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TEA [102]
3 years ago
5

Pls awnser if awnser is wrong I will unmark brainliest

Computers and Technology
1 answer:
choli [55]3 years ago
3 0

Answer: i think 1, 3, and 4

Explanation:

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On the first line, place your name in a comment. Create a program that does the following: Take in three integer numbers from th
Helen [10]

Answer and Explanation:

//buchi

Var firstNumber=prompt("please enter first number");

Var secondNumber=prompt("please enter second number");

Var thirdNumber=prompt("please enter third number");

Var numberTotal=firstNumber+secondNumber+thirdNumber;

Function calculateNumbers(){

return numberTotal;

return int(numberTotal/3);

return firstNumber*secondNumber*thirdNumber;}

Console.log(calculateNumbers());

8 0
3 years ago
You are a visitor at a political convention with delegates; each delegate is a member of exactly one political party. It is impo
Fed [463]

Answer:

The algorithm is as follows:

Step 1: Start

Step 2: Parties = [All delegates in the party]

Step 3: Lent = Count(Parties)

Step 4: Individual = 0

Step 5: Index = 1

Step 6: For I in Lent:

Step 6.1: If Parties[Individual] == Parties[I]:

Step 6.1.1: Index = Index + 1

Step 6.2: Else:

Step 6.2.1 If Index == 0:

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

Step 7: Else

Step 7.1: Index = Index - 1

Step 8: Print(Party[Individual])

Step 9: Stop

Explanation:

The algorithm begins here

Step 1: Start

This gets the political parties as a list

Step 2: Parties = [All delegates in the party]

This counts the number of delegates i.e. the length of the list

Step 3: Lent = Count(Parties)

This initializes the first individual you come in contact with, to delegate 0 [list index begins from 0]

Step 4: Individual = 0

The next person on the list is set to index 1

Step 5: Index = 1

This begins an iteration

Step 6: For I in Lent:

If Parties[Individual] greets, shakes or smile to Party[i]

Step 6.1: If Parties[Individual] == Parties[I]:

Then they belong to the same party. Increment count by 1

Step 6.1.1: Index = Index + 1

If otherwise

Step 6.2: Else:

This checks if the first person is still in check

Step 6.2.1 If Index == 0:

If yes, the iteration is shifted up

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

If the first person is not being checked

Step 7: Else

The index is reduced by 1

Step 7.1: Index = Index - 1

This prints the highest occurrence party

Step 8: Print(Party[Individual])

This ends the algorithm

Step 9: Stop

The algorithm, implemented in Python is added as an attachment

<em>Because there is an iteration which performs repetitive operation, the algorithm running time is: O(n) </em>

Download txt
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10. Two technicians are discussing recycling specifications for refrigerant. Technician A says that refrigerant oil with 5,000 p
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a

Explanation:

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In cell j6, enter a formula to determine the cost of gas if the customer chose that option as indicated in column i. if the cust
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a t t i c

Explanation:

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Who wants to be my tutor?
anzhelika [568]

Answer:

i would but im not big brain

Explanation:

8 0
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