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3 years ago

Which of the following is productivity strategy for collaboration?

Computers and Technology

1 answer:

mixer [17]3 years ago

8 0

Which of the following is a productivity strategy for collaboration?

A. Saving focused work for a time of day when you feel most creative.
B. Employing the 80/20 rule to prioritize tasks.
C. Posting files to a web-based shared site.
D. Using white noise to block distractions in the office.

Answer:

Posting files to a web-based shared site is a productivity strategy for collaboration.

C. Posting files to a web-based shared site.

Explanation:

Collaborative software or groupware is application programming intended to help individuals taking a shot at a typical errand to achieve their objectives. This permits individual to impart thoughts and their abilities to different individuals with the goal that the assignment can be done both proficiently and adequately.

Joint effort stages ordinarily incorporate an email customer, Web conferencing, internet based life sharing, video capacities, report sharing abilities, texting and that's just the beginning. Endeavor joint effort stages are intended to be introduced on-premises or conveyed by means of the Web as cloud-based administrations.

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Which of the following is a hardware component used to hold the BitLocker encryption key and ensures encrypted data is not acces

Vadim26 [7]

Answer:

TPM(Trusted Platform Module) is the correct answer to the following question.

Explanation:

Because TPM is the computer chip which protects your system data and store all keys and access of data that is encrypted by these keys. So that's why TPM is that hardware component which is used to contain the encryption keys and that data is not accessible to any other person's whether your hard drive will be lost or stolen by anyone.

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2 years ago

14. The heart of a computer is a CPU b. Memory c. I/O Unit d. Disks

NNADVOKAT [17]

Answer:

CPU

I dont know a whole ton about computers. But I'm confident this is correct. Without the CPU the other options on this list wouldnt work so the CPU would be the "heart".

3 0

2 years ago

(10 points) 5.12. Discuss how the following pairs of scheduling criteria conflict in certain settings. a. CPU utilization (effic

Novosadov [1.4K]

Answer:

Check the explanation

Explanation:

CPU utilization and response time: CPU utilization is amplified if the overheads that are connected with context switching or alternating are minimized. The context switching outlay can be reduced by performing context switches occasionally. This could on the other hand lead to increasing the response time for processes.
The Average turnaround time and maximum waiting time: Average turnaround time is reduced by implementing the shortest or simple tasks first. Such a scheduling and arrangement strategy could nevertheless starve long-running tasks and in so doing boost their overall waiting time.
I/O device utilization and CPU utilization: CPU utilization is maximized by executing a list of long-running CPU-bound tasks not including the performing context switches. This is maximized by setting up I/O-bound tasks as early as they become ready to run, thus sustaining the overheads of context switches.

3 0

3 years ago

If an ______ is caught dumping hazardous materials, that person can be prosecuted. Answer choices A) Employee B) Employer C) Bot

Citrus2011 [14]

Answer:

Both A and B

Explanation:

I think it is right

hope this helps :)

5 0

2 years ago

Using Amdahl’s Law, calculate the speedup gain of an application that has a 60 percent parallel component for (a) two processing

Nina [5.8K]

Answer:

a) Speedup gain is 1.428 times.

b) Speedup gain is 1.81 times.

Explanation:

in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:

$t=\frac{1 }{(S + (1- S)/N) }$

Where S is the portion of the application that must be performed serially, and N is the number of processing cores.

(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t = \frac{1}{(0.4 + (1-0.4)/2)} =1428671$

Speedup gain is 1.428 times.

(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4

Thus, the speedup is:

$t=\frac{1}{(0.4 + (1-0.4)/4)} = 1.8181$

Speedup gain is 1.81 times.

8 0

3 years ago