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3 years ago

Freeeee BRAINLIESTHBJHBJHB

Computers and Technology

2 answers:

andre [41]3 years ago

4 0

Answer:A lot of people cry when they cut an onion. The trick is not to form an emotional bond.lol

nalin [4]3 years ago

4 0

Answer:

jidagfldfeiowjsd whrj

Explanation:

You might be interested in

A buffer storage that improve computer performance by reducing access time is

kvasek [131]

Cache memory

Hope it helps

4 0

3 years ago

A student is helping a friend with a home computer that can no longer access the Internet. Upon investigation, the student disco

Alexxandr [17]

Answer:

Due to unreachable DHCP server

Explanation:

As we know that Dynamic Host Configuration Protocol, in short DHCP is the network protocol which is used to central and automatic management of IP address with in the network.

So due to unreachable DHCP server a student could not able to connect with the internet.When the network connection break it means that DHCP server is offline.

So the answer is Due to unreachable DHCP server .

8 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 $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

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)

From the given parameters; 3 out of 15 is detective;

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;

Probability that at least one is defective + Probability that at none is defective = 1

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

What makes for a good algorithm compatibility complexity humor correctness efficiency?

Svetlanka [38]

Correctness is key. The other ones are optimizations at most. Although I would always include humor... ;-)

5 0

3 years ago

Using Task Manager, you discover an unwanted program that is launched at startup. Of the following items, which ones might lead

Harrizon [31]

Answer: Option (a) ,(c) and (e) can become a solution for the problem.

Option(b) and (d) are not exactly a solution to the issue rather can be considered on temporary basis

Explanation: Windows Task Manager is used for processing task taking place in the system and monitors the performance of that task or program that is running.To stop the unnecessary program execution on the system. methods like searching the source of the program from Task scheduler, startup files or registry key that initiated the execution that unwanted task.

Methods like disabling the program by using Msconfig and Task manager will only deactivate them but they are not eliminated from the source .Thus the appropriate answers would be option(a), (c) and (e).

7 0

3 years ago