Mathematics Extension 1 HSC thoughts (1 Viewer)

Trebla

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Wait so scaling obs applies on the corresponding raw of your average HSC mark right?
Correct. Your raw mark branches off into two different types of transformations:
- Aligning (for your HSC certificate)
- Scaling (for your ATAR)
 

mmmmmmmmaaaaaaa

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do the regular projectile motion things to find time of flight (let y=0 etc), solve simultaneously for x and rearrange, you're left with:



Then, optimise the function on the LHS and you get the maximum value of is equal to 0.369..., so then all other d must be below
The one before that, the inequality one
 

tgone

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\
The one before that, the inequality one
Perpendicular distance is minimised distance between a line and a point, any other magnitude between the point (u) and the line (lambda(v)) willl be greater or equal to, just draw a diagram to prove the given is perpendicular distance with projection formulae, and construct the line lambda(v)— the approach mentioned above is the same thing in algebraic terms
 

Daedalus13

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I redid this and got 402. can anyone concur?
I did the approach that 011235 mentions, but got the same answer as you the first time, 358. 402 seems much too large to be plausible if it's 95% chance someone will turn up.
 

tgone

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I redid this and got 402. can anyone concur?
I haven’t looked at your solutions so I’m not sure your methods, but I just defined X~Bin(n,0.95) where n is number of tickets sold, and X is the event where the passenger DOES go on the flight.

By the managements decision:
P(X>350)<0.01
P(X=<350)<0.9900
taking the maximum case for maximum seats, use the normal approximarion for Z=<2.33 (which has probability 0.9901)

Solve a quadratic for the case p-hat=350/n for the Z-score = 2.33, and you get 358 (rounded)

Apologies if this is already the solution, just typing it out here to remind myself haha.
 

tgone

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Here are my typeset solutions to this year's MX1 HSC exam. I wrote these in a great rush so there are probably mistakes lol
11e I believe R=2rt3, I don’t think (1)^2 and (2)^2 were added properly
 

notme123

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Here are my typeset solutions to this year's MX1 HSC exam. I wrote these in a great rush so there are probably mistakes lol
Why is the range of f in 13 c the reals

i thought they were inverses but some people here think otherwise. idk what the reason is tho

EDIT: wait i know f cannot have an inverse since its not monotonic but is that enough?
 

tgone

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Why is the range of f in 13 c the reals

i thought they were inverses but some people here think otherwise. idk what the reason is tho
the question gave that f had domain in all real x
so the range of inverse should be in all real x, which it isn’t for g, so it can’t be the inverse.
 

sneak11

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the question gave that f had domain in all real x
so the range of inverse should be in all real x, which it isn’t for g, so it can’t be the inverse.
in canteen’s solutions, he has stated that the angle for 11. e) is -pi/3. would it also be true that the angle could be 2pi/3 for that?
 

Longbottom

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The multis were straight traps I could think of a lot of people doing 23pi/12 for 1 and myself doing 9 as A because brain fart city from the rest. Everything except all the first year actuarial stuff was good though.
 

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