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nikklg [1K]
3 years ago
14

1. If an F# function has type 'a -> 'b when 'a : comparison, which of the following is not a legal type for it? Select one:

Computers and Technology
1 answer:
Stella [2.4K]3 years ago
7 0

Answer:

B. string -> (int -> int)

Explanation:

We are going to perform comparison operations '->'. It is important to notice that the comparison operation gives us a bool value (True or False) and the comparison operation is legal if and only if the data types to be compared are the same.

Example:

int(4)->int(5) False

int(4)->int(4) True

int(4)->string(4) Error, data types don't match

For this reason:

  • A. Is legal because float -> float evaluates to True, True is a boolean value and bool -> bool is legal because both are the same data type.
  • B. Is illegal because int -> int evaluates to True, True is a boolean value and string is not a boolean (string -> bool).
  • C. Is legal because int is the same type than int.
  • D. Is legal because the list is the same type than list regardless it's content.

Note:

The operations inside parentheses are evaluated first.

List is a type by itself regardless of its content.

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Alenkasestr [34]

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<h3>Writing a pseudocode we have that:</h3>

<em>while </em>

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<em>break</em>

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<em> If hours >36 : </em>

<em>print ( "Name is " ,name)</em>

<em>print( "No of hours are ",hours)</em>

<em>print("Congratulaion! Your working hours are more than 36")</em>

<em>If hours <30 : #</em>

<em>print ( "Name is " ,name)</em>

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<em>print("Warning !!!")</em>

<em>End loop</em>

See more about pseudocode at brainly.com/question/13208346

#SPJ1

3 0
2 years ago
Before you ever buy your first stock or bond, it's important to understand what type of investor you are. This depends on a numb
Oksi-84 [34.3K]

Answer:

Moderate investor

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barxatty [35]

Answer:

#include <string>

#include <iostream>

using namespace std;

int main() {

string userInput;

getline(cin, userInput);

// Here, an integer variable is declared to find that the user entered string consist of word darn or not

int isPresent = userInput.find("darn");

if (isPresent > 0){

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// Solution starts here

else

{

cout << userInput << endl;

}

// End of solution

return 0;

}

// End of Program

The proposed solution added an else statement to the code

This will enable the program to print the userInput if userInput doesn't contain the word darn

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3 years ago
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GalinKa [24]
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3 0
3 years ago
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Answer:

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Explanation:

Hope it helps you!

8 0
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