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victus00 [196]
3 years ago
10

Which CIDR network has the smallest possible number of hosts? a. 123.0.0.0/8 b. 123.45.0.0/16 c. 123.45.67.0/24 d. All have the

same possible number of hosts
Computers and Technology
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

CIDR is based on a variable-length subnet masking technique, which allows a new method of representation for IP addresses. Routing prefix is written with a suffix number of bits of the name, such as 123.0.0.0/8, as the CIDR network has the smallest possible amount of hosts.

Explanation:

CIDR ( Class Inter-Domain Routing ) It is a method that is allocating IP addresses and routing the IP. CIDR is introduced in 1933 and replace the architecture of network design on the internet. CIDR slows down the growth of the routing across the web and helps to slow the IP addresses such as IPv4 addresses.

CIDR consists of two groups of bits in the address. In the new age, the network prefix identifies the whole network. This is used as the basis of routing between IP networks and allocation policies.

IPv4 in-network prefix is 8-bit groups.

A typical IPv4 address is 192.168.0.5 the lowest value is 0, and the highest value is 255

In the given choices, 123.0.0.0 / 8 of the possible smallest number of hosts.

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2 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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