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3 years ago

7

In the C-SCAN disk scheduling algorithm, the disk arm is required to move in one direction only until it reaches the last track

or there are no more requests to service in that direction, then it reverses direction and the scan proceeds in the opposite direction in the same fashion.
True / False.

Computers and Technology

1 answer:

dezoksy [38]3 years ago

3 0

Answer:

False.

Explanation:

In the C-SCAN disk scheduling algorithm, the disk arm is required to move in one direction only until it reaches the last track or there are no more requests to service in that direction, then it reverses direction and the scan proceeds in the opposite direction in the same fashion is a False statement.

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Suppose you are an ISP that owns a / 22 IPv4 address block. Can you accommodate requests from six customers who need addresses f

lisabon 2012 [21]

Answer:

It is not possible.

Explanation:

In this example, we need to accommodate 473 computers for six clients that are 473 IP addresses.

For this request just we have /22 IPv4 address blocks, this mean

22 red bits 11111111111111111111110000000000 <--- 10 host bits

We must increase red bits to 25, we need these 3 bits to create 6 sub red, in this case, 2^3 = 8 sub red.

Why did we ask 3 bits? Because if we ask only 2, 2^2 = 4, and we need 6 sub red.

25 red bits 11111111111111111111111110000000 7 host bits

In this case, we need more than 260 computers, but just we have 7 bits, this means.

2^7 = 128 and just one customer needs 260, for that is impossible.

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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

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Answer:

A microprocessor is a computer processor where the data processing logic and control is included on a single integrated circuit. A computer need microprocessors to perform the functions of a computer's central processing unit.

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Answer:

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