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grigory [225]
2 years ago
10

Jim has excellent oral and written communication skills. He enjoys public speaking and wants a job in which he will interact wit

h the press and other communication professionals. Which career would allow him to use his skills, talents, and abilities the most? A. public relations specialist B.budget analyst C.marketing specialist D.freight forwarder E.transportation manager
Computers and Technology
2 answers:
fredd [130]2 years ago
7 0
The best job for Jim would be public relations specialist.
asambeis [7]2 years ago
7 0

Based on Jim’s capabilities and his preference, the job that would fit him best would be (A) public relations specialist.

A public relations specialist is <em>responsible for communicating with the public on behalf of a company, organization, or individual</em>. Their job includes maintaining relationship with the media, creating press releases, holding corporate events for external parties.  

You might be interested in
Write an application that determines which, if any, of the following files are stored in the folder where you have saved the exe
Elza [17]

Answer:

Here is the JAVA application that determines which if any of the following files are stored in folder where you have saved the exercises created.

import java.nio.file.*;   // used to manipulate files and directories

import java.nio.file.attribute.*;  //used to provide access to file and file attributes

import static java.nio.file.AccessMode.*;  // provides access modes to test the accessibility of a file

import java.io.IOException;  //for exception handling

class FindSelectedFiles {

public static void main(String[] args) { //start of main function

   Path f1 = Paths.get("/root/sandbox/autoexec.bat"); //creates Path object f1 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f2 = Paths.get("/root/sandbox/CompareFolders.java");  //creates Path object f2 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f3 = Paths.get("/root/sandbox/FileStatistics.class");   //creates Path object f3 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   Path f4 = Paths.get("/root/sandbox/Hello.doc"); //creates Path object f4 using Paths.get() method and this instance contains the file name and directory list used to construct the specified path

   int count=0; //used to count the number of files found  

try{  // used to define a code chunk to be tested for errors

 f1 = Paths.get("/root/sandbox/autoexec.bat");  // Converts a path string when joined form a path string to a Path

 System.out.println(f1);  //prints the file

     /* getFileSystem method is used to return the file system that created this Path object f1 and checkAccess Verifies access and throws exception if something is wrong

 f1.getFileSystem().provider().checkAccess(f1);  

 System.out.println(f1 +" File Exists\n");  //if file exists then displays this message

 count++;  }  //adds 1 to the count of number of files found

 catch (IOException e) {  //a code chunk to be executed if error occurs in the try part

  System.out.println("File Doesn't Exist\n"); }  //if file not found displays this message

   

try{  //works same as above try block but for 2nd folder and object f2

 f2 = Paths.get("/root/sandbox/CompareFolders.java");

 System.out.println(f2);

 f2.getFileSystem().provider().checkAccess(f2);

 System.out.println(f2 +" File Exists\n");

 count++;  }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

   

try{  //works same as above try block but for 3rd folder and object f3

 f3 = Paths.get("/root/sandbox/FileStatistics.class");

 System.out.println(f3);

 f3.getFileSystem().provider().checkAccess(f3);

 System.out.println("File Exists\n");

 count++; }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

   

try{  //works same as above try block but for 4th folder and object f4

    f4 = Paths.get("/root/sandbox/Hello.doc");

    System.out.println(f4);

 f4.getFileSystem().provider().checkAccess(f4);

 System.out.println(f4 +" File Exists\n");

 count++; }

 catch (IOException e) {

  System.out.println("File Doesn't Exist\n"); }

   

   System.out.println(count+" of the files exist");  }  } //displays the number of files found

Explanation:

If you want to directly pass the file name as argument to Paths.get() method then you can use the following code:

   Path f1 = Paths.get("autoexec.bat");

   Path f2 = Paths.get("CompareFolders.java");

   Path f3 = Paths.get("FileStatistics.class");

   Path f4 = Paths.get("Hello.doc");

The program determines if any, of the files from autoexec.bat, CompareFolders.java, FileStatistics.class, and Hello.doc. are stored in the folder. If a file is found then the program displays the message File exists and add 1 to the count variable each time a file is found. If all the files are found then the program returns the count as 4 otherwise the program display the number of files found by displaying the computed final value of count in the output. Lets say all the files are found then the output of the above program is:

autoexec.bat                                                                                                                                                   autoexec.bat File Exists  

                                                                                                                                                      CompareFolders.java                                                                                                                                            CompareFolders.java File Exists                                                      

                                                                                                                                                         FileStatistics.class                                                                                                                                           File Exists                                                                                        

                                                                                                                                                     Hello.doc                                                                                                                                                      Hello.doc File Exists  

   

4 of the files exist    

8 0
2 years ago
Steps needed to deposit cash on cash deposit machine
Artyom0805 [142]
Honestly, just google this please... but because I'm eating a cookie here you go...
<span>Deposit in your own account using your debit card:</span><span>Insert debit card and enter PIN for validation.Select account type (Saving or Current).<span>Place the money in the cash deposit slot and click “Continue”.</span><span>Machine will sort the cash and will show denomination-wise amount to be deposited.</span><span>If correct, click “Deposit”.
or this website
http://www.uandblog.com/How-To-Deposit-Money-In-A-Cash-Deposit-Machine

</span></span>
7 0
3 years ago
A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0
2 years ago
Write a full class definition for a class named Counter, and containing the following members: A data member counter of type int
vfiekz [6]

// making the class

class Counter {

int counter;

int limit;

// Constructor

Counter(int a, int b){

counter = a;

limit = b;

}

// static function to increment

static increment(){

if(counter<limit)

nCounter+=1;

}

// Decrement function

void decrement(){

if(counter>0)

nCounter-=1;

}

int getValue(){

return counter;

}

static int nCounter;

int getNCounters(){

return nCounter;

}

};

// Initializa the static

int Counter::nCounter = 0;

4 0
3 years ago
Desinated for a network connection within a largecity or multiple small citys
frozen [14]

I think it is WAN (Wide Area Network).

6 0
2 years ago
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