Subcategories

• Module 1 Day 15 Challenge Part 1

  Distance, Rate, Time

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  Hi! The question says,"You bike up a hill at 4 miles per hour and come back down the same way at 12 miles per hour. What is your overall average speed for the entire roundtrip?"

  I'm confused of why the average speed isn't equal to all the speeds mentioned / the number of speeds. Like, if you're driving on a highway and for the 1st second your speed is 100 km/h, for the 2nd second your speed is 102km/h and for the 3rd second your speed is 106 km/h, your average speed would be 102km/h which you get by doing (100+102+106)/3 =102 km/h. Following this logic, wouldn't the average speed equal (12+4)/2= 8 miles per hour?

• Module 1 Day 15 Challenge Part 2

  Average Speed

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  The posts here are repeated...

• Module 1 Day 15 Challenge Part 3

  Nested Rational Equation, Reciprocals

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• Module 1 Day 15 Challenge Part 4

  Alternative Formula for Harmonic Mean

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  Re: Mysteries - Who invented the harmonic mean and what is its purpose and why does it work?

  @thomas This was a really good explanation! I really like how you explained why this is "weird", and why the usual arithmetic mean doesn't work - that because 12 mph is greater than 8 mph, so when traveling the same distance, you spend less time going at 12 mph, so the answer should be closer to 8 mph instead of right in between 8 and 12. Still, I don't really get why the harmonic mean works. Why do I want to find the average time it takes ((1/x+1/y)/2) to get the average speed?

  In the last problem about the houses you included, is the answer supposed to be 1/(1/2+1/3)=6/5 hours? I don't understand why here the harmonic mean is not applied. Thanks!

• Module 1 Day 15 Your Turn Part 1

  Lap Rate Problem

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• Module 1 Day 15 Your Turn Part 2

  Unit Rates, Application of Harmonic Mean

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• Module 1 Day 15 Your Turn Part 3

  Unit Rates, Application of Harmonic Mean

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• Module 1 Day 15 Your Turn Part 4

  Unit Rates, Application of Harmonic Mean

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  @quacker88 Hi! Thanks for the helpful explanation because I had the same question victorioussheep did. It's just when I tried to solve the problem you gave, I got the answer B/(B-A).

  (1/A-1/B)t=1 Simplifying (1/A-1/B), I get (B-A/AB)t=1 and so t= AB/B-A. t is the time it takes for the first runner to catch up with the second runner. So the number of laps the first runner runs is AB/B-A divided by A which is B/B-A.

  Also, I think how Prof. Loh got A/A-B was by solving for the number of laps the trainer or the second runner was running.

  I don't know if this is correct, but I hope you can look over my work. Thanks!