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Anarel [89]
2 years ago
5

10) What is the BEST way to rewrite sentence (1) to make it more

Computers and Technology
2 answers:
Nina [5.8K]2 years ago
8 0

Answer: The answer is D

Explanation:

I took the test.

givi [52]2 years ago
3 0
I believe the answer is D. It makes more sense and is the most correct grammatically.

Hope this helped and pls mark as brainliest!

~ Luna
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Daniel [21]

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Option  A is the correct answer.

Explanation:

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Software that people commonly use in the workplace to make their lives easier is called
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Software that people commonly use in the workplace to make their lives easier is called system software.
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Write a function in MATLAB that takes as an input a matrix of coefficients for a system of linear equations (A), the solution ve
Illusion [34]
Answer is A all the way to the right
7 0
3 years ago
Develop a C program that calculates the final score and the average score for a student from his/her (1)class participation, (2)
Ghella [55]

Answer:

#include <iomanip>

#include<iostream>

using namespace std;

int main(){

char name[100];

float classp, test, assgn, exam, prctscore,ave;

cout<<"Student Name: ";

cin.getline(name,100);

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; }

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){  cout<<"Test: "; cin>>test; }

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; }

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; }

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; }

ave = (int)(classp + test + assgn + exam + prctscore)/5;

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

return 0;}

Explanation:

The required parameters such as cin, cout, etc. implies that the program is to be written in C++ (not C).

So, I answered the program using C++.

Line by line explanation is as follows;

This declares name as character of maximum size of 100 characters

char name[100];

This declares the grading items as float

float classp, test, assgn, exam, prctscore,ave;

This prompts the user for student name

cout<<"Student Name: ";

This gets the student name using getline

cin.getline(name,100);

This prompts the user for class participation. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Class Participation: "; cin>>classp; </em>

<em> while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; } </em>

This prompts the user for test. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Test: "; cin>>test; </em>

<em> while(test <0 || test > 100){  cout<<"Test: "; cin>>test; } </em>

This prompts the user for assignment. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Assignment: "; cin>>assgn; </em>

<em> while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; } </em>

This prompts the user for examination. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Examination: "; cin>>exam; </em>

<em> while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; } </em>

This prompts the user for practice score. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Practice Score: "; cin>>prctscore; </em>

<em> while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; } </em>

This calculates the average of the grading items

ave = (int)(classp + test + assgn + exam + prctscore)/5;

This prints the calculated average

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

8 0
3 years ago
Suppose there are 69 packets entering a queue at the same time. Each packet is of size 7 MiB. The link transmission rate is 1.7
svp [43]

Answer:

69.08265412 milliseconds

Explanation:

Lets first convert 7 MiB to bits

7*1024*1024*8=58720256 bits

Now convert bits to Gbits

58720256/10^{9}  =0.058720256 Gbits

Queuing Delay = Total size/transmission link rate

Queuing Delay= \frac{0.058720256}{1.7} =0.03454132706 seconds

Delay of packet number 3 = 0.03454132706*2=0.06908265412 seconds

or 0.06908265412= 69.08265412 milliseconds

7 0
3 years ago
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