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2 years ago

What is the new subnet mask in prefix format?

Computers and Technology

1 answer:

vodomira [7]2 years ago

3 0

In the case above, the the new subnet mask in prefix format is seen in IPv6 and in IPv4.

<h3>What is subnet prefix mask?</h3>

A lot or a single computer that is known to be linked to a subnet is said to shares a kind of an identical part of the IP address.

Note that this shared information is said to be called the routing prefix, and in IPV4 (Internet Protocol Version 4), the routing prefix is said to be called a subnet mask.

Hence, The the new subnet mask in prefix format is seen in IPv6 and in IPv4 because that is the format that it always comes in.

Learn more about subnet from

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Rasek [7]

Answer:

I think that part C correct option

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3 years ago

Word processing software allows users to do which of the following: format text design pages share documents mail merge document

Sergio039 [100]

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3 years ago

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

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3 years ago

The 169.254.78.9 ip address is an example of what type of ip address

scoundrel [369]

As specified in RFC5735, this is an address from the "link local" block. It is assigned to a network interface as a temporary address, for instance if no static address is configured and the DHCP server is not found.

If you boot your PC without a network cable, you'll probably end up with a 169.254.*.* address.

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3 years ago

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alukav5142 [94]

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