sky-light

@egreg I tried to make the table as simple as I can and at the same time raise my question.
The central column is for the data shared for all case studies

@StefanKottwitz OK.
I have another question please..

Hi,
I would like to know how many antennas are inside an iPhone 7 for the 4G connection?
Is it possible to have multiple antennas on an iPhone to make the received signal stronger and give better data throughput?

The impact of the A and the B are considered.
or
The impact of the A and the B is considered.

Using "the" in the following case is right or wrong:
The authors of this article investigated the characteristic of ...
Or
Authors of this article investigated the characteristic of ...

Could anyone tell me if "lose" is the correct word here:
Moreover, the glue scheme does not lose its advantage despite the presence of water

Hi, which of these is better concerning formal writing:

The mentioned works gave us a good background understanding
in the area of pipe design ...

A good background understanding in the area of pipe design has been gained from the works mentioned above ...

Hi,
Can someone try to solve this integral with Mathematica 11.0?

\int_a^{\infty } e^{-b t} (t+1)^{-c} t^{d-1} \, dt

Thanks in advance

@MichaelE2 Hi
Please see this example
http://www.wolframalpha.com/input/?i=Solve%5Ba%2F(1+%2B+b+x)+-+Log%5B1+%2B+b+x%5D+%2B+1+%3D%3D+0,+x%5D

In Mathematica 9.0, how do I set the precision back to the default value?
I used this line to increase the precision

$PreRead = (# /.
s_String /;
StringMatchQ[s, NumberString] &&
Precision@ToExpression@s == MachinePrecision :>
s <> "`30." &);

However, it ruined my calculations. Now I want to move back to the default settings.

Dear friends,
I have the following code; I appreciate if someone helps me what is wrong with it?

KK = 3;
limm = 4;
yall = {};
For[j = 0, j <= limm, j = j + 1, pt = j;
px = 2^pt;
outloop = 0;
For[i = 1, i <= KK, i = i + 1, ntest = i;
Which[ntest == 1,
y1 = px + 1,
ntest == 2,
y1 = px + 2,
ntest == 3,
y1 = px + 3]
outloop = outloop + y1;
]
y2 = N[(outloop)];
yall = Append[yall, outloop];
]