Physics, 17.07.2019 23:30 cabrerafreddy780

An electronics company is hiring a part-time programmer to better calibrate one of its electronic devices. four candidates for the position have stated how long it would take them to complete the project and what their hourly fee would be. which programmer would be the most cost-effective choice for completing this project?

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If you mean the cheapest, it's the candidate for whom the product
of his (estimated total hours) times his (hourly fee) is a minimum.

But gosh ! The quality, reliability, and usability of the final product,
based somehow on the candidate's past experience with similar
projects, surely ought to figure into the calculation somehow.
It won't be all that "cost effective" if it takes twice as long to calibrate
one of the devices, or if it freezes up in the middle of the calibration
half the time.

Answer from: LJOrozco

The most economical option would be the programmer who could complete the job in less time and with the lowest hourly rate.

Explanation:

As you know, we live in an extremely competitive capitalist world. This can be seen in the question above, since a single company has four candidates for the same position and will have to choose the candidate that is most profitable for the company.

The most profitable candidate will be the one who is able to complete the activity requested by the company in the shortest time, in a highly efficient manner and charging a lower hourly rate. This will allow the company to get its activity done quickly, with a high quality standard and a low cost.

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

Domain eukarya; kingdom plantae should be the answer.

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