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AysviL [449]
3 years ago
15

Which scenario depicts an ethical workplace practice by a business owner

Computers and Technology
2 answers:
antoniya [11.8K]3 years ago
5 0

Answer:

This question is incomplete; the full one is the following:

Which scenario depicts an ethical workplace practice by a business owner?

a. sharing personal information of its employees with marketing agencies

b. sharing statutory personal information of its employees with law agencies

c. sharing internet browsing details of an employee with other employees

d. blaming human errors on technology bugs

e. allowing employees to make errors

The correct answer is letter "e. allowing employees to make errors".

Explanation:

Letter "e" is the most reasonable answer in terms of ethics, whether because it is conscious about human conditions and the possibility of always making mistakes, whether because it shows that a business owner has to think carefully about his/her employees. All the other options do not apply to the question because they are not ethical at all; all of them depict a business owner as someone who is intrusive and disrespectful, therefore, the last option is the most ethical among all the others.

(ps: mark as brainliest, please?!)

viva [34]3 years ago
4 0

The answer is D. It's asking for the most UNETHICAL not ETHICAL, the answer before me was ETHICAL. Sharing personal data is UNETHICAL.

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In Java, it is possible to create an infinite loop out of while and do loops, but not for-loops. true or false
soldier1979 [14.2K]

Answer:

False is the correct answer for the above question.

Explanation:

  • In java programming language or any other programming language, any loop can be infinite.
  • It is because the infinite loop is called for that loop which is not run in a finite number of times.
  • The loop is used to repeat some lines in a finite number of times. Any loop has three things- first is the initial value which tells the loop to start, The second is the condition check which states when the loop will stop and the third is an operation which directs some variable, so that the condition may be false after some finite amount of time.
  • If the condition will not false in any iteration of the loop, then the loop can proceed for an infinite amount of time.
  • The above-question states that the while loop and do while loop can be infinite which is a true statement.
  • But it also states that the 'for' loop can not be infinite which is not a correct statement which is described above. Hence false is the correct answer to the above question.
8 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

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