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Makovka662 [10]

3 years ago

15

Core component of the linux operating system is the linux kernel. if you were a linux systems administrator for a company, when

would you need to upgrade your linux kernel

Computers and Technology

1 answer:

iragen [17]3 years ago

5 0

You would only need to upgrade your kernel via linux if u were a mlg hacker and knoushs how to speakith the englith

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A cpu with an external clock speed of 2 ghz and a 64-bit data bus can (theoretically) transfer how much data per second?

GREYUIT [131]

The CPU could cycle the bits 2000000000 a second. That means theoretically it could process 16 gigabytes of data a second.

5 0

3 years ago

Explain drawing and painting package.

Murljashka [212]

Enhances drawing experiences, adds more features and adds better quality

8 0

3 years ago

which endpoint application runs on an endpoint device that only detects an attack in an endpoint device? chqgg

WARRIOR [948]

A host-based intrusion detection system works similarly to a network-based intrusion detection system in that it can monitor and analyze both the internal workings of a computer system and the network packets on its network ports.

<h3>What is Host-Based IPS?</h3>

A host-based intrusion detection system works similarly to a network-based intrusion detection system in that it can monitor and analyze both the internal workings of a computer system and the network packets on its network ports.
The Host-based Intrusion Prevention System (HIPS) guards against malicious software and other activities that aim to harm your computer. HIPS uses sophisticated behavioral analysis in conjunction with network filtering's detection capabilities to keep track of active programs, files, and registry keys.
The integrated endpoint security system known as endpoint detection and response (EDR), also referred to as endpoint threat detection and response (ETDR), combines real-time continuous monitoring and gathering of endpoint data with rules-based automated reaction and analysis capabilities.

To learn more about Host-Based IPS refer to:

brainly.com/question/20490376

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5 0

1 year ago

What two statements about IPv6 addresses are true? This task contains the radio buttons and checkboxes for options. The shortcut

Oliga [24]

Answer:

A and C

Explanation:

Option A:

In IPv6 there is a rule to reduce an IPv6 address when there are two or more consecutive segments of zeros just one time. This rule says that you can change the consecutive zeros for “::”

Here is an example

How to reduce the following IPv6 address?

ff02:0000:0000:0000:0000:0000:0000:d500

Ans: ff02::d500

Example 2:

2001:ed02:0000:0000:cf14:0000:0000:de95

Incorrect Answer -> 2001:ed02::cf14::de95

Since the rule says that you can apply “::” just one time, you need to do it for a per of zero segments, so the correct answer is:

Correct Answer -> 2001:ed02::cf14:0:0:de95

Or

2001:ed02:0:0:cf14::de95

Option C:

Since in IPv6 there are $2^{128}$ available addresses which means 340.282.366.920.938.463.463.374.607.431.768.211.456 (too many addresses), there is no need of NAT solution, so each device can have its own IP address by the same interface to have access through the internet if needed. If not, you can block the access through internet by the firewall.

4 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