Answer:
Yes the answer is flexible.
Explanation:
I took the test and got it right.
Consider the following code segment, where num is an integer variable.
int [][] arr = {{11, 13, 14 ,15},
{12, 18, 17, 26},
{13, 21, 26, 29},
{14, 17, 22, 28}};
for (int j = 0; j < arr.length; j++)
{
for (int k = 0; k < arr[0].length; k++)
{ if (arr[j][k] == num){System.out.print(j + k + arr[j][k] + " ");
}
}
}
What is printed when num has the value 14?
Answer:
Following are the complete code to the given question:
public class Main//main class
{
public static void main(String[] args) //main method
{
int [][] arr = {{11, 13, 14 ,15},{12, 18, 17, 26},{13, 21, 26, 29},{14, 17, 22, 28}};//defining a 2D array
int num=14;//defining an integer variable that holds a value 14
for (int j = 0; j < arr.length; j++)//defining for loop to hold row value
{
for (int k = 0; k < arr[0].length; k++)//defining for loop to hold column value
{
if (arr[j][k] == num)//defining if block that checks num value
{
System.out.print(j + k + arr[j][k] + " ");//print value
}
}
}
}
}
Output:
16 17
Explanation:
In the question, we use the "length" function that is used to finds the number of rows in the array. In this, the array has the 4 rows when j=0 and k=2 it found the value and add with its index value that is 0+2+14= 16.similarly when j=3 and k=0 then it found and adds the value which is equal to 3+0+14=17.So, the output is "16,17".10. This question refers to the chart from the previous question.
Which of the following conclusions is BEST supported by the scatter plot?
OOOO
A. Most dogs breeds have a maximum weight of 100 lbs or more
B. Most dog breeds have a maximum life span of 10 or fewer years
C. All dog breeds that weigh less than 50 pounds have a maximum lifespan of more than 10 years
D. No dog breeds that weigh more than 150 pounds are expected to live more than 10 years.
Answer: C. All dog breeds that weigh less than 50 pounds have a maximum lifespan of more than 10 years.
Explanation:
Looking at the scatter plot, it is shown that all dogs that weigh less than 50 pounds have a maximum lifespan that is above 10 years with a number of them even approaching 20 years.
This means that on average, dogs that weigh less tend to live longer than dogs that weigh more which as shown in the scatter plot have a lower lifespan the heavier they are.
Landing pages in a foreign language should never be rated fully meets?
Answer:
if the landing page provides all kind information of information as to that site people usually like it or will most likely enjoy it
BRAINLIEST?????
Explanation:
When parameters are passed between the calling code and the called function, formal and actual parameters are matched by: a.
Answer:
their relative positions in the parameter and argument lists.
Explanation:
In Computer programming, a variable can be defined as a placeholder or container for holding a piece of information that can be modified or edited.
Basically, variable stores information which is passed from the location of the method call directly to the method that is called by the program.
For example, they can serve as a model for a function; when used as an input, such as for passing a value to a function and when used as an output, such as for retrieving a value from the same function. Therefore, when you create variables in a function, you can can set the values for their parameters.
A parameter can be defined as a value that must be passed into a function, subroutine or procedure when it is called.
Generally, when parameters are passed between a calling code and the called function, formal and actual parameters are usually matched by their relative positions in the parameter and argument lists.
A formal parameter is simply an identifier declared in a method so as to represent the value that is being passed by a caller into the method.
An actual parameter refers to the actual value that is being passed by a caller into the method i.e the variables or values that are passed while a function is being called.
You are in the process of implementing a network access protection (NAP) infrastructure to increase your network's security. You are currently configuring the remediation network that non-compliant clients will connect to in order to become compliant. The remediation network needs to be isolated from the secure network. Which technology should you implement to accomplish this task
Answer:
Network Segmentation
Explanation:
The best technology to implement in this scenario would be Network Segmentation. This allows you to divide a network into various different segments or subnets. Therefore, this allows you to isolate the remediation network from the secure network since each individual subnet that is created acts as its own network. This also allows you to individually implement policies to each one so that not every network has the same access or permissions to specific data, traffic, speeds, etc. This would solve the issues that you are having and add multiple benefits.
Popular periodicals might include newspapers
True or false?
Answer:
False
Explanation:
Because there is no way that would be correct!
Answer:
True
Explanation:
It makes sense
~Plz tap the crown ~
~Thank you~
Given positive integer n, write a for loop that outputs the even numbers from n down to 0. If n is odd, start with the next lower even number.
if(n % 2 == 0){
for(int i = n; i >= 0; i-=2){
System.out.println(i);
}
}
else{
for(int i = n - 1; i >= 0; i-=2){
System.out.println(i);
}
}
Sample output
Output when n = 12
12
10
8
6
4
2
0
Output when n = 21
20
18
16
14
12
10
8
6
4
2
0
Explanation:The above code is written in Java.
The if block checks if n is even by finding the modulus/remainder of n with 2. If the remainder is 0, then n is even. If n is even, then the for loop starts at i = n. At each cycle of the loop, the value of i is reduced by 2 and the value is outputted to the console.
If n is odd, then the else block is executed. In this case, the for loop starts at i = n - 1 which is the next lower even number. At each cycle of the loop, the value of i is reduced by 2 and the value is outputted to the console.
Sample outputs for given values of n have been provided above.
To iterate through (access all the entries of) a two-dimensional arrays you
need for loops. (Enter the number of for loops needed).
Answer: You would need two loops to iterate through both dimensions
Explanation:
matrix = [[1,2,3,4,5],
[6,7,8,9,10],
[11,12,13,14,15]]
for rows in matrix:
for numbers in rows:
print(numbers)
The first loop cycles through all the immediate subjects in it, which are the three lists. The second loop calls the for loop variable and iterates through each individual subject because they are lists. So the first loop iterates through the 1st dimension (the lists) and the seconds loop iterates through the 2nd dimension (the numbers in the lists).
To iterate through (access all the entries of) a two-dimensional arrays you need two for loops.
What is an array?A collection of identically typed items, such as characters, integers, and floating-point numbers, are kept together in an array, a form of data structure.
A key or index that can be used to retrieve or change the value kept in that location identifies each element in the array.
Depending on how many indices or keys are used to identify the elements, arrays can be one-dimensional, two-dimensional, or multi-dimensional.
Two nested for loops are often required to iterate over a two-dimensional array: one to loop through the rows, and the other to loop through the columns.
As a result, two for loops would be required to access every entry in a two-dimensional array.
Thus, the answer is two for loops are required.
For more details regarding an array, visit:
https://brainly.com/question/19570024
#SPJ6
How to use the RANK Function in Microsoft Excel
Answer:
=RANK (number, ref, [order])
See Explanation
Explanation:
Literally, the rank function is used to rank values (i.e. cells) in a particular order (either ascending or descending).
Take the following instances:
A column used for total sales can use rank function to rank its cells from top sales to least.
A cell used for time can also use the rank function to rank its cells from the fastest time to slowest.
The syntax of the rank function is:
=RANK (number, ref, [order])
Which means:
[tex]number \to[/tex] The rank number
[tex]ref \to[/tex] The range of cells to rank
[tex]order \to[/tex] The order of ranking i.e. ascending or descending. This is optional.
A pharmaceutical company is going to issue new ID codes to its employees. Each code will have three letters followed by one digit. The letters and and the digits , , , and will not be used. So, there are letters and digits that will be used. Assume that the letters can be repeated. How many employee ID codes can be generated
Answer:
82,944 = total possible ID's
Explanation:
In order to find the total number of combinations possible we need to multiply the possible choices of each value in the ID with the possible choices of the other values. Since the ID has 3 letters and 1 digit, and each letter has 24 possible choices while the digit has 6 possible values then we would need to make the following calculation...
24 * 24 * 24 * 6 = total possible ID's
82,944 = total possible ID's
Write a program second.cpp that takes in a sequence of integers, and prints the second largest number and the second smallest number. Note that in the case of repeated numbers, we really mean the second largest and smallest out of the distinct numbers (as seen in the examples below). You may only use the headers: and . Please have the output formatted exactly like the following examples: (the red is user input)
Answer:
The program in C++ is as follows:
#include <iostream>
#include <vector>
using namespace std;
int main(){
int n;
cout<<"Elements: ";
cin>>n;
vector <int>num;
int input;
for (int i = 1; i <= n; i++){ cin>>input; num.push_back(input); }
int large, seclarge;
large = num.at(0); seclarge = num.at(1);
if(num.at(0)<num.at(1)){ large = num.at(1); seclarge = num.at(0); }
for (int i = 2; i< n ; i ++) {
if (num.at(i) > large) {
seclarge = large;;
large = num.at(i);
}
else if (num.at(i) > seclarge && num.at(i) != large) {
seclarge = num.at(i);
}
}
cout<<"Second Largest: "<<seclarge<<endl;
int small, secsmall;
small = num.at(1); secsmall = num.at(0);
if(num.at(0)<num.at(1)){ small = num.at(0); secsmall = num.at(1); }
for(int i=0; i<n; i++) {
if(small>num.at(i)) {
secsmall = small;
small = num.at(i);
}
else if(num.at(i) < secsmall){
secsmall = num.at(i);
}
}
cout<<"Second Smallest: "<<secsmall;
return 0;
}
Explanation:
See attachment for explanation