Answer:
See attachment for program source file
Explanation:
The source code in the question is not incorrect; just that, it was poorly formatted.
So, what I did it that:
I edited the source code to be in the right formats.numeric_limits needs the <limits> header file to work properly. The <limits> header file was not included in the original source file; so, I added #include<limits> at the beginning of the source fileI corrected the wrongly declared function prototypes.See attachment for the modified program source file
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.
Which of the following describes a characteristic of organic light-emitting diodes (OLEDs) used in clothing?
uniform
flexible
transparent
sizeable
Answer:
Yes the answer is flexible.
Explanation:
I took the test and got it right.
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.
Write functions to compute a subset, find member, union, and intersection of sets. Follow the steps below:
1. Read two integers from the user.
2. Suppose one of the integers is 2311062158. The binary equivalent of this integer stored in a register will be 1000 1001 1100 0000 0000 0010 1000 1110. This data should be regarded as bit strings representing subsets of the set {1, 2, … 32}. If the bit string has a 1 in position i, then element i is included in the subset. Therefore, the string: 1000 1001 1100 0000 0000 0010 1000 1110 corresponds to the set: {2, 3, 4, 8, 10, 23, 24, 25, 28, 32}.
3. Print out members of the set from smaller to larger. You can do a loop from 1 to 32. Load a masking bit pattern that corresponded to the position number of the loop counter (0x00000001 for 1). Isolate the bit in the operand by using the AND operation. If the result of the AND is not 0 then the loop counter is in the set and should be displayed. Increment the counter and shift the masking bit pattern to the left.
4. Read a number from the user. Determine if that element is a member of the given sets.
5. Determine the union of two sets.
6. Determine the intersection of two sets.
7. Implement a loop back to the main function. See the prompts below: "Enter the first number:" "Enter the second number:" "Members of Set 1:" "Members of Set 2:" "Enter an element to find:" "It is a member/ not a member of set 1" "It is a member/ not a member of set 2" "Union of the sets:" "Intersection of the sets:" "Do you want to compute set functions again?"
8. Test the program using the following data:
Enter the first number: 99999
Enter the second number: 111445
Members of set 1: 1 2 3 4 5 8 10 11 16 17
Members of set 2: 1 3 5 7 9 10 13 14 16 17
Enter an element to find: 7
It is not a member of set 1
It is a member of set 2
Union of the sets: 1 2 3 4 5 7 8 9 10 11 13 14 16 17
Intersection of the sets: 1 3 5 10 16 17
Explanation:
Suppose one of the integers is 2311062158. The binary equivalent of this integer stored in a register will be 1000 1001 1100 0000 0000 0010 1000 1110. This data should be regarded as bit strings representing subsets of the set {1, 2, … 32}. If the bit string has a 1 in position i, then element i is included in the subset. Therefore, the string: 1000 1001 1100 0000 0000 0010 1000 1110 corresponds to the set: {2, 3, 4, 8, 10, 23, 24, 25, 28, 32}.
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:
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
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Answer:
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