1. A network has 10 nodes and 17 branches. The number of different node pair are





Write Comment

Type in
(Press Ctrl+g to toggle between English and the chosen language)

Comments

Tags
Show Similar Question And Answers
QA->In a Program Graph, ‘X’ is an if-then-else node. If the number of paths from start node to X is ‘p’ number of paths from if part to end node is ‘q’ and from else part to end node is ’r’, the total number of possible paths through X is :....
QA->If n represents the dimension of cube and k, the radix(no. of nodes along each dimensions. then the number of nodes N of a K-ary n –cube network is :....
QA->The number of null branches for a binary tree with n nodes is:....
QA->A program Graph has ‘start’ and ‘end’ nodes. The total number of paths from start to end is equivalent to the -----------set of test data required to test the software.....
QA->Which nuclides having the same number of neutrons but different number of protons or mass number?....
MCQ->What will be the output of the program in 16-bit platform (under DOS)? #include<stdio.h> int main() { struct node { int data; struct node link; }; struct node p, q; p = (struct node ) malloc(sizeof(struct node)); q = (struct node ) malloc(sizeof(struct node)); printf("%d, %d\n", sizeof(p), sizeof(q)); return 0; }....
MCQ->In a six-node network, two nodes are connected to all the other nodes. Of the remaining four, each is connected to four nodes. What is the total number of links in the network?....
MCQ->A network has 10 nodes and 17 branches. The number of different node pair are....
MCQ->The network has 10 nodes and 17 branches. The number of different node power voltages would be....
MCQ->Statements: All branches are flowers. All flowers are leaves. Conclusions: All branches are leaves. All leaves are branches. All flowers are branches. Some leaves are branches.

....
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
DMCA.com Protection Status Powered By:Omega Web Solutions
© 2002-2017 Omega Education PVT LTD...Privacy | Terms And Conditions