in your notes books and in your vopy
Answer:
Number of packets ≈ 5339
Explanation:
let
X = no of packets that is not erased.
P ( each packet getting erased ) = 0.8
P ( each packet not getting erased ) = 0.2
P ( X ≥ 1000 ) = 0.99
E(x) = n * 0.2
var ( x ) = n * 0.2 * 0.8
∴ Z = X - ( n * 0.2 ) / 
~ N ( 0.1 )
attached below is the remaining part of the solution
note : For the value of <em>n</em> take the positive number
Answer:
There are multiple critical paths
Explanation:
The critical path method (CPM), or critical path analysis (CPA), is an algorithm for scheduling a set of project activities. It is commonly used in conjunction with the program evaluation and review technique (PERT). A critical path is determined by identifying the longest stretch of dependent activities and measuring the time required to complete them from start to finish.
The essential technique for using CPM is to construct a model of the project that includes the following:
- A list of all activities required to complete the project (typically categorized within a work breakdown structure),
- The time (duration) that each activity will take to complete,
- The dependencies between the activities and,
- Logical end points such as milestones or deliverable items.
Using these values, CPM calculates the longest path of planned activities to logical end points or to the end of the project, and the earliest and latest that each activity can start and finish without making the project longer. This process determines which activities are "critical" (i.e., on the longest path) and which have "total float" (i.e., can be delayed without making the project longer).
considering the above function of the cpm analysis because you have multiple path, there is tendency that more than path through the project network will have zero slack values.
Answer:
import java.util.Scanner;
class Main {
public static int calcSeries(int n) {
int sum = 0;
for(int i=10; i>=n; i--) {
sum += i;
}
return sum;
}
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
int n = 0;
do {
System.out.print("Enter n: ");
n = reader.nextInt();
if (n >= 10) {
System.out.println("Please enter a value lower than 10.");
}
} while (n >= 10);
reader.close();
System.out.printf("sum: %d\n", calcSeries(n));
}
}
Answer:
int x;
indata.open("lottowins");
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata.close();
