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astraxan [27]
3 years ago
8

Joe, an end-user, reports that the Windows PC he is using automatically locks when he leaves his desk and walks to a printer to

retrieve documents. Joe is then required to type in his username and password to unlock the computer. The technician looks at the settings on the PC and notices that the screensaver and screen-lock options are grayed out on the computer and cannot be changed. Which of the following is the MOST likely cause of this issue?
A. Domain-level group policies
B. Antivirus domain-level policies
C. Corrupted registry settings
D. Incorrect local-level user policies
Computers and Technology
1 answer:
ivann1987 [24]3 years ago
6 0

Answer:

A. Domain-level group policies

Explanation:

Because domain-level polices are automatically implement all users that are login on the PC and PC is connect to the domain. From domain polices you can set what user can open and what can not open, Joe Windows PC he is using automatically locks also through domain policy, Administrator set the time when any end-user leave their PC automatically after the time set administrator screen saver run and asked password of that user to return back in the windows.

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Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working​ alone, one machine would take one hour long
Alborosie

<em><u>Answer</u></em>

5 hours

<em><u>Explanation</u></em>

The two working together can finish a job in

\frac{20}{9}  \: hours

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

\frac{1}{x}    +  \frac{1}{x - 1}

This should be equal to 20/9 hours.

\frac{1}{x}  +  \frac{1}{x - 1}  =  \frac{9}{20}

Multiply through by;

20x(x - 1) \times \frac{1}{x}  +20x(x - 1) \times   \frac{1}{x - 1}  =  20x(x - 1) \times \frac{9}{0}

20(x - 1)  +20x = 9x(x - 1)

20x - 20+20x = 9{x}^{2}  - 9x

9{x}^{2}  - 9x - 20x - 20x + 20= 0

9{x}^{2}  - 49x  + 20= 0

Factor to get:

(9x - 4)(x - 5) = 0

x =  \frac{4}{9}  \: or \: x = 5

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

8 0
3 years ago
Write an application that allows a user to enter the names and birth dates of up to 10 friends. Continue to prompt the user for
mylen [45]
Sorry I don’t know the answer I am really sorry
5 0
3 years ago
All changes
Vsevolod [243]

You should get up and move every 30 minutes

Hope  this helps

-scav

7 0
3 years ago
An incurred cost that cannot be recovered, which is irrelevant for all decisions about the future, is included in the projected
iragen [17]

Answer:

An incurred cost that cannot be recovered, which is irrelevant for all decisions about the future, is included in the projected cost of a project. According to "Thinking Like an Economist," this an example of:<u> Failing to ignore sunk costs</u>

Explanation:

A sunk cost is a cost that cannot be recovered or changed and is independent of any future costs a business may incur. Since decision-making only affects the future course of business, sunk costs should be irrelevant in the decision-making process

3 0
3 years ago
In the original UNIX operating system, a process executing in kernel mode may not be preempted. Explain why this makes (unmodifi
Elena L [17]

Answer:

the preemption is -> The ability of the operating

system to preempt or stop a currently

scheduled task in favour of a higher priority

task. The scheduling may be one of, but not

limited to, process or 1/0 scheduling etc.

Under Linux, user-space programs have always

been preemptible: the kernel interrupts user

space programs to switch to other threads,

using the regular clock tick. So, the kernel

doesn't wait for user-space programs to

explicitly release the processor (which is the

case in cooperative multitasking). This means

that an infinite loop in an user-space program

cannot block the system.

However, until 2.6 kernels, the kernel itself was

not preemtible: as soon as one thread has

entered the kernel, it could not be preempted to

execute an other thread. However, this absence

of preemption in the kernel caused several

oroblems with regard to latency and scalability.

So, kernel preemption has been introduced in

2.6 kernels, and one can enable or disable it

using the cONFIG_PREEMPT option. If

CONFIG PREEMPT is enabled, then kernel code

can be preempted everywhere, except when the

code has disabled local interrupts. An infinite

loop in the code can no longer block the entire

system. If CONFIG PREEMPT is disabled, then

the 2.4 behaviour is restored.

So it suitable for real time application. Only

difference is we don't see many coders using it

5 0
3 years ago
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