Why is it important to try to make financial decisions without emotions.

Signal sets company contracts to deliver one hundred 52-inch plasma high-definition television sets to a new retail customer, tuner tv store, on may 1, with payment to be made on delivery. signal tenders delivery in its own truck. tuner's manager notices that some of the cartons have scrape marks. tuner's owner phones signal's office and asks whether the sets might have been damaged as they were being loaded. signal assures tuner that the sets are in perfect condition. tuner tenders signal a check, which signal refuses, claiming that the first delivery to new customers is always for cash. tuner promises to pay the cash within two days. signal leaves the sets with tuner, which stores them in its warehouse pending its "grand opening sale" on may 15. two days later, tuner's stocker opens some of the cartons and discovers that a number of the sets are damaged beyond ordinary repair. signal claims tuner has accepted the sets and is in breach by not paying on delivery. will signal succeed on these claims? explain.

Match the following. 1. show grouping of word processing tasks that can be performed quick access toolbar 2. shortcut location for commonly used elements scroll bars 3. organized commands used to modify documents ribbon 4. used to align and measure content in a word screen zoom bar 5. vertical and horizontal bars that are used to navigate through a document contextual tabs 6. displays the name of the document in use ruler 7. allows users to enlarge or shrink a visual of a word document title bar

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; }