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matrenka [14]
4 years ago
15

Meaningful decisions are important to sustaining immersion, but it’s generally considered poor game design to constantly give th

e player “critical” decisions. Describe a game you know and how it asks the player to make a variety of decisions from the different levels of Tracy Fullerton’s Decision Scale.
Computers and Technology
1 answer:
bezimeni [28]4 years ago
7 0

Answer:

Explanation:

Meaningful decisions are important to sustaining immersion, but it's generally considered poor game design to constantly give the player "critical" decisions. Describe a game you know and how it asks the player to make a variety of decisions from the different levels of Tracy Fullerton's Decision Scale.

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mihalych1998 [28]

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Ok so the power of sussy is hard to find because you have to be sus which the impostor sus sussy then you sus more of the sus then you get the sus sus sus sus sus sus sus sus sus sus sus amogus.

Explanation:

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Halcyon, an e-publisher, has recently decided to use an information system that administers the way its customers access its onl
puteri [66]

Answer:

b) digital rights management

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Halcyon, an e-publisher, has recently decided to use an information system that administers the way its customers access its online publications. The system assigns each customer with a unique ID, maintains records of the books purchased by them, encrypts electronic documents for transmission, and includes options to order hard copies of the electronic documents they read online. The set of technology being used here is digital rights management.

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3 years ago
Is it true that if the user log out the computer will turn off automatically​
sergiy2304 [10]
Yes that is a true statement
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Write a program that takes paragraph from the user and prints all unique words in that paragraph using strings in c++
Bezzdna [24]

Answer:

void printUniquedWords(char filename[])

void printUniquedWords(char filename[]) {

void printUniquedWords(char filename[]) {     // Whatever your t.text file is

    fstream fs(filename);

    fstream fs(filename);   

    fstream fs(filename);       

    fstream fs(filename);           map<string, int> mp;

    fstream fs(filename);           map<string, int> mp;   

    fstream fs(filename);           map<string, int> mp;       // This serves to keep a record of the words

    string word;

    string word;     while (fs >> word)

    string word;     while (fs >> word)     {

    string word;     while (fs >> word)     {         // Verifies if this is the first time the word appears hence "unique word"

        if (!mp.count(word))

        if (!mp.count(word))             mp.insert(make_pair(word, 1));

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }   

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();   

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;     }

        if (!mp.count(word))             mp.insert(make_pair(word, 1));         else             mp[word]++;     }       fs.close();       // Traverse map and print all words whose count     //is 1     for (map<string, int> :: iterator p = mp.begin();          p != mp.end(); p++)     {         if (p->second == 1)             cout << p->first << endl;     } }

5 0
3 years ago
Which best explains the process he would follow?​
Anni [7]

The first option.

You gotta find the format tab, and change the color from there.

4 0
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