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MariettaO [177]
2 years ago
11

10. There is repeated code in these onevent blocks. Choose the correct code for the updateScreen() function which would be calle

d in each of the onEvent blocks.
1

var counter

0;

2

3

onEvent("upButton", "click",

function

4

counter

counter + 1;

5

setText ("counter_label", counter);

set Property ("counter_label", "text-color",

6

"red");

7

if

counter

8

set Property ("counter label",

"font-size", 24);

9

10

11

onEvent ("downButton",

"click",

function

12

counter

counter
Computers and Technology
1 answer:
Lorico [155]2 years ago
6 0

Answer:

front zise24

Explanation:

on event down botton

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Assign test_stat_72 to the value of the test statistic for the years 1971 to 1973 using the states in death_penalty_murder_rates
Alika [10]

Answer:

Find attached below the solution and explanation to the problem.

Explanation:

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2 years ago
A computer’s memory is composed of 8K words of 32 bits each. How many bits are required for memory addressing if the smallest ad
victus00 [196]

Answer:

The minimum number of bits necessary to address 8K words is 13.

Explanation:

You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.  

8000 words can be addressed with 2^{n}  units. Now you have to find the value of n that approximates to the number of words.  

2^2=4\\2^4=16\\2^7=128\\2^{10}=1024\\2^{12}=4096\\2^{13}=8192

So you can see that 13 bits are needed to address 8K words.

7 0
2 years ago
You are given an array x of string elements along with an int variable n that contains the number of elements in the array. You
Slav-nsk [51]

Answer:

The code to this question can be given as:

Code:

int i,j,count_previous=0,count_next=0; //define variable

for (j=0; j<n; j++) //loop for array

{

if (x[0]==x[j]) //check condition

{

count_previous++; //increment value by 1.

}

}

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

{

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

{

if (x[i]==x[j]) //check condition

{

count_next++; //increment value by 1.

}

}

if (count_previous>count_next) //check condition

{

mode=x[i-1];

}

else

{

mode=x[i];

count_previous=count_next; //change value.

count_next= 0 ; //assign value.

}

}

Explanation:

In the question it is define that x is array of the string elements that is already define in the question so the code for perform operation in the array is given above. In the code firstly we define the variable that is i, j, count_previous, count_next. The variable i,j is used in the loop and variable count_previous and count_next we assign value 0 that is used for check the values of array. Then we define the for loop in this loop we use conditional statement in the if block we check that array of zero element is equal to array of j value then the count_previous is increase by 1. Then we use nested loop It is also known as loop in a loop. In this first loop is used for array and the second loop is used for check array element.In this we use the condition that if array x of i value is equal to array x of j then count_next will increment by 1.Then we use another condition that is if count_previous is greater then count_next then mode of x is decrement by 1. else mode equal to array and count_previous holds the value of count_next and count_next is equal to 0.

8 0
2 years ago
PLEASE HELP PROGRAMMING WILL GIVE BRAINLIEST
JulijaS [17]
The fourth choice is correct.
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2 years ago
Write a method for the Customer class that that will return a string representing a bill notice when passed a double value repre
Nesterboy [21]

Answer:

Following are the method definition to this question:

public String notice_bill(double amount) //defining method

{

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}

Explanation:

In the given question some information is missing, that is example So, method definition to this question can be described as follows:

  • In the above method definition a string method "notice_bill" is declared, which accepts a double value in its parameter, that is "amount".
  • Inside the method, this keyword is used, that hold values and return its value as a message.  
5 0
2 years ago
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