Answer:
Communication path basically define the path in which the information and messages can be exchange by using the efficient communication path.
There are simple formula for calculating the total number of communication channel that is :
Where, n is the number of stack holder.
Now, the maximum number of communication paths for a team of twenty people can be calculated as:
n=20
=\frac{20(20-1)}{2} = 190
Therefore, 190 is the total number of communication path.
medical charts and assinging codes
Answer:
M2 is the answer for the above question.
Explanation:
- The above question states the scenario, where firstly main function calls the M1 function.
- Then M1 function calls the function M2.
- Then M2 function calls the function M3.
- Then M3 function calls the function M5.
- Then Again M2 function will resume and calls the function M4.
- Then Again M2 function will resume because when the M4 will ends it return to the statement of the M2 function from which the M4 function has been called.
- It is because when any function ends its execution it returns to that line of a statement from which it has been called. This is the property of programming language to call the function.
- So when the M4 will terminate its execution then the M2 function will resume again.
Answer:
public boolean equals(Object other)
{
// check correct instance
if (other instanceof IntTree)
{
try
{
IntTree otherTree = (IntTree) other;
return equals(overallRoot, otherTree.overallRoot);
}
// caught exceptions
catch (Exception e)
{
return false;
}
}
return false;
}
public boolean equals(IntTreeNode tree1, IntTreeNode tree2)
{
// if reach ends
if (tree1 == null && tree2 == null)
{
// they are equal
return true;
}
else if (tree1 == null || tree2 == null)
{
// not equal
return false;
}
// check left part
boolean leftPart = equals(tree1.left, tree2.left);
// check current element is same
boolean currentPart = tree1.element.equals(tree2.element);
// check right part
boolean rightPart = equals(tree1.right, tree2.right);
// all are equal
return (leftPart && currentPart && rightPart);
}
Explanation:
