What are the two reasons for a clear cut

Which of the following conditions is a good sign of minor

Kryger [21]

Answer:

Explanation:

d

7 0

3 years ago

If the density of states function in the conduction band of a particular semiconductor is a constant equal to K, derive the expr

s2008m [1.1K]

Answer:

full details of the answer is attached

5 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

7 0

3 years ago

In a black box experiment, when the amount of material exiting a closed system is less than the amount of material entering the

Oksi-84 [34.3K]

When the material that exits is lesser in amount than that of the entering material in a black box experiment, the parts of the system need to be changed.

<h3>What happens in a black box experiment?</h3>

In a black box experiment, the experimenters need to make assumptions regarding the drawing of conclusions. One such conclusion is the amount of material that exits.

If such amount is lesser than the one that enters the system, such experiment concludes that it is the time to change the parts of the system.

Hence, option D holds true regarding the black box experiment.

Learn more about black box experiment here:

brainly.com/question/13403296

#SPJ1

4 0

2 years ago

An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service

Nadusha1986 [10]

A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents

B) The number of agents required on duty to reduce cost = 9 agents

Given data :

Arrival rate of customers ( β ) = 220 per hour

Service rate ( mu ) = 60 minutes / 2 minutes = 30 customer per hour

utilization ( rho ) = 220 / 30 ≈ 7

at least 8 server personnel are required for stability of the queue

A) Determine the number of agents required to achieve a wait time of 10 minutes or less per customer

waiting time = 10 - 2 = 8 minutes

number of customers waiting ( ∝ ) = 7 and required server = 8

assuming Lq = 5.2266

Hence the waiting time in line = Lq / arrival rate

= 5.2266 / 220 = 0.0238 hour

= 0.0238 * 60 = 1.428 minutes

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

B) Determine the number of ticket agents that should be on duty to minimize cost

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

i) When 8 agents are used

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

= £ 122.18

ii) When 9 agents are used

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

Learn more about cost reduction : brainly.com/question/14115944

6 0

3 years ago