seo – Why did my site votes (stars) disappeared from Google SERP?

From the Structured Data Testing Tool. You can see that there is no review markup present on your website.

This link is specific to your website –
https://search.google.com/structured-data/testing-tool/u/0/#url=khedmatazma.com

You need to change the plugin that you are using or change the config to render the markup correctly.

This is an example of how it would look like if implemented correctly.
https://developers.google.com/search/docs/data-types/review. Click on embed markup in the example.

How can I sort Quora answers by votes?

The more upvoted an answer, the higher I desire it to appear. If possible, please include screenshots to illume where to click.

Be honest -the number one reason anyone votes for Donald Trump is because they love racism and that’s what he stands for right?

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[ Politics ] Open Question : Will Biden get any votes at all? Outside of corporations, china and wallstreet, who would vote for biden?

[ Politics ] Open Question : Will Biden get any votes at all? Outside of corporations, china and wallstreet, who would vote for biden?

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equation solving: find the number of votes in my reddit post

I posted a video on reddit that has more negative votes than positive votes. Instead of displaying a negative symbol, it displays a zero. It says that I have 29% of votes in favor, which I suppose was rounded, which means that we must estimate the votes for and against.

Yes $ x $ is the number of votes in favor and $ and $ is the number of negative votes, I guess we have to solve

$$ frac {x} {x + y} =. 29 $$
$$ frac {1} {2} text {sign} (x-y) + frac {1} {2} = 0 $$

I tried to solve in Wolfram Mathematica

In(24):= NSolve(x/(x + y) == .29 && 1/2*Sign(x - y) + 1/2 == 0, {x, y, z})

During evaluation of In(24):= NSolve::ifun: Inverse functions are being used by NSolve, so some solutions may not be found; use Reduce for complete solution information.

Out(24)= {{x -> 0.361186, y -> 0.884283}}

But this does not make sense. x is almost zero and y is approximately 1. 0 / (0 + 1) = 0.

So I tried to reduce

In(25):= Reduce(
 x/(x + y) == .29 && 1/2*Sign(x - y) + 1/2 == 0, {x, y, z})

During evaluation of In(25):= Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.

Out(25)= Re(x) > 0 && Im(x) == 0 && y == 2.44828 x

Nothing works

How do we solve this? Could we use graphics?

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. (tagsToTranslate) quora (t) votes in favor (t) seo (t) construction (t) readers (t) views

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Combinatorial – Voting Counting: How many possible paths exist in which candidate A never has 2 votes ahead?

This question is inspired by Bertrand's ballot theorem. I want to check if I understood the counting method correctly. I changed the initial problem of the ballot theorem a bit.

Suppose we have two candidates, A and B. After counting the votes, we have a tie. How many paths exist in which candidate A is never? $ geq2 $ votes ahead?

This is my approach:

I imagine the election as a path in the $ x $-axis that begins in $ (0, 0) $ and ends in $ (2p, 0) $, where $ 2p $ denotes the total number of votes. In the test of the voting theorem they use a second path that is constructed reflecting in part the original paths on the $ x $-axis. I will try to follow this idea.

First, I change the beginning of all my paths to $ (- 2.2) $ and the first two votes should always be in favor of B. Second, I will extend the way to $ (2p + 2, 2) $ while the last two votes will count for A. Of those paths I will only consider those that begin with two consecutive votes for B and end with two consecutive votes for A (otherwise, we would include paths that are definitely not allowed). Denote $ M $ as the set of those paths. Thus, $ | M | = {2p choose p} $.

Then i build assistant roads as follows:

Allow $ P $ a path of $ M $. While $ P $ does not touch the horizontal line that passes $ (0.2) $ (see red line in the image) Reflect your values ​​on the horizontal line that crosses $ (0.2) $. Those values ​​are the first points of the assistant routes. When $ P $ play the $ (0.2) $-line the assistant paths will follow the rest of $ P $. Building those assistant roads is a bijection in the set of those roads of $ M $ that touch or cross the $ (0, 2) $ line. So I just have to subtract everything assistant paths of $ | M | $.

_ auxiliary route_

Now I count all assistant paths (I will do this with a little more detail):

I have added $ 4 $ votes for $ 2p $ votes from the beginning (see green lines). $ 4 $ votes of each assistant the route always has the same meaning as the first two votes and the last two votes always count for A. Therefore, all assistant the roads add up: $ {2p choose p-2} $. The total number of roads where A is never $ geq 2 $ votes ahead is $ {2p choose p} – {2p choose p-2} = frac {2} {p + 2} {2p choose p} $.

Is this correct?

I appreciate any comments or suggestions and let me know if I should be clearer at some step.

What is the name of the UX pattern to show an ambiguous count of likes, votes, etc.?

While browsing several social media platforms, I find this vote button in favor, which shows an ambiguous count of the things mentioned above.

Is there a name for such a pattern?

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