explain the errors in each error in spreadsheet and how to correct each
Answer:
###### error. Problem: The column is not wide enough to display all the characters in a cell. ...
# Div/0! error. ...
#Name? error. ...
#Value! error. ...
#REF! error. ...
#NUM! error. ...
#NULL error.
Explanation:
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Answer:
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Populations are effected by the birth and death rate, as well as the number of people who move in and out each year. The birth rate is the percentage increase of the population due to births and the death rate is the percentage decrease of the population due to deaths. Write a program that displays the size of a population for any number of years. The program should ask for the following data:
• The starting size of a population
• The annual birth rate
• The annual death rate
• The number of years to display
Write a function that calculates the size of the population for a year. The formula is
N = P + BP − DP
where N is the new population size, P is the previous population size, B is the birth rate, and D is the death rate.
Input Validation: Do not accept numbers less than 2 for the starting size. Do not accept negative numbers for birth rate or death rate. Do not accept numbers less than 1 for the number of years.
#include
using namespace std;
double calcOfPopulation(double, double, double);
double inputValidate(double, int);
int main()
{
double P, // population size
B, // birth rate %
D, // death rate %
num_years;
cout << "Starting size of population: ";
P = inputValidate(P, 2);
cout << "Annual birth rate: %";
B = inputValidate(B, 0);
cout << "Annual death rate: %";
D = inputValidate(D, 0);
cout << "Number of years to display: ";
num_years = inputValidate(num_years, 1);
cout << "Population size for "
<< num_years << " years "
<< " = "
<< (calcOfPopulation(P, B, D) * num_years)
<< endl;
return 0;
} // END int main()
double inputValidate(double number, int limit)
{
while(!(cin >> number) || number < limit)
{
cout << "Error. Number must not be "
<< " 0 or greater:";
cin.clear();
cin.ignore(numeric_limits::max(), '\n');
}
return number;
}
double calcOfPopulation(double P, double B, double D)
{
B *= .01; // 3.33% = .0333
D *= .01; // 4.44% = .0444
return P + (B * P) - (D * P);
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
pleaseeeeeeee tellllllllllllllllllllll
Answer:
visual communication
hope it helps
stay safe healthy and happy.Answer:
visual communication will be the correct answer
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
Please answer it’s timed
Answer:
product safety
Explanation:
brainly:)^
Code a Python program that uses a graphical user interface (GUI) to address a real-world problem. Within reason, you may choose to address any real-world problem that you would like; however, please only choose a real-world problem that is both safe and legal.
Answer:
Explanation:
The following is a simple GUI Python program that solves for the different variables in a Pythagorean theorem equation, which is a real-world problem that many students have to face on a daily basis. The output can be seen in the attached picture below. Due to technical difficulties, I have added the code as a txt file attachment.
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~
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.
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
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:
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.
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}.
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.