【2012.10.20】大學研究所博覽會 名校同場較勁
Published October 22, 2012
At a glance
Views
251
Likes
2
Comments
0
Performance vs. channel
All-time outlier score
Engagement rate
More features coming
We're working on new analytics and tools. Stay tuned.
About
英聽的成績，可以作為大學甄選入學的參考，同時，高三的學生如果還不知道自己對哪個科系有興趣，這個周末，可以到台大綜合體育館，參觀大學暨研究所博覽會。
- Published
- October 22, 2012
- Made for kids
- No
More features coming
We're working on new analytics and tools. Stay tuned.
How 【2012.10.20】大學研究所博覽會 名校同場較勁 is doing
It has 251 views in total. It ranks #14,283,845 among tracked YouTube videos by views.
Key facts about 【2012.10.20】大學研究所博覽會 名校同場較勁
- Total views
- 251
- Total likes
- 2
- Total comments
- 0
- Published
- October 22, 2012
- Global rank by views
- #14,283,845
Frequently asked questions
How many views does 【2012.10.20】大學研究所博覽會 名校同場較勁 have?
【2012.10.20】大學研究所博覽會 名校同場較勁 has 251 views, 2 likes, ranking #14,283,845 among tracked YouTube videos by views.
How accurate are these YouTube video analytics?
View, like, and comment counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what the creator sees privately in YouTube Studio.
What is the outlier score?
How this video's views compare to the channel's usual Shorts or long-form video. A 2x score means double the usual views for that channel, 0.5x means half.
What is engagement rate and how is it calculated?
Likes and comments per view, compared to this channel's typical engagement rate — shown as above or below what's usual for the channel.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view count — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Is this YouTube video analytics checker free?
Yes. You get 10 free channel and video analytics lookups per 24 hours without an account — logging in, which is also free, removes that limit entirely.