its finaly comming down

Sure...is that the way the Smithsonian counts its attendance?

Yes, the Smithsonian figures add together the attendance at each of the 19 museums (they might not double count those going to the two co-located ones, though) to get the total number (see, for example: http://www.washingtonpost.com/wp-dyn/content/article/2006/09/19/AR2006091901720.html). So using that counting mechanism, WDW has about twice the attendance of the Smithsonian, not less. People go visit multiple WDW parks per trip (so one person might be counted 10 times in that figure), but they also visit multiple Smithsonian museums. People also visit the same park/museum over and over. Overall, I would guess that these things tend to inflate the WDW numbers more than the Smithsonian ones, but neither is counting a unique individual only once. If you were to guess that the average visitor visits 4 WDW parks, and 2 Smithsonian museums, the numbers come out about equal.

The main point that was being made, though, is that lots of people still enjoy going to museums. And, that seems reflected in the fact that a lot of people go to the Smithsonian each year (though, as that (old) article points out, attendance seems to be dropping, overall).
 
If an individual pays for 4 admissions, than they should be counted 4 times. In the case of WDW, that would mean they should be counted at least every day, perhaps even twice per day if they buy hopping privileges since they indirectly paid for 2 parks in 1 day.
It wouldn't make sense to count an entire vacation as 1 admission. That person could spend 6 out of 7 days at Universal, or all 7 at WDW. :smokin:

MG
 


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