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Answers to General Maths Exam (1 Viewer)

henry08

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Kiim2507 said:
Do you mean the body mass index question?

B = m/h^2
21 = m/1.625^2
Therefore m = 21 x 1.625^2
= Whatever the answer is
Of course. I feel stupid now as I didn't use the graph.
 

lozjoz

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hey for question 22 isn't the answer B)
i could be completely wrong but just wondering? :uhoh:
please someone answer
 
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Darthelokity

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savio23q said:
I'm still not quite sure about answers to the 4 mark cheese question. Four markers are usually trickier than they look, and although this DID look tricky, the answers posted up seem a bit too simplistic to me. When I first looked at the question, I got those answers, but after thinking about it for a bit, it didn't look right to me.

Length – Circumference of a sector multiplied by 4.
40/360 x 2 x П x 15 x 4 = 41.9cm (to 1 dec. Place)

Depth – Depth of the sector multiplied by 3.

7x 3 = 21cm

Height – For a 4 mark question, I don’t think height of the box is not merely the radius of the sector shown. I reckon it’s far too simple to be that. How can the height of the box be 15cm if the radius is slopping up? This is my working out.

Circumference of the sector divided by 2 (to break the sector into a right angle triangle(ignoring its curve))

40/360 x 2 x П x 15 = 41.889...

41.889/2 = 5.235987..

Now use the radius of the sector (15) and Pythagoras’ theorem to work out the height of the sector.

15^2 – 5.235987^2... = 197.5845...

Square root of 197.5845 = 14.1 cm (to 1 dec. place)

That's what I got...I don't know if it's right though :(
I agree, looked for along time at this question convinced they were asking for something more than simplistic. However the main amount of my concern now is that, i thought technicaly the length of the box isnt the prodnct of the length of the 4 sectors, becaue there curved. I used the idea of the next section and used the length as if there was the bottom cut off.

the width can only b 7 x 3

My method for the height was spliting the sector down the middle 2 make a right angle traingle and using 20 degrees and the length of radius to use trig to find the perpendicular height. now thinking about it i dont think its rite but looked ok at the time.

Darth
 

Kiim2507

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lozjoz said:
hey for question 22 isn't the answer B)
i could be completely wrong but just wondering? :uhoh:
please someone answer
Hey nah it's not

A better way of explaining:

6 appeared 450 times which means the other numbers appeared 750 times
There were 5 other numbers thus they each appeared 150 times
The probability of 1,2,3,4 or 5 appearing is 150/1200 = 1/8
The probability of 6 appearing is 450/1200 = 3/8
Therefore a 6 is expected to appear 3 times as often.
 
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JosephW

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Guys, did anyone get this for the last simpsons rule:

h/3(0+4a+b)+h/3(a+4b+0)
 

savio23q

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JoseshW said:
Guys, did anyone get this for the last simpsons rule:

h/3(0+4a+b)+h/3(a+4b+0)

I simplified it further. The shape is symmetrical, so the two rules you're adding together have exactly the same numbers. This is what I wrote down.

2(h/3(4a+b))

P.S I don't think it's "4b" for the second formula..
Should be: h/3(0+4a+b)+h/3(b+4a+0)
 
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Kelly47

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Kiim2507 said:
Hey nah it's not

A better way of explaining:

6 appeared 450 times which means the other numbers appeared 750 times
There were 5 other numbers thus they each appeared 150 times
The probability of 1,2,3,4 or 5 appearing is 150/1200 = 1/8
The probability of 6 appearing is 450/1200 = 3/8
Therefore a 6 is expected to appear 3 times as often.
I actually disagree with that. That answer said it is expected to happen, but it's all chance, and those numbers would not appear each time, you can't expect it to happen in probability. I think that was just there to confuse people. I think the answer was whichever one said 5 would appear 1/7 times, as the die was biased, meaning 6 would appear 2/7 times, twice the amount of the other numbers. Anyone agree with me?
 

35173

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hanna91 said:
did anyone get the answer right to question 27 b(i) no one in my class could prove it was $2178.67?
Apparently you had to use the present value of an annuity formula.....?
I didnt do it, but thats what I heard after
 

JosephW

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35173 said:
Apparently you had to use the present value of an annuity formula.....?
I didnt do it, but thats what I heard after
You use present value formula, then move the bracket with (n and r) and divide iby the Annuity, and u would get $2178.67
 

Kiim2507

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Kelly47 said:
I actually disagree with that. That answer said it is expected to happen, but it's all chance, and those numbers would not appear each time, you can't expect it to happen in probability. I think that was just there to confuse people. I think the answer was whichever one said 5 would appear 1/7 times, as the die was biased, meaning 6 would appear 2/7 times, twice the amount of the other numbers. Anyone agree with me?
I don't think it can be a or b because they are essentially the same answer
Because if 5 appears 1/7 times and 6 appears 2/7 times then a and b would both be correct
 

shell667

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could anyone explain question 28 ii 2?
i got 82.5% and i dont understand how you got 84?
thanks
 
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D: shyyyyt

okay, i barely got any of those answers

i guess i know my mark ><"

prob a band 2 if im lucky D:

*sigh*, ah wells - shit happens
 

Kelly47

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Kiim2507 said:
I don't think it can be a or b because they are essentially the same answer
Because if 5 appears 1/7 times and 6 appears 2/7 times then a and b would both be correct
After what you said I thought you might be right, so I went to check the options again.
A) "The probability of rolling the number 5 is expected to be 1/7"
B) "The number 6 is expected to appear twice as often as any other number"

Both options look very similar, I admit, but the wording is different. A specifically uses the word 'probability' for 5 appearing 1/7 times. That is correct as that is what the expected probability would be, not the expected times a number would occur.
B implies that 6 will occur twice as often as all of the rest, not talking about the probability.
I admit I could be wrong, but I still think it's A. I remember a past paper or test, it had a similar question about a biased die with six appearing twice as often, and it said it's probability was 2/7.
And I don't think that C can be correct, as just because it occured 450/1200 times, does not mean it always will, that number doesn't really have anything to do with its probability, you can't expect those numbers to occur each time.
I hope I've made sense lol.
 
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bao28

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PC said:
2008 GENERAL MATHS HSC EXAM
QUESTION 27
(a) (i) 8377 km (or 8334 km)
(ii) 9 hours
(iii) 12 pm Wednesday
they said to ignore time zones? so wudnt it still be 7am wednesday? (Ignore time zones.) unless the time zones have nothing to do with the additional "5" meaning i failed the wording part of maths ~~
 

John-G-B-2008

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Kelly47 said:
After what you said I thought you might be right, so I went to check the options again.
A) "The probability of rolling the number 5 is expected to be 1/7"
B) "The number 6 is expected to appear twice as often as any other number"

Both options look very similar, I admit, but the wording is different. A specifically uses the word 'probability' for 5 appearing 1/7 times. That is correct as that is what the expected probability would be, not the expected times a number would occur.
B implies that 6 will occur twice as often as all of the rest, not talking about the probability.
I admit I could be wrong, but I still think it's A. I remember a past paper or test, it had a similar question about a biased die with six appearing twice as often, and it said it's probability was 2/7.
And I don't think that C can be correct, as just because it occured 450/1200 times, does not mean it always will, that number doesn't really have anything to do with its probability, you can't expect those numbers to occur each time.
I hope I've made sense lol.
The answer is definetely that 6 occurs 3 times more

450/1200 X 100 = 37.5.
100 - 37.5 = 62.5
62.5 divided by 5 = 12.5

37.5 % of rolling a 6 and 12.5% chance for every other number. 37.5 is three time 12.5.
 

Kelly47

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John-G-B-2008 said:
The answer is definetely that 6 occurs 3 times more

450/1200 X 100 = 37.5.
100 - 37.5 = 62.5
62.5 divided by 5 = 12.5

37.5 % of rolling a 6 and 12.5% chance for every other number. 37.5 is three time 12.5.
Maybe I'm wrong, maybe I'm being stubborn. But I still disagree. Because 450/1200 is not its probability, that's just the results from 1200 throws. 1200 throws again would likely result in totally different numbers. 450/1200 is not its probability, that's what its statistics were. The probability is the likely-hood of an event occurring, not how many times an event will occur.
That's my opinion anyway.
 

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