1; co endThe final possible values of 'X' are:a)3,5b)3,4,5c)3,5,6d)5,3,2
3 votes
37
Ace operating system
A user process can become blocked only if it is. : (a) In the ready state only. (b) In the running state only (c) In the blocked (or waiting) state. (d) In the Ready or Blocked State.
A user process can become blocked only if it is. :(a) In the ready state only. (b) In the running state only (c) In the blocked (or waiting) state. (d) In the Ready or Bl...
0 votes
38
Practice Question
How to solve this problem?
How to solve this problem?
0 votes
39
ISRO APRIL 2018
How many lines of output does the following C code produce? #include<stdio.h> main() { float sum=0.0,j=1.0,i=2.0; while (i/j > 0.001) { j=j+1; sum=sum+(i/j); printf("%f\n", sum); } } 0-9 line of output 10-19 lines of output 20-29 lines of output More than 29 lines of output
How many lines of output does the following C code produce?#include<stdio.h main() { float sum=0.0,j=1.0,i=2.0; while (i/j 0.001) { j=j+1; sum=sum+(i/j); printf("%f\n", ...
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