Plagiarism is the answer to the blank
Answer:
See Explanation
Explanation:
Given
See attachment 1 for proper table definition
To answer this question, one of the basic sql commands (the "create" command) will be used to create the required table, followed by the fields of the table.
<em>I could not submit my answer directly. So, I added two additional attachments which represent the answer section and the explanation section, respectively.</em>
1.
#include <iostream>#include <string>
using namespace std;
int main(){ string chars; // This is where we will put our @ signs and print them for(int x=0;x < 5; x++){
chars = chars + '@'; // This will concatenate an @ sign at the end of the variable cout << chars << "\n"; }}
2.
#include <iostream>#include <string>
using namespace std;
int main(){ string name; // Our variable to store the name cout << "What is your name? \n"; // Asks the user for their name cin >> name; cout << "\nWell, hello " << name << "!";}
3.
#include <iostream>#include <string>
using namespace std;
int main(){ int number; // Our variable cout << "Enter a number\n"; // Asks for a number cin >> number; cout << "You entered " << number << "%!";}
4.
#include <iostream>#include <string>
using namespace std;
int main(){ int number; // Our variable cout << "Enter a number\n"; cin >> number;
int check = number % 2; // The modulo operator (the percent sign) gets the remainder of the quotient if (check == 0) { cout << number << " is even!"; // If the remainder is 0 then it prints out "x is even" } else { cout << number << " is odd!"; // If the remainder is not 0 then it prints out "x is odd" }}
5.
#include <iostream>#include <string>
using namespace std;
int main(){ float r; // Our variable cout << "Enter a radius\n"; cin >> r; if (r < 0){ cout << "Lol. No."; // If the radius is less than zero print out that message } float circumference=2*3.14*r; float area=r*r*3.14; cout << "\n\n Circumference of circle: " << circumference; cout << "\n Area of circle: " << area;}
Answer:
a. Utilization = 0.00039
b. Throughput = 50Kbps
Explanation:
<u>Given Data:</u>
Packet Size = L = 1kb = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
RTT = 20 msec
<u>To Find </u>
a. Sender Utilization = ?
b. Throughput = ?
Solution
a. Sender Utilization
<u>As Given </u>
Packet Size = L = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec
Utilization = Transmission Time / RTT + Transmission Time
= 8 micro-sec/ 20 msec + 8 micro-sec
= 0.008 sec/ 20.008 sec
Utilization = 0.00039
b. Throughput
<u>As Given </u>
Packet Size = 1kb
RTT = 20ms = 20/100 sec = 0.02 sec
So,
Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps
So, the system has 50 kbps throughput over 1 Gbps Link.
Answer:
b. The Upcoming Lunch & Learn Program Is in March.
Explanation:
A subject line identifies the e-mail intent. The subject line displayed the user or recipient when they look at the list of messages.
For informational e-mail, some points take into consideration.
- Write a subject line
- keep it short
- place important words
- eliminate filler words
- clear and specify the topic
- keep the subject line simple
- Set a deadline in the subject line
- Highlight the value that offered
I think option b is suitable for the informational subject line in an e-mail.
