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yulyashka [42]
3 years ago
10

Which statements describe OneDrive?

Computers and Technology
2 answers:
Vesna [10]3 years ago
6 0

Answer:

OneDrive is a storage location on your personal computer.

Explanation:

Its gives u extra storage as well as helps u to take backup for all ur files that are important to u

Hope this helps...

Pls mark my ans as brainliest

If u mark my ans as brainliest u will get 3 extra points

andrezito [222]3 years ago
5 0
One drive is a storage location on your personal computer
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If there are 8 opcodes and 10 registers, a. What is the minimum number of bits required to represent the OPCODE? b. What is the
Nimfa-mama [501]

Answer:  

For 32 bits Instruction Format:

OPCODE   DR               SR1                   SR2      Unused bits

a) Minimum number of bits required to represent the OPCODE = 3 bits

There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.

Ceil (log2 (8)) = 3

b) Minimum number of bits For Destination Register(DR) = 4 bits

There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value.  4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.  

Ceil (log2 (10)) = 4

c) Maximum number of UNUSED bits in Instruction encoding = 17 bits

Total number of bits used = bits used for registers + bits used for OPCODE  

     = 12 + 3 = 15  

Total  number of bits for instruction format = 32  

Maximum  No. of Unused bits = 32 – 15 = 17 bits  

OPCODE                DR              SR1             SR2              Unused bits

  3 bits              4 bits          4 bits           4 bits                17 bits

7 0
3 years ago
[1] Please find all the candidate keys and the primary key (or composite primary key) Candidate Key: _______________________ Pri
AVprozaik [17]

Answer:

Check the explanation

Explanation:

1. The atomic attributes can't be a primary key because the values in the respective attributes should be unique.

So, the size of the primary key should be more than one.

In order to find the candidate key, let the functional dependencies be obtained.

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing ->  Emp_ID

Course_ID -> Course Name

Course_Name ->  Course_ID

Date_Completed -> Course_Name

Closure of attribute { Emp_ID, Date_Completed } is { Emp_ID, Date_Completed , Name, DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Name , Date_Completed } is { Name, Date_Completed , Emp_ID , DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { DeptID, Date_Completed } is { DeptID, Date_Completed , Emp_ID,, Name, , Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Marketing, Date_Completed } is { Marketing, Date_Completed , Emp_ID,, Name, DeptID , Salary, Course_Name, Course_ID}.

So, the candidate keys are :

{ Emp_ID, Date_Completed }

{ Name , Date_Completed }

{ DeptID, Date_Completed }

{ Marketing, Date_Completed }

Only one candidate key can be a primary key.

So, the primary key chosen be { Emp_ID, Date_Completed }..

2.

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing ->  Emp_ID

Course_ID -> Course Name

Course_Name ->  Course_ID

Date_Completed -> Course_Name

3.

For a relation to be in 2NF, there should be no partial dependencies in the set of functional dependencies.

The first F.D. is

Emp_ID -> Name, DeptID, Marketing, Salary

Here, Emp_ID -> Salary ( decomposition rule ). So, a prime key determining a non-prime key is a partial dependency.

So, a separate table should be made for Emp_ID -> Salary.

The tables are R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

and R2( Emp_ID , Salary)

The following dependencies violate partial dependency as a prime attribute -> prime attribute :

Name -> Emp_ID

DeptID -> Emp_ID

Marketing ->  Emp_ID

The following dependencies violate partial dependency as a non-prime attribute -> non-prime attribute :

Course_ID -> Course Name

Course_Name ->  Course_ID

So, no separate tables should be made.

The functional dependency Date_Completed -> Course_Name has a partial dependency as a prime attribute determines a non-prime attribute.

So, a separate table is made.

The final relational schemas that follows 2NF are :

R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

R2( Emp_ID , Salary)

R3 (Date_Completed, Course_Name, Course_ID)

For a relation to be in 3NF, the functional dependencies should not have any transitive dependencies.

The functional dependencies in R1(Emp_ID, Name, DeptID, Marketing, Date_Completed) is :

Emp_ID -> Name, DeptID, Marketing

This violates the transitive property. So, no table is created.

The functional dependencies in R2 (  Emp_ID , Salary) is :

Emp_ID -> Salary

The functional dependencies in R3 (Date_Completed, Course_Name, Course_ID) are :

Date_Completed -> Course_Name

Course_Name   ->  Course_ID

Here there is a transitive dependency as a non- prime attribute ( Course_Name ) is determining a non-attribute ( Course_ID ).

So, a separate table is made with the concerned attributes.

The relational schemas which support 3NF re :

R1(Emp_ID, Name, DeptID, Course_ID, Marketing, Date_Completed) with candidate key as Emp_ID.

R2 (  Emp_ID , Salary) with candidate key Emp_ID.

R3 (Date_Completed, Course_Name ) with candidate key Date_Completed.

R4 ( Course_Name, Course_ID ).  with candidate keys Course_Name and Course_ID.

6 0
3 years ago
Rocks created from compaction and cementation of sediments are called:
Effectus [21]

Answer:

Sedimentary Rocks

Explanation:

Sedimentary rocks are essentially the Frankenstein monsters of the rock world. They're made up of pieces of igneous and metamorphic rocks, sand, clay, and other sedimentary rocks.

4 0
3 years ago
Programming CRe-type the code and fix any errors. The code should convert non-positive numbers to 1.
LUCKY_DIMON [66]

Answer:

Given

The above lines of code

Required

Rearrange.

The code is re-arrange d as follows;.

#include<iostream>

int main()

{

int userNum;

scanf("%d", &userNum);

if (userNum > 0)

{

printf("Positive.\n");

}

else

{

printf("Non-positive, converting to 1.\n");

userNum = 1;

printf("Final: %d\n", userNum);

}

return 0;

}

When rearranging lines of codes. one has to be mindful of the programming language, the syntax of the language and control structures in the code;

One should take note of the variable declarations and usage

See attachment for .cpp file

Download cpp
5 0
3 years ago
What is the name of html command? ​
Makovka662 [10]

Here are some.

  • <html></html> This is the root element tag. ...
  • <head></head> ...
  • <title></title> ...
  • <body></body> ...
  • <h1></h1> ...
  • <p></p> ...
  • <a></a> ...
  • <img></img>
<h2>hope it helps.</h2><h2>stay safe healthy and happy....</h2>
3 0
2 years ago
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