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Molodets [167]
3 years ago
7

What is the pennsylvania law on cyber bullying

Computers and Technology
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0
Convicted of a criminal offense may face fines,imprisonment.hope that and you got it right.
You might be interested in
- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -> {D}, {B} -&
Olenka [21]

Answer:

The key of R is {A, B}

Explanation:

A key can be seen as a minimal set of attributes whose closure includes all the attributes in R.

Given that the closure of {A, B}, {A, B}+ = R, one key of R is {A, B} But in this case, it is the only key.

In order for us to to normalize R intuitively into 2NF then 3NF, we have to make use of these approaches;

First thing we do is to identify partial dependencies that violate 2NF. These are attributes that are

functionally dependent on either parts of the key, {A} or {B}, alone.

We can calculate

the closures {A}+ and {B}+ to determine partially dependent attributes:

{A}+ = {A, D, E, I, J}. Hence {A} -> {D, E, I, J} ({A} -> {A} is a trivial dependency)

{B}+ = {B, F, G, H}, hence {A} -> {F, G, H} ({B} -> {B} is a trivial dependency)

To normalize into 2NF, we remove the attributes that are functionally dependent on

part of the key (A or B) from R and place them in separate relations R1 and R2,

along with the part of the key they depend on (A or B), which are copied into each of

these relations but also remains in the original relation, which we call R3 below:

R1 = {A, D, E, I, J}, R2 = {B, F, G, H}, R3 = {A, B, C}

The new keys for R1, R2, R3 are underlined. Next, we look for transitive

dependencies in R1, R2, R3. The relation R1 has the transitive dependency {A} ->

{D} -> {I, J}, so we remove the transitively dependent attributes {I, J} from R1 into a

relation R11 and copy the attribute D they are dependent on into R11. The remaining

attributes are kept in a relation R12. Hence, R1 is decomposed into R11 and R12 as

follows: R11 = {D, I, J}, R12 = {A, D, E} The relation R2 is similarly decomposed into R21 and R22 based on the transitive

dependency {B} -> {F} -> {G, H}:

R2 = {F, G, H}, R2 = {B, F}

The final set of relations in 3NF are {R11, R12, R21, R22, R3}

4 0
4 years ago
1. Write a program that declares an array named alpha with 50 components of the type double. Initialize the array so that the fi
Aleksandr [31]

Answer:

Explanation:

1. Write a program that declares an array named alpha with 50 components of the type double. Initialize the array so that the first 25 components are equal to the square of the counter (or index) variable and the last 25 components are equal to three times the index variable.  

  double alpha[50];

   for (int i=0;i<25;i++)

   {

       alpha[i]=i*i;

       alpha[i+25]=(i+25)*3;

  }

2. Output the array so that exactly ten elements per line are printed.  

   for (int i=0;i<50;i++)

   {

       cout<<i+1<<". "<<alpha[i]<<" ";

       if (((i+1)%10)==0)

       {

           cout<<endl;

       }

   }

3. Run your program again, but this time change the code so that the array is filled with random numbers between 1 and 100.  

   double alpha[50];

   for (int i=0;i<50;i++)

   {

       alpha[i]=rand()%101;

   }

   for (int i=0;i<50;i++)

   {

       cout<<i+1<<". "<<alpha[i]<<" ";

       if (((i+1)%10)==0)

       {

           cout<<endl;

       }

   }

4. Write the code that computes and prints the average of elements of the array.  

   double alpha[50],temp=0;

   for (int i=0;i<50;i++)

   {

       alpha[i]=rand()%101;

       temp+=alpha[i];

   }

   cout<<"Average :"<<(temp/50);

5. Write the code that that prints out how many of the elements are EXACTLY equal to 100.

   double alpha[50],temp=0;

   for (int i=0;i<50;i++)

   {

       alpha[i]=rand()%101;

       if(alpha[i]==100)

       {

           temp++;

       }

   }

   cout<<"Elements Exacctly 100 :"<<temp;

Please note:  If you put  each of above code to the place below comment  it will run perfectly after compiling

#include <iostream>

using namespace std;

int main()

{

   // If you put  each of above code here it will run perfectly after compiling

   return 0;

}

8 0
3 years ago
Which type of navigation involves multiple frames that are linked to a number of other frames?
sesenic [268]

Answer:

1. b)

2. c)

3. c)

4. a)

5. b)

Explanation:

1. and 5. Linear kind of navigation is a system with a sequential manner web pages that are perfect for some sorts of sites that are having information that has to be viewed as a book (5) and when we are talking about that view we are considering one page after another page like we are reading a book. It is also the simplest navigation. This is the explanation for question 1 and question 5.

2. The most well-designed navigation system is an intuitive one because in this design of the website we have website traffic that is easy because it flows from one web page to another web page. It is showing us where to go to find and look for something and even where to go if there is no concrete options for what are we looking for.

3. A Sitemap is referring to the organized hierarchy of links and it is the protocol that is allowing us to search through many links. A Sitemap is having a listing of the URLs for some site and that is why this is the correct answer.

4. In using liner reciprocal navigation the interface should include how frames are left and how many of them are there. The more the frames, the more times the user will spend on them and the site.

8 0
3 years ago
What is sexual intercourse <br><br>​
n200080 [17]

Answer:

immmm ....

I know the answer but I won't tell..lolllll

7 0
3 years ago
Read 2 more answers
What is an 8086 microprocessor used for
Liula [17]
A sixteen bit microprocessor chip used in early IBM PCs. The Intel 8088 was a version with an eight-bit externaldata bus.

The Intel 8086 was based on the design of the Intel 8080 <span>and </span>Intel 8085 (it was source compatible with the 8080)with a similar register set, but was expanded to 16 bits. The Bus Interface Unit fed the instruction stream to theExecution Unit through a 6 byte prefetch queue, so fetch and execution were concurrent - a primitive form ofpipelining (8086 instructions varied from 1 to 4 bytes).

It featured four 16-bit general registers, which could also be accessed as eight 8-bit registers, and four 16-bit indexregisters (including the stack pointer). The data registers were often used implicitly by instructions, complicatingregister allocation for temporary values. It featured 64K 8-bit I/O (or 32K 16 bit) ports and fixed vectored interrupts.There were also four segment registers that could be set from index registers.

The segment registers allowed the CPU to access 1 meg of memory in an odd way. Rather than just supplyingmissing bytes, as most segmented processors, the 8086 actually shifted the segment registers left 4 bits and addedit to the address. As a result, segments overlapped, and it was possible to have two pointers with the same valuepoint to two different memory locations, or two pointers with different values pointing to the same location. Mostpeople consider this a brain damaged design.

Although this was largely acceptable for assembly language, where control of the segments was complete (it couldeven be useful then), in higher level languages it caused constant confusion (e.g. near/far pointers). Even worse, thismade expanding the address space to more than 1 meg difficult. A later version, the Intel 80386, expanded thedesign to 32 bits, and "fixed" the segmentation, but required extra modes (suppressing the new features) forcompatibility, and retains the awkward architecture. In fact, with the right assembler, code written for the 8008 canstill be run on the most <span>recent </span>Intel 486.

The Intel 80386 added new op codes in a kludgy fashion similar to the Zilog Z80 and Zilog Z280. The Intel 486added full pipelines, and clock doubling (like <span>the </span>Zilog Z280).

So why did IBM chose the 8086 series when most of the alternatives were so much better? Apparently IBM's own engineers wanted to use the Motorola 68000, and it was used later in the forgotten IBM Instruments 9000 Laboratory Computer, but IBM already had rights to manufacture the 8086, in exchange for giving Intel the rights to its bubble memory<span> designs.</span> Apparently IBM was using 8086s in the IBM Displaywriter word processor.

Other factors were the 8-bit Intel 8088 version, which could use existing Intel 8085-type components, and allowed the computer to be based on a modified 8085 design. 68000 components were not widely available, though it could useMotorola 6800 components to an <span>extent.
</span>
Hope this helps
7 0
3 years ago
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