In laissez-faire styleof group leadership, members of the group are not given inactivity plans by the leader only when it is requested.
true
false

Determine which of the four levels of measurement (nominal, ordinal, interval, ratio) is most appropriate for the data below. internet speed levels of "fast comma " "medium comma " and "slow" choose the correct answer below. a. the ordinal level of measurement is most appropriate because the data can be ordered comma but differences left parenthesis obtained by subtraction right parenthesis cannot be found or are meaningless. nothing nothing nothing nothing nothing b. the interval level of measurement is most appropriate because the data can be ordered comma differences left parenthesis obtained by subtraction right parenthesis can be found and are meaningful comma and there is no natural starting point. c. the nominal level of measurement is most appropriate because the data cannot be ordered. nothing d. the ratio level of measurement is most appropriate because the data can be ordered comma differences left parenthesis obtained by subtraction right parenthesis can be found and are meaningful comma and there is a natural starting point.

Technician a says that if a valve is open when a piston rises to the very top of a cylinder, the piston may actually strike the valve head and cause serious engine damage. technician b says if the camshaft is located in the engine block, then the engine is called an overhead valve engine, ohv engine, or an in-block camshaft. who is right? a. b only b. both a and b c. a only d. neither a nor b

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }