Intro to communication disorders paper

The midterm for this course is a 3-5 page double-spaced reflection paper of learning in course. Specifically, you will discuss what you have learned about the classification of communication disorder (5 points) and the Speech-Language Pathology and Audiology profession and professionals (10 points). Please includes specifically your understanding of the scope and practice of speech-language pathologists and audiologists and where they work and with whom they work (10 points).  Additionally, you will reflect on the developmental basis of speech and language including bodily systems involved; and theories, stages and cultural and linguistic of speech and language development (15 points).  Writing mechanics such as diction, grammar, sentence structure, etc. will account for 10 points.  If you have any questions or concerns about this paper, contact me.

In 750-1,000 words, do the following:Analyze the substantive differences between a vocational rehabilitation clinician and a standard mental health clinician, include treatment ideology for both.Descr

Analyze the substantive differences between a vocational rehabilitation clinician and a standard mental health clinician, include treatment ideology for both.

 

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Advanced operating systems project | Computer Science homework help

 There are 4 parts for the project. The question may be long to read but it’s not a heavy work because there are many examples and explanations for the each parts.*Part 1.  The first part of this project requires that you implement a class that will be used to simulate a disk drive. The disk drive will have numberofblocks many blocks where each block has blocksize many bytes. The interface for the class Sdisk should include :

Class Sdisk
{
public :

Sdisk(string diskname, int numberofblocks, int blocksize);

int getblock(int blocknumber, string& buffer);
int putblock(int blocknumber, string buffer);
int getnumberofblocks(); // accessor function
int getblocksize(); // accessor function

private :

string diskname;        // file name of software-disk

int numberofblocks;     // number of blocks on disk
int blocksize;          // block size in bytes
};

An explanation of the member functions follows :

  • Sdisk(diskname, numberofblocks, blocksize) This constructor incorporates the creation of the disk with the “formatting” of the device. It accepts the integer values numberofblocks, blocksize, a string diskname and creates a Sdisk (software-disk). The Sdisk is a file of characters which we will manipulate as a raw hard disk drive. The function will check if the file diskname exists. If the file exists, it is opened and treated as a Sdisk with numberofblocks many blocks of size blocksize. If the file does not exist, the function will create a file called diskname which contains numberofblocks*blocksize many characters. This file is logically divided up into numberofblocks many blocks where each block has blocksize many characters. The text file will have the following structure :  

                                                            -figure 0 (what I attached below)              

  • getblock(blocknumber,buffer) retrieves block blocknumber from the disk and stores the data in the string buffer. It returns an error code of 1 if successful and 0 otherwise.
  • putblock(blocknumber,buffer) writes the string buffer to block blocknumber. It returns an error code of 1 if successful and 0 otherwise.

IMPLEMENTATION GUIDELINES : It is essential that your software satisfies the specifications. These will be the only functions (in your system) which physically access the Sdisk. NOTE that you must also write drivers to test and demonstrate your program.*Part 2.  The second part of this project requires that you implement a simple file system. In particular, you are going to write the software which which will handle dynamic file management. This part of the project will require you to implement the class Filesys along with member functions. In the description below, FAT refers to the File Allocation Table and ROOT refers to the Root Directory. The interface for the class should include :

Class Filesys: public Sdisk
{
Public :
Filesys(string diskname, int numberofblocks, int blocksize);
int fsclose();
int fssynch();
int newfile(string file);
int rmfile(string file);
int getfirstblock(string file);
int addblock(string file, string block);
int delblock(string file, int blocknumber);
int readblock(string file, int blocknumber, string& buffer);
int writeblock(string file, int blocknumber, string buffer);
int nextblock(string file, int blocknumber);

Private :



int rootsize;           // maximum number of entries in ROOT

int fatsize;            // number of blocks occupied by FAT
vector<string> filename;   // filenames in ROOT
vector<int> firstblock; // firstblocks in ROOT
vector<int> fat;             // FAT
};

An explanation of the member functions follows :

  • Filesys() This constructor reads from the sdisk and either opens the existing file system on the disk or creates one for an empty disk. Recall the sdisk is a file of characters which we will manipulate as a raw hard disk drive. This file is logically divided up into number_of_blocks many blocks where each block has block_size many characters. Information is first read from block 1 to determine if an existing file system is on the disk. If a filesystem exists, it is opened and made available. Otherwise, the file system is created.The module creates a file system on the sdisk by creating an intial FAT and ROOT. A file system on the disk will have the following segments:                                                           -figure 1 (what I attached below)           
  • consists of two primary data objects. The directory is a file that consists of information about files and sub-directories. The root directory contains a list of file (and directory) names along with a block number of the first block in the file (or directory). (Of course, other information about the file such as creation date, ownership, permissions, etc. may also be maintained.) ROOT (root directory) for the above example may look something like                                                   -figure 2 (what I attached below)  The FAT is an array of block numbers indexed one entry for every block. Every file in the file system is made up of blocks, and the component blocks are maintained as linked lists within the FAT. FAT[0], the entry for the first block of the FAT, is used as a pointer to the first free (unused) block in the file system. Consider the following FAT for a file system with 16 blocks.  

                                                        -figure 3 (what I attached below)

  • In the example above, the FAT has 3 files. The free list of blocks begins at entry 0 and consists of blocks 6, 8, 13, 14, 15. Block 0 on the disk contains the root directory and is used in the FAT for the free list. Block 1 and Block 2 on the disk contains the FAT. File 1 contains blocks 3, 4 and 5; File 2 contains blocks 7 and 9; File 3 contains blocks 10, 11, and 12. Note that a “0” denotes the end-of-file or “last block”. PROBLEM : What should the value of FAT_size be in terms of blocks if a file system is to be created on the disk? Assume that we use a decimal numbering system where every digit requires one byte of information and is in the set [0..9]. Both FAT and ROOT are stored in memory AND on the disk. Any changes made to either structure in memory must also be immediately written to the disk.  
  • fssynch This module writes FAT and ROOT to the sdisk. It should be used every time FAT and ROOT are modified.
  • fsclose This module writes FAT and ROOT to the sdisk (closing the sdisk).
  • newfile(file) This function adds an entry for the string file in ROOT with an initial first block of 0 (empty). It returns error codes of 1 if successful and 0 otherwise (no room or file already exists).
  • rmfile(file) This function removes the entry file from ROOT if the file is empty (first block is 0). It returns error codes of 1 if successful and 0 otherwise (not empty or file does not exist).
  • getfirstblock(file) This function returns the block number of the first block in file. It returns the error code of 0 if the file does not exist.
  • addblock(file,buffer) This function adds a block of data stored in the string buffer to the end of file F and returns the block number. It returns error code 0 if the file does not exist, and returns -1 if there are no available blocks (file system is full!).
  • delblock(file,blocknumber) The function removes block numbered blocknumber from file and returns an error code of 1 if successful and 0 otherwise.
  • readblock(file,blocknumber,buffer) gets block numbered blocknumber from file and stores the data in the string buffer. It returns an error code of 1 if successful and 0 otherwise.
  • writeblock(file,blocknumber,buffer) writes the buffer to the block numbered blocknumber in file. It returns an appropriate error code.
  • nextblock(file,blocknumber) returns the number of the block that follows blocknumber in file. It will return 0 if blocknumber is the last block and -1 if some other error has occurred (such as file is not in the root directory, or blocknumber is not a block in file.)IMPLEMENTATION GUIDELINES : It is essential that your software satisfies the specifications. These will be the only functions (in your system) which physically access the sdisk.

*Part 3.   The third part of this project requires that you implement a simple shell that uses your file system. This part of the project will require you to implement the class Shell along with member functions. The interface for the class should include :

class Shell: public Filesys
{
Public :

Shell(string filename, int blocksize, int numberofblocks);

int dir();// lists all files
int add(string file);// add a new file using input from the keyboard
int del(string file);// deletes the file
int type(string file);//lists the contents of file
int copy(string file1, string file2);//copies file1 to file2
};

An explanation of the member functions follows :

  • Shell(string filename, int blocksize, int numberofblocks): This will create a shell object using the Filesys on the file filename.
  • int dir(): This will list all the files in the root directory.
  • int add(string file): add a new file using input from the keyboard
  • int del(string file): deletes the file
  • int type(string file): lists the contents of file
  • int copy(string file1, string file2): copies file1 to file2

IMPLEMENTATION GUIDELINES :See the figure 4  (what I attached below) for the ls function of Filesys.See the figure 5 (what I attached below) for dir function of Shell. See the figure 6 (what I attached below)  for main program of Shell.*Part 4.  In this part of the project, you are going to create a database system with a single table which uses the file system from Project II. The input file will consist of records associated with Art History. The data file you will use as input consists of records with the following format: The data (180 records) is in date.txt file (what I attached below)

  • Date : 5 bytes
  • End : 5 bytes
  • Type : 8 bytes
  • Place : 15 bytes
  • Reference : 7 bytes
  • Description : variable

In the data file, an asterisk is also used to delimit each field and the last character of each record is an asterisk. The width of any record is never greater than 120 bytes. Therefore you can block the data accordingly. This part of the project will require you to implement the following class:

Class Table : Public Filesys
{
Public :

Table(string diskname,int blocksize,int numberofblocks, string flatfile, string indexfile);

int Build_Table(string input_file);
int Search(string value);

Private :

string flatfile;
string indexfile;

int IndexSearch(string value);
};

The member functions are specified as follows :

  • Table(diskname,blocksize,numberofblocks,flatfile,indexfile) This constructor creates the table object. It creates the new (empty) files flatfile and indexfile in the file system on the Sdisk using diskname.
  • Build_Table(input_file) This module will read records from the input file (the raw data file described above), add the records to the flatfile and create index records consisting of the date and block number, and then add the index records to the index file. (Note that index records will have 10 bytes .. 5 bytes for the date and 5 bytes for the block number.)
  • Search(value) This module accepts a key value, and searches the index file with a call to IndexSearch for the record where the date matches the specified value. IndexSearch returns the blocknumber of the block in the flat file where the target record is located. This block should then be read and the record displayed.
  • IndexSearch(value) This module accepts a key value, and searches the index file indexfile for the record where the date matches the specified value. IndexSearch then returns the block number key of the index record where the match occurs.

See the figure 7 (what I attached below) for the main program of Shell which includes a search command. 

Wk 4 discussion [no word count requirement, i expect quality work

Review the “Chapter Eleven Case: Zappos Is Passionate for Customers” at the end of Ch. 11 of Business Driven Technology.

Discuss whether you are for or against the following statement in the electronic age: Customer relationships are more important than ever, and Zappos provides the new benchmark that all corporations should follow.

Select a side in the argument and justify your answer. Also, discuss the following:

  • Does Zappos use a CRM?
  • How can using a CRM support business operations?
  • What are the benefits and challenges of using a CRM in a business?
  • What would you suggest as best practices that should be used for a CRM to be successful? Explain your answer.

Respond to at least three of your peers. In your response, address any thoughts you have on CRMs supporting business organizations, benefits/challenges of CRMs, and best practices.

Students need to contribute three substantive posts in this discussion by the due date indicated. The substantive posts can be any combination of responses and replies.

Peer 1 

Good evening Class

Zappos I would say does use CRM now, but back in 1996 I sure that it was not as in depth as now and was not really a think of large use. You have to think the technology of 1996 was not that of today and did not have such a large impact on society. Now I’m sure that they have the best CRM software available on the market now. Small business operations using CRM’s allow for the business to learn there customer, manage administrative needs, gives business personal the most of there time by managing customer orders, supply needs, and logistics, along with HR needs. Benefits of using a CRM system are that it is upstreamed and can be viewed real time, marketing solutions are also a plus and customer management. The downside to the CRM system is that it is very expensive and training an entire staff is very challenging across the field. In my opinion when implementing a new CRM system it is always best to first research different makes and see with best works with your network system. Keep track of your training of employees on your system so that everyone is trained and up to standard and there is never a time where anyone says that they do not understand the system or in the event that training needs to be conducted new employees must be the first to attend this training. Ensure that your IT team is current on any updates that the system requires and make sure that your help desk personal and IT team can answer any questions at a moments notice without hesitation.

Peer 2

Good afternoon,

There are a couple of advantages and disadvantages to CRM. But these disadvantages can be reduced with the support of an appropriate CRM person (internal or external) which will work as the driving force within the organization and should provide the right and ideal practice for optimum utilization and implementation.

Advantages:

  • One place for all customer data.
  • Manages growth
  • Automates sending auto emails and SMS to customers.
  • Builds long term relationships with customers.
  • Seeks out how and where to make improvements

Disadvantages:

  • The transition from manual to automated.
  • Not selecting the right system
  • Poor user adoption
  • Integration with existing software
  • Incorrect implementation

If need be. Take a minute to consider if customer relationship management software is the right fit for your business. Which likely will be a majority of the time. In my opinion it is a very needed part of today’s business.

Peer 3

I agree customer relationship are more important than ever I believe to get ahead in business a lot of companies offer the same thing for example apple music google play and other streaming services some offer exclusive music from artist but they have to be competitive the way Zappos stay competitive is Zappos’s passionate customer service strategy encourages customers to order as many sizes and styles of products as they want, ships them for free, and offers free return shipping. Does Zapo use CRM? I would say yes is a means of managing all aspects of a customer’s relationship with an organization to increase customer loyalty and retention and an organization’s profitability and I say Zappos does this with the free shipping and free return I think CRM can support business operations by 24 hour customer service checking online orders shipping online orders up to 11 pm providing next day shipping and responding to customer inquiries which is the call center that sits in the corporate building as well. The benefits and challenges of using CRM I think one of the benefits like in the reading what if a customer have a problem with one of the products and the sells representative from the company may try to go down to the company to make a sell and the customer already having the product that could be a problem but with the customer service representative the make a ticket log a call the sales representative already should know about the issue before he reaches out the customer maybe he or she can do some damage control and make an up sell or keep the customer happy to be a returning customer. I think customer services is extremely helpful and in house tech support because I know when I worked for OnStar we had user wanted to cancel before we get them and when we get the Onstar System back up and running give them a couple free minutes or 3 months free we was able to keep customers and make them happy and maybe we can up sell the user in the futures.

https://phoenix.vitalsource.com/#/books/9781259852275/cfi/6/52!/4/2/8/[email protected]:100