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salantis [7]
3 years ago
5

An important ethical concern for behavior analysts involves special cautions in interactions with a client in any capacity outsi

de of the role of behavioral practitioner. For example, it may be difficult to decide what to do when a client offers a small kindness or gift of appreciation for services received, or invites the behavior analyst to attend social, nonprofessional events. However, one must take care to avoid violation of ethical codes regarding conflicts of interest and ______ relationships. a. dual b. multiple c. friendly d. Both A and B are correct.
Engineering
1 answer:
Phoenix [80]3 years ago
4 0

Answer:A and B are correct. one must take care to avoid violation of ethical codes regarding conflicts of interest and dual or multiple relationships

Explanation:

A conflict of interest (COI) is a situation in which a person or organization is involved in multiple interests, financial or otherwise, and serving one interest could involve working against another. 

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Problem a) – c): Use the method of joints, the method of sections, or both to solve the following trusses. Draw F.B.Ds for all y
Ghella [55]

Answer:

This is confusing sorry

Explanation:

7 0
3 years ago
This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons
telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, d_{prop} = m/sec (s)

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet,  d_{trans} = L/R

(c)

As we know, end-to-end delay or total no delay,

\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}

Here,  $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }

(d)

The expression, time time $t=d_{\text {trans }}$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at time $t=d_{\text {trans }}$ the last bit of the packet has been pushed out or transmitted.

(e)

If \ d_{prop} >d_{trans}

Then, at time $t=d_{\text {trans }}$ the bit has been transmitted from host A, but to condition (1),  the first bit has not reached B.

(f)

If \ d_{prop}

Then, at time $t=d_{\text {trans }}$, the first bit has reached destination on B

Here,s=2.5 \times 10^{8} \mathrm{sec}

\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}

It's given that \ d_{prop} =d_{trans}

Hence,

        \begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}

5 0
3 years ago
Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str
stiks02 [169]

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

}

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

}

6 0
3 years ago
Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo c
brilliants [131]

Answer:

The algorithm is as follows:

1. Declare Arr1 and Arr2

2. Get Input for Arr1 and Arr2

3. Initialize count to 0

4. For i in Arr2

4.1 For j in Arr1:

4.1.1 If i > j Then

4.1.1.1 count = count + 1

4.2 End j loop

4.3 Print count

4.4 count = 0

4.5 End i loop

5. End

Explanation:

This declares both arrays

1. Declare Arr1 and Arr2

This gets input for both arrays

2. Get Input for Arr1 and Arr2

This initializes count to 0

3. Initialize count to 0

This iterates through Arr2

4. For i in Arr2

This iterates through Arr1 (An inner loop)

4.1 For j in Arr1:

This checks if current element is greater than current element in Arr1

4.1.1 If i > j Then

If yes, count is incremented by 1

4.1.1.1 count = count + 1

This ends the inner loop

4.2 End j loop

Print count and set count to 0

<em>4.3 Print count</em>

<em>4.4 count = 0</em>

End the outer loop

4.5 End i loop

End the algorithm

5. End

6 0
3 years ago
What are the coventional representative of automation
Oxana [17]

Answer:

ere el merjor 5iyer

Explanation:

yyhh espero ayuder 8 mucho 666

5 0
3 years ago
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