You are the network administrator for your organization. Your DHCP server (Server1) has a scope of 10.10.16.0 to 10.10.16.254 with a subnet mask of /20.

How would the Get-DhcpServerv4Scope ensure that all of the client computers obtain an IP address from Server1.

python3

Let x0,...,xn−1 be n numbers stored in the list x. The median of x, denoted median(x) is the “middle”-value of the numbers in x in the sense that half of the numbers in x are less than the median and the other half of the numbers in x are greater than the median. Let y denote a copy of x sorted in ascending order. Then median(x)=(y[n+1 2 −1] if n is oddy [n 2−1]+y[n 2] 2 if n is even. Complete the function my_median with the following specifications: • It that takes in one parameter, x, which is a list of numbers. • It returns median(x). • Do not alter x. Evaluating your function my_median(x) should not change the value of the list x. • Do not use Python’s built-in median function.

Raintree Cosmetic Company sells its products to customers on a credit basis. An adjusting entry for bad debt expense is recorded only at December 31, the company’s fiscal year-end. The 2017 balance sheet disclosed the following:

During 2018, credit sales were $1,750,000, cash collections from customers $1,830,000, and $35,000 in accounts receivable were written off. In addition, $3,000 was collected from a customer whose account was written off in 2017. An aging of accounts receivable at December 31, 2018, reveals the following:

Required:

1. Prepare summary journal entries to account for the 2018 write-offs and the collection of the receivable previously written off.
2. Prepare the year-end adjusting entry for bad debts according to each of the following situations:

1. Bad debt expense is estimated to be 3% of credit sales for the year.
2. Bad debt expense is estimated by computing net realizable value of the receivables. The allowance for uncollectible accounts is estimated to be 10% of the year-end balance in accounts receivable.
3. Bad debt expense is estimated by computing net realizable value of the receivables. The allowance for uncollectible accounts is determined by an aging of accounts receivable.

3. For situations (a)–(c) in requirement 2 above, what would be the net amount of accounts receivable reported in the 2018 balance sheet?

Please answer question #3, thank you!

1. The larger the calculated χ² value in a goodness-of-fit test, the _______ likely the data fits the hypothesized distribution.

2.

If the observed frequencies are exactly equal to the expected frequencies in a chi-square test, the value of χ² is:

a. close to 1

b. a large positive value

c. 0

d. a very small positive value

3. A chi-square statistic can be used to analyze contingency tables:

True or False

4. The chi-square statistic for a 6x4 contingency table will have ____ degrees of freedom.

5. The chi-square test can be used to determine if there is a significant difference between two sample proportions:

True or False

6. Which of the following null hypotheses cannot be tested with a chi-squared test?

a. H_o: the two variables defining the contingency table are independent.

b. H_o: the population distribution is a normal distribution.

c. H_o: the true category proportions are the same for all of the population.

d. H_o: the true population means are the same for all of the populations

7. To test the null hypothesis that a population proportion is p = 0.40 using a sample size of n = 12, we use the ____________ distribution.

a. t

b. normal

c. binomial

d. chi-square

8. The null hypothesis that two processes produce the same proportion of defectives can be written:

a. p = 0

b. x₁/n₁ = x₂/n₂

c. p₁ - p₂ = 0

d. x/n = 0

9. The chi-square distribution is symmetrical:

True or False

10. The test statistics for a small-sample test concerning proportion is the ___________. [hint: use all lowercase letters]

tcp 16115

#include <iostream>
#include <fstream>
#include <vector>
#include <string>
using namespace std;

// Define a Person class, including age, gender, and yearlyIncome.
class Person {
public:
Person();
void Print();
void SetData(int a); // FIXME Also set gender and yearly income
void SetGender(string gender);
void SetIncome(int income);
int GetAge();
string GetGender();
int GetIncome();

private:
int age;
string gender;
int yearlyIncome;
};

// Constructor for the Person class.
Person::Person() {
age = 0;
gender = "default";
yearlyIncome = 0;
return;
}

// Print the Person class.
void Person::Print() {
cout << "Age = " << this->age
<< ", gender = " << this->gender
<< ", yearly income = " << this->yearlyIncome
<< endl;
return;
}

// Set the age, gender, and yearlyIncome of a Person.
void Person::SetData(int a) { // FIXME Also set gender and yearly income
this->age = a;
return;
}

void Person::SetGender(string gender){
this->gender = gender;
}

void Person::SetIncome(int income){
this->yearlyIncome = income;
}
// Get the age of a Person.
int Person::GetAge() {
return this->age;
}

string Person::GetGender(){
return this->gender;
}

int Person::GetIncome() {
return this->yearlyIncome;
}

// Get a filename from program arguments, then make a Person for each line in the file.
bool ReadPeopleFromFile(int argc, char* argv[], vector<Person> &people) {
Person tmpPrsn;
ifstream inFS;
int tmpAge = 0;
string tmpGender = "";
int tmpYI = 0;

if (argc != 2) {
cout << "\nUsage: [EXECUTABLE FILE] [TEXT DATA FILE], e.g. myprog.exe dev_people.txt" << endl;
return true; // indicates error
}

cout << "Opening file " << argv[1] << ".\n";
inFS.open(argv[1]); // Try to open file
if (!inFS.is_open()) {
cout << "Could not open file " << argv[1] << ".\n";
return true; // indicates error
}

while (!inFS.eof()) {
inFS >> tmpAge;
inFS >> tmpGender;
inFS >> tmpYI;
tmpPrsn.SetData(tmpAge); // FIXME Also set gender and yearly income
tmpPrsn.SetGender(tmpGender);
tmpPrsn.SetIncome(tmpYI);
tmpPrsn.Print();
people.push_back(tmpPrsn);
}
inFS.close();
cout << "Finished reading file." << endl;

return false; // indicates no error
}

// Ask user to enter age range.
void GetUserInput(int &ageLowerRange, int &ageUpperRange, string &gender, int &yILowerIncome,int &yIHigherIncome) {
cout<<"\nEnter lower range of age: ";
cin >> ageLowerRange;

cout << "Enter upper range of age: ";
cin >> ageUpperRange;

cout << "Enter the gender: ";
cin >> gender;

cout << "Enter the lower range of yearlyIncome: ";
cin >> yILowerIncome;

cout << "Enter the higher range of yearlyIncome: ";
cin >> yIHigherIncome;

return;
}

// Return people within the given age range.
vector<Person> GetPeopleInAgeRange(vector<Person> ppl, int lowerRange, int upperRange) {
unsigned int i = 0;

vector<Person> pplInAgeRange;
int age = 0;
for (i = 0; i < ppl.size(); ++i) {
age = ppl.at(i).GetAge();
if ((age >= lowerRange) && (age <= upperRange)) {
pplInAgeRange.push_back(ppl.at(i));
}
}

return pplInAgeRange;
}

// Return people with same Gender.
vector<Person> GetPeopleWithSpecificGender(vector<Person> ptntlCstmrs,string gender){
vector<Person> pplWithSameGender;
string gndr;
for (int i=0;i<ptntlCstmrs.size();i++){
gndr = ptntlCstmrs.at(i).GetGender();
if (gndr.compare(gender) == 0){
pplWithSameGender.push_back(ptntlCstmrs.at(i));
}
}
return pplWithSameGender;
}

// Return people within the given income range.
vector<Person> GetPeopleInIncomeRange(vector<Person> ptntlCstmrs,int lowerRange, int higherRange){
vector<Person> pplInIncomeRange;
int range = 0;
for (int i=0;i<ptntlCstmrs.size();i++){
range = ptntlCstmrs.at(i).GetIncome();
cout << range << endl;
if ((range >= lowerRange) && (range <= higherRange)){
pplInIncomeRange.push_back(ptntlCstmrs.at(i));
}
}
return pplInIncomeRange;
}

int main(int argc, char* argv[]) {
vector<Person> ptntlCstmrs;
bool hadError = false;
int ageLowerRange = 0;
int ageUpperRange = 0;
string gender;
int yILowerIncome = 0;
int yIHigherIncome = 0;

hadError = ReadPeopleFromFile(argc, argv, ptntlCstmrs);
if( hadError ) {
return 1; // indicates error
}

GetUserInput(ageLowerRange, ageUpperRange, gender, yILowerIncome, yIHigherIncome );
ptntlCstmrs = GetPeopleInAgeRange(ptntlCstmrs, ageLowerRange, ageUpperRange);
ptntlCstmrs = GetPeopleWithSpecificGender(ptntlCstmrs,gender);
ptntlCstmrs = GetPeopleInIncomeRange(ptntlCstmrs,yILowerIncome,yIHigherIncome);
cout << "\nNumber of potential customers = "<<ptntlCstmrs.size() << endl;

return 0;
}

this is my code for my assignment, my professor said i'm missing

Based on that rubric, it looks like the code is missing two things:
* Person class in separate files

* Output persons with user-entered gender works for "any"

can someone help me with this?

Draw and label a schematic of an Ag/AgCl reference electrode. As part of your answer, include the ½ reaction that is utilized by this electrode and include a description of how reference electrodes are utilized in electrochemicalmeasurements

Garrett Kelly. Corporation's identified their current production capacity to range from 6,500 units to 22,000 units. when it produced and sold 12,000 units, its average costs per unit are as follows:

1) If 15,000 units are produced, the total amount of Prime Costs incurred is closest to: ______
2) Determine the Conversion Costs for 11,000 units.
3) Determine the Total Product Costs for 12,000 units.
4) Detemine Total Period Costs when 12,000 units are sold.

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 12.0 pC . The inner cylinder has a radius of 0.500 mm , the outer one has a radius of 7.60 mm , and the length of each cylinder is 24.0 cm .

Part A

What is the capacitance?

Use 8.854×10⁻¹² F/m for the permittivity of free space.

C=___ F

Part B

What applied potential difference is necessary to produce these charges on the cylinders?

V=___ V

In radical chlorination of alkanes, non-equivalent hydrogens react with chlorine atoms at different rates. At 35

State whether each of the following events will result in a movement
along the demand curve for McDonald’s Big Mac hamburger or whether
it will cause the curve to shift. If the demand curve shifts indicate
whether it will shift to the right or to the left, and draw the graph to
illustrate the shift. Briefly explain your answers.

a. The price of Burger King’s Whopper hamburger declines.

b McDonald’s distributes coupons that offer a $3 discount on the
purchase of a Big Mac.

c Because of the shortage of potatoes, the price of French fries
increases.

Need someone to fix my code: based on this version, give me another version without adding struct student.

#include <iostream>
#include <iomanip>
using namespace std;

struct student
{
double firstQuizz;
double secondQuizz;
double midTerm;
double finalTerm;
double overallScore;
char gradeLetter;
string name;
};

void getStudentData(student &s);
void calcPercentage(student &s);
void gradeLetter(student &s);
void display(student s[],int n);

int main()
{

int n;
cout<<"enter the number of students"<<endl;
cin>>n;
struct student students[n];

int i; struct student istudent;
for(i=0;i<n;i++)
{
   cout<<":: Student#"<<(i+1)<<" ::"<<endl;
getStudentData(students[i]);
}

for(i=0;i<n;i++)
{
   calcPercentage(students[i]);
   gradeLetter(students[i]);
}

display(students,n);

return 0;
}

void getStudentData(student &s)
{
   cin.ignore();
   cout<<"Student name?";
getline(cin,s.name);
cout<<"Enter marks in first quizz :";
cin>>s.firstQuizz;
cout<<"Enter marks in second quizz :";
cin>>s.secondQuizz;
cout<<"Enter marks in mid term :";
cin>>s.midTerm;
cout<<"Enter marks in final term :";
cin>>s.finalTerm;
}

Programming language: C++
void calcPercentage(student &s)
{
double overAllScore = (s.firstQuizz + s.secondQuizz)*5 * 0.25 + s.midTerm * 0.25 + s.finalTerm* 0.50;
s.overallScore=overAllScore;
}
void gradeLetter(student &s)
{
   double average=s.overallScore;
   char gradeLetter;
  
if (average >= 90 && average<=100)
gradeLetter = 'A';
else if (average >= 80 && average < 90)
gradeLetter = 'B';
else if (average >= 70 && average < 80)
gradeLetter = 'C';
else if (average >= 60 && average < 70)
gradeLetter = 'D';
else if (average < 60)
gradeLetter = 'F';

s.gradeLetter=gradeLetter;
}
void display(student s[],int n)
{
       //setting the precision to two decimal places
   std::cout << std::setprecision(2) << std::fixed;

   cout<<setw(15)<<left<<"Name"<<setw(15)<<right<<"Overall Score"<<setw(15)<<right<<"Grade Letter"<<endl;
   cout<<setw(15)<<left<<"----"<<setw(15)<<right<<"-------------"<<setw(15)<<right<<"------------"<<endl;
   for(int i=0;i<n;i++)
   {
   cout<<setw(15)<<left<<s[i].name<<setw(15)<<right<<s[i].overallScore<<setw(15)<<right<<s[i].gradeLetter<<endl;
   }
  
}

Programming language: C++

Requirement: based on this version, give me another version without adding struct student.

Given a value of the money supply that the Fed chooses, the equilibrium interest rate can be read off of the money demand schedule. The quantity of money demanded (Md) depends negatively on the interest rate. An increase in P and/or Y shifts the money demand curve to the right. Assume that the inverse money demand is given as r = (10+Y) -0.05 Md. Y =$19.1 (trillion) and Ms=581.1. Using Excel create a spreadsheet with the column headings Ms, r, Md, and M*P. Let’s sstart with no inflation, i.e., P=1.00. Fill in the spreadsheet’s cells for r= 0.035 to r=0.060 in increments of 0.001. (Note: All your answers should be rounded to the nearest thousandth (third digits after the decimal point. Ex: 0.0143 =>0.014, 0.1866=> 0.187) What is the equilibrium interest rate? ______ How much of the money is demanded by the economy at the equilibrium interest rate? ______ Assume that the nation’s GDP increased from 19.1 to 19.2. What happens in the money market at the existing interest rate? ______ (shortage or surplus) of money by $ ______ What is the eventual interest rate as a result of increase in GDP? ______ Assume that the price level increased by 0.5% during the year. GDP remains at $19.1 (trillion). What happens in the money market at the existing interest rate? ______ (shortage or surplus) of money by $ ______.

Tukey's Post test. Alpha = .05. Group I: 0, 2, 2, 0, 1 Group II: 4, 6, 1, 5, 4 Group III: 1, 3, 0, 1, 0. Step by Step how to solve