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salantis [7]
1 year ago
15

which one of the following would not normally be found in an organization's information security policy?

Computers and Technology
1 answer:
kotegsom [21]1 year ago
4 0

Requirement to use AES-256 encryption would not normally be found in an organization's information security policy.

When it comes to information security, organizations have a variety of different policies and requirements that they adhere to. One of those requirements is the use of AES-256 encryption.

However, not every organization has AES-256 encryption as a requirement in their information security policy. In fact, many organizations don't even know what AES-256 encryption is.

Why is AES-256 encryption relevant?

Well, AES-256 encryption is a strong form of encryption that is used to protect data. It is often used by governments and organizations to protect sensitive data.

AES-256 encryption is a requirement for many organizations because it provides a high level of security for data. Without AES-256 encryption, data could be compromised.

Learn more about information security policy here:

brainly.com/question/14292882

#SPJ4

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Answer:

Nelson proposed a system where copying and linking any text excerpt, image or form was possible.

Explanation:

Ted Nelson is one of the theoretical pioneers of the world wide web who is best known for inventing the concept of hypertext and hypermedia in the 1960s. As one of the early theorists on how a networked world would work.

How I know:

I goggle it.

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All of the following are forms of verbal communication except
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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
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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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You are trying to access WiFi network at a coffee shop. What protocol will this type of wireless networking most likely use?
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