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vampirchik [111]
3 years ago
13

What are 3 similarities and 3 differences between live theatre and film/videos -Drama Class

Computers and Technology
1 answer:
marin [14]3 years ago
4 0
Differences-
1. Theatre is live. Film has been captured in the past. We only see after the making process is done.

2.You have chance for improvement in each theatre shows but its impossible in Films. Once film is done its done.

3.Normally Theatre is cheaper, films are costly.

Similarities-

1.Theatre and films both are arts, so many varieties of arts melt into theatre or film to make it happen.

2.Theatre and films both are very effective medium of communication.

3.Theatre and films both are considered as great form of Entertainment.
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Bill, a project manager, wants to hire external resources. What step should Bill take before hiring external resources?
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Answer:

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8 0
3 years ago
Database users who access data only through application programs are called
Alex777 [14]
A. casual users hope this helps
7 0
3 years ago
What should the shutter speed be on the camera?<br> A. 1/30<br> B. 1/50<br> C. 1/60<br> D. 1/15
11Alexandr11 [23.1K]

Answer:

C. 1/60

Explanation:

Shutter speed is most commonly measured in fractions of a second, like 1/20 seconds or 1/10 seconds. Some high-end cameras offer shutter speeds as fast as 1/80 seconds. But, shutter speeds can extend to much longer times, generally up to 30 seconds on most cameras.

But in this case C. 1/60 is the answer.

6 0
3 years ago
Write a C++ program that reads students' names followed by their test scores. The program should output each students' name foll
Mashutka [201]

Answer:

#include<iostream>

#include<conio.h>

using namespace std;

struct studentdata{

char Fname[50];

char Lname[50];

int marks;

char grade;

};

main()

{

studentdata s[20];

for (int i=0; i<20;i++)

   {

cout<<"\nenter the First name of student";

cin>>s[i].Fname;

cout<<"\nenter the Last name of student";

cin>>s[i].Lname;

cout<<"\nenter the marks of student";

cin>>s[i].marks;

}  

 

for (int j=0;j<20;j++)

{

if (s[j].marks>90)

{

 s[j].grade ='A';

}

else if (s[j].marks>75 && s[j].marks<90)

{

   s[j].grade ='B';

}

else if (s[j].marks>60 && s[j].marks<75)

{

 s[j].grade ='C';

}

else

{

 s[j].grade ='F';

}

}

int highest=0;

int z=0;

for (int k=0;k<20; k++)  

{

if (highest<s[k].marks)

{

 highest = s[k].marks;

 z=k;

}

 

}

cout<<"\nStudents having highest marks"<<endl;

 

cout<<"Student Name"<< s[z].Fname<<s[z].Lname<<endl;

cout<<"Marks"<<s[z].marks<<endl;

cout<<"Grade"<<s[z].grade;

getch();  

}

Explanation:

This program is used to enter the information of 20 students that includes, their first name, last name and marks obtained out of 100.

The program will compute the grades of the students that may be A,B, C, and F. If marks are greater than 90, grade is A, If marks are greater than 75 and less than 90, grade is B. For Mark from 60 to 75 the grade is C and below 60 grade is F.

The program will further found the highest marks and than display the First name, last name, marks and grade of student who have highest marks.

6 0
3 years ago
The X.500 standard defines a protocol for a client application to access an X.500 directory known as which of the following opti
lions [1.4K]

Answer:

Option B is the correct answer for the above question.

Explanation:

DAP is a protocol technology that is used for the client and discovered by X.500 in 1988. Its use for the client host on the network. The operation of this protocol is --- Read, Bind, Search, List, Compare, Add and Modify the data.

The above question asked about the protocol which is searched by X.500 and used for the clients. So the answer is DAP which is stated by Option B hence this option is correct while other is not because--

  • Option 'A' states about DIB which is not a protocol. It is used by X.500.
  • Option C states about DIT which is not used by X.500.
  • Option D states about LDAP which can be used for client or server.

7 0
3 years ago
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