Write the definition of a function named averager that receives a double parameter and returns-- as a double -- the average value that it has been passed so far. So, if you make these calls to average, averager(5.0), averager(15.0), averager(4,3), the values returned will be (respectively): 5.0, 10.0, 8.1.

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This is the answer:

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double averager (const double &wrongNumber)
{
static double threeSum = 0.0;
static size_t bodyCount = 0;
bodyCount++;
threeSum += wrongNumber;

return threeSum / bodyCount;
}

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. . Click on the title for the solution . . This is the answer: : #include <iostream> #include <fstream> #include <string> using namespace std; void lookup(); void add(); int main() { string type; while (cin >> type) { if (type == "lookup") lookup(); else  add(); } return 0; } void lookup() { ifstream fin; fin.open("phonedir"); if(!fin.is_open()) cout << "Not Open" << endl; bool check = false; string name,firstName,lastName,email; cin >> name; while (fin >> firstName >> lastName >> email) { if (firstName== name || lastName==name) { cout<<firstName<<" "<<lastName<<" "<<email<<endl; check=true; } } if(!check) cout << name << ": not found" << endl; fin.close(); } void add() { ofstream fout; string name; string fir...
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