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maksim [4K]
2 years ago
6

which type of encoding involves relating new information to existing knowledge that you already have stored in long-term memory?

Computers and Technology
1 answer:
rjkz [21]2 years ago
7 0

A type of encoding that involves relating new information to existing knowledge that you already have stored in long-term memory is: Semantic encoding.

Encoding can be defined as a communication process in which the sender of a message transforms an information into an encrypted message, in the form of notable symbols that represent ideas or concepts.

Generally, a message that has been encoded by a sender requires a decoding action by the recipient, in order for the communication process to be complete.

Semantic encoding is a type of encoding which typically involves the relation of new information with respect to an existing knowledge that has been stored in long-term memory.

In conclusion, semantic encoding deals with the encoding of meaning from new information rather than perceptual characteristics such as the meaning of words in English language.

Read more: brainly.com/question/24113575

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Public class Bird
Anna007 [38]

Answer:

Answer below.

Explanation:

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6 0
3 years ago
Finish and test the following two functions append and merge in the skeleton file:
avanturin [10]

Answer:

Explanation:

#include <iostream>

using namespace std;

int* append(int*,int,int*,int);

int* merge(int*,int,int*,int);

void print(int*,int);

int main()

{ int a[] = {11,33,55,77,99};

int b[] = {22,44,66,88};

print(a,5);

print(b,4);

int* c = append(a,5,b,4); // c points to the appended array=

print(c,9);

int* d = merge(a,5,b,4);

print(d,9);

}

void print(int* a, int n)

{ cout << "{" << a[0];

for (int i=1; i<n; i++)

cout << "," << a[i];

cout << "}\n";

}

int* append(int* a, int m, int* b, int n)

{

int * p= (int *)malloc(sizeof(int)*(m+n));

int i,index=0;

for(i=0;i<m;i++)

p[index++]=a[i];

for(i=0;i<n;i++)

p[index++]=b[i];

return p;

}

int* merge(int* a, int m, int* b, int n)

{

int i, j, k;

j = k = 0;

int *mergeRes = (int *)malloc(sizeof(int)*(m+n));

for (i = 0; i < m + n;) {

if (j < m && k < n) {

if (a[j] < b[k]) {

mergeRes[i] = a[j];

j++;

}

else {

mergeRes[i] = b[k];

k++;

}

i++;

}

// copying remaining elements from the b

else if (j == m) {

for (; i < m + n;) {

mergeRes[i] = b[k];

k++;

i++;

}

}

// copying remaining elements from the a

else {

for (; i < m + n;) {

mergeRes[i] = a[j];

j++;

i++;

}

}

}

return mergeRes;

}

4 0
3 years ago
Select all that apply.
Valentin [98]

Answer:

1, 3, & 4.

Is the way to do so. Good luck.

6 0
3 years ago
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