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Andreyy89
3 years ago
8

When a user utilizes another computer to communicate with a third party, with the result that the third party cannot recognize t

he IP address of the originating communication is called____________.
Computers and Technology
1 answer:
avanturin [10]3 years ago
4 0

Answer:

Online or Web Proxy

Explanation:

Web or online proxy allows you to hide your Internet Protocol (IP) address from the website you are trying to visit or access. It is like a shield between the website you are accessing and you. They can be seen as a middleman between you and the site you are visiting and as such, the site sees that a specific IP address is accessing its server but the address is not yours as all requests between your computer and the web server are first passed through the proxy server.

<em>Hope this helps!</em>

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How does the dns help the world wide web scale so that billions of users can access billions of web pages?.
valentina_108 [34]

Answer:

domain name systems allow web users to choose where they want to go and to have many different sites

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2 years ago
What did Francis Ford Coppola and George Lucas create?
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. 3d animation markers
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President Roosevelt's Fireside Chats were:
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C. Entertaining radio shows that families listened to in the evening. He did these chats to inform the public on what he was going to do about the problems facing the public.
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Read 2 more answers
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
3 years ago
NIST recommends selecting cloud providers that support strong encryption, have appropriate redundancy mechanisms in place, emplo
WITCHER [35]

Answer:

The answer is "Option a".

Explanation:

In cloud computing, it is also known as the model, that enables you for accessible, convenient, through the-demand network access to global computer resources, which can be rapid to get and published via low administrative effort.  

Its recommending selection for the cloud providers support for the robust encryption, that has adequate replication processes in place, use user authentication, or provide ample clarity to customers regarding mechanisms that defend subscriptions against other subscriptions and the supplier.

8 0
3 years ago
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