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dlinn [17]
3 years ago
11

The best recruitment source, that is, one which tends to result in attracting employees who experience greater loyalty, tenure,

and job satisfaction than other recruiting sources,is:________
Mathematics
1 answer:
Anna11 [10]3 years ago
6 0

Answer:

The best recruitment source, that is, one which tends to result in attracting employees who experience greater loyalty, tenure, and job satisfaction than other recruiting sources,is **referrals from current employees**

Step-by-step explanation:

This is an HR question.

Referrals given to potential employees by current employees refer to the glowing references about the firm provided by those current employees to the potential employees.

The best set of people to talk about work conditions, pay, etc., at a workplace are the actual workers of that workplace.

Whenever they provide such privy information to a potential employee, it leaves no massive room for unpleasant surprises when the potential employee takes the job. The referrals basically groin the employee for the the best and the worst of such a workplace, that even qualified personnel, before taking up jobs, like to speak to current employees of that firm.

In HR, there's the popular saying that a firm is as good as the way they treat their employees.

So, such total information before taking a job leads to gaining employees that know all about what to expect and which realistic job satisfaction to target. Their realities after starting the job is often already calculated.

Hence, anyone that interacts with current employees of a firm and still takes up job with the firm, has a huge tendency to stay longer, be more loyal and have a bigger job satisfaction.

Hope this Helps!!!

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Step-by-step explanation:

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You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han
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Answer:

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Step-by-step explanation:

If air resistance is neglected, the energy is constant. That means that in every point in the trajectory the energy is the same.

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E=mgh

At point of height h/4, some of the potential energy has transformed in kinetic energy and the velocity is no longer zero. But we know the total amount of energy is the same as the maximum height point.

With that information, we can calculate the velocity v:

E=mg(\frac{h}{4})+m\frac{v^2}{2}=mgh\\\\m\frac{v^2}{2}=mg(h-\frac{h}{4})\\\\\frac{v^2}{2}=\frac{3gh}{4}\\\\v^2=\frac{3gh}{2}\\\\v=\sqrt{\frac{3gh}{2}}

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How would I solve 1 and 2?
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Answer:

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3 years ago
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