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kiruha [24]
3 years ago
15

The use of public wireless connections can increase a user's vulnerability to monitoring and compromise. ____________ software c

an be used to encrypt transmissions over public networks, making it more difficult for a user's PC to be penetrated. DDos CAPTCHa VPN Rootkit Keylogging
Computers and Technology
1 answer:
GalinKa [24]3 years ago
3 0

The use of public wireless connections can increase a user's vulnerability to monitoring and compromise. <u>VPN </u> software can be used to encrypt transmissions over public networks, making it more difficult for a user's PC to be penetrated.

Explanation:

  • A VPN, or Virtual Private Network, allows you to create a secure connection to another network over the Internet.
  • VPNs can be used to access region-restricted websites, shield your browsing activity from prying eyes on public Wi-Fi, and more. .
  • Virtual Private Network, is a private network that encrypts and transmits data while it travels from one place to another on the internet. VPNs aren't just for desktops or laptops, you can set up a VPN on your iPhone, iPad or Android phone.
  • A VPN encrypts the traffic from your machine to the exit point of the VPN network. A VPN isn't therefore likely to protect you from an adversary like Anonymous.
  • VPNs add another layer of encryption to your internet traffic, your latency will go up and speeds will go down.

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Both UDP and TCP use port numbers to identify the destination entity when delivering a message. Give at least one reason for why
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Follows are the solution to this question:

Explanation:

The process ID is not static because this can't be used to identity, therefore, it includes excellent service providers like HTTP since it is allocated dynamically only to process whenever a process is initiated.

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Explanation:

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How does classless inter-domain routing (cidr) help reduce waste of ip addresses?
Scorpion4ik [409]
It allows for more accurate sizing of networks.
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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
3 years ago
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