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Research results indicate over 80 percent of PS3 owners watch Blu-ray titles and the combined player sales are meaningful, but comparing DVD and Blu-ray player sales head to head does not paint a very precise picture. Disc sales, perhaps, tell a more consequential story. In the first two years, somewhere between 16.3 and 30.6 million DVDs were sold in the US, depending on who is counting. It seems only fair to combine Blu-ray and HD DVD sales for the period since the two formats were competing for the same customers and were spread fairly equally across the studios. High-definition discs of both flavors sold somewhere between 6.1 and 11.25 million units in the US between June of 2006 and the end of 2007. That is more or less half of DVD sales during the equivalent period. In October 2000, the ten millionth US DVD player shipped. Three-and-a-half years after their US introduction, DVD players achieved the mark that VCRs and CD players each took eight years to reach. By the end of 2007, over 98 million US households (84 percent) owned a DVD player. This compares to 112 million US households (97 percent) in 2007 that owned a TV. The total number of DVD playback devices including players, DVD computers, and DVD game consoles passed the one billion mark in 2007. DVD disc sales reached their peak from 2004 to 2006, generating over $24 billion in US sales and rentals each year. Worldwide DVD sales and rentals in 2006 were over $50 billion. In June 2008, Futuresource Consulting released a report that projected combined sales of Blu-ray players and PS3 consoles in Western Europe would reach 10.48 million units by the end of the year. The report pointed out that only 1.58 million DVD players sold during the similar time period (1997 to 1999). However, DVD was not officially launched in Europe until the middle of 1998, so it was not a fair comparison. And, of course, there is the PS3 effect to consider.
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11. Convert the ERD shown in Figure 6.CP11 into tables. List the conversion rules used and the re sulting changes to the tables.
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Which nerve can be injured during delivery, with subsequent urinary retention Following a 300-minute second stage of labor, a G1P1 is unable to flex her hips against tension and has difficulty walking. What type of neuropathy is likely responsible Which episiotomy is associated with perineal or pelvic hematoma How soon should an Rh(D-) mother be given Rhogam (anti-D immune globulin), following delivery of an Rh+ fetus Where is the fundus during the postpartum period
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5. Learning style of learner (auditory, visual, or kinesthetic) Information is best communicated using the learner s preferred learning style. Learners with an auditory style prefer to hear the information; visual individuals want to see information, often preferring graphs and illustrations to words; kinesthetic learners want direct and thorough discussions of the information and prefer written materials and conversations that include the implications for action but not a prescription for action they can take as a result of the information. Verbal cues often indicate an individual s primary mode, especially the nouns and verbs the person most commonly uses. Following are three examples of how one person might describe the same successful work project, depending on which modality he or she uses.
Since null-terminated character arrays are not technically data types in their own right, the C++ operators cannot be applied to them. This makes even the most rudimentary string operations clumsy. More than anything else, it is the inability to operate on null-terminated strings using the standard C++ operators that has driven the development of a standard string class. Remember, when you define a class in C++, you are defining a new data type that can be fully integrated into the C++ environment. This, of course, means that the operators can be overloaded relative to the new class. Therefore, by adding a standard string class, it becomes possible to manage strings in the same way as any other type of data: through the use of operators. There is, however, one other reason for the standard string class: safety. In the hands of an inexperienced or careless programmer, it is very easy to overrun the end of an array that holds a null-terminated string. For example, consider the standard string copy function, strcpy( ). This function contains no provision for checking the boundary of the target array. If the source array contains more characters than the target array can hold, then a program error or system crash is possible (likely). As you will see, the standard string class prevents such errors. In the final analysis, there are three reasons for the inclusion of the standard string class: consistency (a string now defines a data type), convenience (you can use the standard C++ operators), and safety (array boundaries will not be overrun). Keep in mind that there is no reason that you should altogether abandon normal, null-terminated strings. They are still the most efficient way in which to implement strings. However, when speed is not an overriding concern, using the new string class gives you access to a safe and fully integrated way to manage strings. Although not traditionally thought of as part of the STL, string is another container class defined by C++. This means that it supports the algorithms described in the previous section. However, strings have additional capabilities. To have access to the string class, you must include <string> in your program. The string class is very large, with many constructors and member functions. Also, many member functions have multiple overloaded forms. For this reason, it is not possible to look at the entire contents of string in this chapter. Instead, we will examine several of its most commonly used features. Once you have a general understanding of how string works, you will be able to easily explore the rest of string s features on your own. The prototypes for three of string s most commonly used constructors are shown here: string( ); string(const char *str); string(const string &str);
Hence, integration, which is the inverse operation to differentiation, results in division by j in the phasor domain. Given a sinusoidal function f (t) with phasor transform F we have the phasor transform pair
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