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2020-08-16, 19:41
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#1 |
Jun 2010
FE16 Posts |
Math knowledge of the general population?
Suppose you picked a whole bunch of American adults at random and asked them the following question:
You're on a beach and see a cruise ship sailing away from you. The top of the ship is 158.4 feet / 0.03 miles above the ocean, and your eyes are 5.28 feet / 0.001 miles above the ocean. Assume that the Earth's diameter is 8,000 miles, that it's a clear day, and that there are no boats, islands, or other obstructions between you and the ship. How many miles will the ship be from you at the moment when the top of it disappears over the horizon and is no longer visible to you? Round your answer to one decimal place. They're not allowed to use the Internet or any other outside help and are given nothing besides pencils and blank papers. No calculators, equations, theorems, or math textbooks. If they can correctly answer that question in less than an hour and can correctly show how they got that answer, they'll get a $100 reward. Under those conditions, what percentage of them do you think would win the $100? My guess is 10%, but I wonder what everyone here thinks. |
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2020-08-16, 20:04
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#2 |
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∂2ω=0
Sep 2002
República de California
103×113 Posts |
Why waste time with idle hypotheticals when you can actually be out gathering some data? I suggest you procure some $100 bills and actually conduct the experiment you propose - find some group of adults who look like they might be willing&able to spare up to an hour for the chance to earn $100, say at the local coffee shop or whatnot, and see how many you can get to try your deal.
Of course there is the chance that you might end up shelling out more $ than you like, but if after, say the initial sample of 10, the numbers are such that yu decide to discontinue the experiment, then your question has been answered at least to a "at the place I selected, more than I would have expected" sense. Last fiddled with by ewmayer on 2020-08-16 at 20:04 |
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2020-08-16, 20:23
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#3 | |
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6809 > 6502
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Aug 2003
101×103 Posts
23×1,223 Posts |
Quote:
And you can't use an equation? So no math to do math..... Last fiddled with by Uncwilly on 2020-08-16 at 20:28 |
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2020-08-16, 20:25
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#4 |
Dec 2012
The Netherlands
170210 Posts |
It depends how many of them know about the iceberg...
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2020-08-16, 21:22
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#5 | |
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∂2ω=0
Sep 2002
República de California
103×113 Posts |
Quote:
circle: x^2 + y^2 = r^2 Taking only + signed solutions, for given r and eps << r, we want to estimate x which yields y = (r-eps). For eps = 0, clearly x = 0. Assume the corresponding x << r, though this needs to be checked post hoc. solve for y = sqrt(r^2 - x^2) = r*sqrt(1 - x^2/r^2). For x << r, this can be approximated as y ~= r*(1 - x^2/(2*r^2)). Setting y = r - eps ~= r*(1 - x^2/(2*r^2)) = r - x^2/(2*r) and solving for x we have x ~= sqrt(2*r*eps). [I believe conventional notation for the so-called horizon problem uses h in place of my eps.] For the stated problem, we have r = 4000 mi = 4000*5280 ft and need to solve for 2 values of eps and add the results: eps1 = 5.28 ft, which = r/4e6, yielding x1 ~= sqrt(2*r^2/4e6) = 2*sqrt(2) mi. Approximating sqrt(2) ~= 1.4, x1 ~= 2.8 mi. eps2 = 158.4 ft = 30*eps1 = 30*r/4e6, yielding x2 ~= sqrt(240) mi. Now 15^2 = 225 and 16^2 = 256, so approximate x2 = sqrt(240) mi ~= 15.5 mi. Thus my answer rounded to the nearest int is that the ship will be (2.8+15.5) ~= 18 mi away from the observer at the point the top of its mast sinks out of sight. That took me the better part of an hour, so I would surmise that of a randm sampling of adults, for sure < 10% would win the $100 ... probably < 1%, but I don't want to overestimate my own abilities. Last fiddled with by ewmayer on 2020-08-16 at 21:25 |
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2020-08-16, 21:29
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#6 | |
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6809 > 6502
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Aug 2003
101×103 Posts
23·1,223 Posts |
Quote:
https://en.wikipedia.org/wiki/Pythagorean_theorem |
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2020-08-16, 21:37
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#7 | |
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∂2ω=0
Sep 2002
República de California
103·113 Posts |
Quote:
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2020-08-16, 22:26
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#8 | |
Sep 2002
Oeiras, Portugal
26×23 Posts |
Quote:
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2020-08-16, 22:37
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#9 |
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6809 > 6502
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Aug 2003
101×103 Posts
23·1,223 Posts |
I had a FE'er explain to me that with a really good specific camera one could continue to see the ship. And when I tried to explain atmospheric distortion and the like, they went off. Since they did not like anything that they could not personally experience, I tried to build off of the idea of defraction and mirages. They would not stay calm to allow me to complete my point.
Their claim was that gravity is not a thing it is just things stratifying based upon their density. |
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2020-08-16, 22:50
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#10 |
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Sep 2002
Oeiras, Portugal
26×23 Posts |
By the way, I did the calculations using very basic geometry results and came to ~ the same value as ewmayer. In my case, 18,32mi, which, according to the OP´s request, was rounded to 18.3
I don´t live in the US, do I qualify for the $100? 
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2020-08-16, 22:57
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#11 | |
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Sep 2002
Oeiras, Portugal
26·23 Posts |
Quote:
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