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Mamont248 [21]
3 years ago
10

IPv6 can use a DHCPv6 server for the allocation of IPv6 addressing to hosts. Another IPv6 addressing option utilizes the IPv6 Ne

ighbor Discovery Protocol (NDP) to discover the first portion of the IPv6 address (network prefix) from local routers, and the host can create its own host ID. What standard is commonly used by an IPv6 host to generate its own 64-bit host ID
Computers and Technology
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

Following are the answer to this question.

Explanation:

It is a mechanism, that is also known as EUI-64, which enables you an automatic generation of its specific host ID. It using the device on the 48-bit MAC address, which helps to construct the special 64-bit host ID. It also helps you to build a DHCP-type IPv6 network, that's why we can say that the above-given standard is widely used to create a 64-bit host ID on IPv6 servers.

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Tone
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7 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

8 0
3 years ago
For a list to be binary search-able, we need the following thing to be true, because... a. The list must be sorted, because chec
rusak2 [61]

Answer:

The answer is option A.

Explanation:

Negative numbers can be found by binary search, this makes option B incorrect.

Unsorted and randomized lists are also not things that support a binary search, options C and D are incorrect.

Binary search uses a technique where the middle element of the list is located and used to determine whether the search should be done within the lower indexed part of the list or the higher. So for a list to be binary search-able, it should be sorted and not randomized.

The answer is A.

I hope this helps.

7 0
3 years ago
Question 4 True or false: The same plaintext encrypted using the same algorithm and same encryption key would result in differen
Cloud [144]

Answer:

False

Explanation:

7 0
2 years ago
What is the binary equivalent of the volume 3F?
kaheart [24]

Answer:

The binary equivalent of the volume 3F is

A. 00111111

Hope it will help. :)❤

6 0
3 years ago
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