HSC Stuff > HSC Mathematics Extension 1
3U Maths Question Thread
jaccs:
consider the unrestricted choices: 18C11 =31 824
if we remove the 3 left-handed from the pool to choose from, that leaves 15 right-handed to select 11
then subtract the case of no left handed: 15C11 = 1365
at least one left handed is 31 824 - 1365 = 30 459
the alternative is to find all the cases: 1, 2, 3, left handed students and add them
3C1 x 15C10 + 3C2 x 15C9 + 3C3 x 15C8 = 30 459
jaccs:
Question 1 from the mock trial in the Atar notes extension 1 topic tests
Is there something missing from the information in this question?
Gerald is conducting an experiment at a golf course. He needs to guess the lowest amount of golf balls on the golf course at any given time. He is given the following information:
– There are 3 holes that a golfer can choose from
– It is guaranteed that one hole contains at least 4 balls.
What is the minimum number of balls Gerald should guess?
the solution says 3 x 3 + 1 = 10
do we assume a certain number of balls in any hole - at least 3 and then one has to have at least 4?
or am i just completely misinterpreting it
thanks
Opengangs:
--- Quote from: jaccs on August 06, 2020, 04:15:13 pm ---Question 1 from the mock trial in the Atar notes extension 1 topic tests
Is there something missing from the information in this question?
Gerald is conducting an experiment at a golf course. He needs to guess the lowest amount of golf balls on the golf course at any given time. He is given the following information:
– There are 3 holes that a golfer can choose from
– It is guaranteed that one hole contains at least 4 balls.
What is the minimum number of balls Gerald should guess?
the solution says 3 x 3 + 1 = 10
do we assume a certain number of balls in any hole - at least 3 and then one has to have at least 4?
or am i just completely misinterpreting it
thanks
--- End quote ---
Hi there!
I admit that I may not have framed the question as well as I should have but the intention was to apply the pigeonhole principle but in reverse! You have 3 holes and you want to guarantee that one hole contains 4 balls. We observe that that if we stack 3 balls in each hole, then the next ball MUST guarantee that one of these holes contains 4 balls. So the minimum number of balls that guarantees 4 balls must be (3 x 3 + 1 = 10).
jaccs:
Thanks for the clarification, that makes sense, if the golfer is sinking balls in the three holes.
Thanks again.
twelftholmes:
hey! with solving this differential equation I'm confused about a step in the worked solution.
Here is the link: https://imgur.com/Js9b5Hz
Why did they move the constant C from the power to the front and turn it into a k? I know this is a log law but I would initially think to take the whole power (x^2/2 +C) to the front and turn it into k, not just C. Is this just something you are able to do (like a rule) or is there a line of working out that they grouped into one (which frustratingly this textbook usually does haha).
Thanks in advance for any assistance!
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