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.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
Hey nah it's notlozjoz said:hey for question 22 isn't the answer B)
i could be completely wrong but just wondering?
please someone answer
JoseshW said:Guys, did anyone get this for the last simpsons rule:
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?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.
Apparently you had to use the present value of an annuity formula.....?hanna91 said:did anyone get the answer right to question 27 b(i) no one in my class could prove it was $2178.67?
You use present value formula, then move the bracket with (n and r) and divide iby the Annuity, and u would get $2178.6735173 said:Apparently you had to use the present value of an annuity formula.....?
I didnt do it, but thats what I heard after
I don't think it can be a or b because they are essentially the same answerKelly47 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?
After what you said I thought you might be right, so I went to check the options again.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
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 ~~PC said:2008 GENERAL MATHS HSC EXAM
(a) (i) 8377 km (or 8334 km)
(ii) 9 hours
(iii) 12 pm Wednesday
The answer is definetely that 6 occurs 3 times moreKelly47 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.
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.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.