Section II (1 Viewer)

[emma211]

Hmm maybe. Then perhaps both would be right...?

hopefully for you

 

[matttayl]

you know what i hate, coming on BOD and reading everyones answers. Then seeing i got questions wrong. It just makes me more depressed..

haha im the same

 

[knel]

for 28c i dived how far he could see out to sea somehting like 4.5k by how high his eyes where. like 1.6 that equalled 2.8 then i times that by 16 or however high it was and bingo there is your answer...... i hope, i did do in no joke like 5seconds

 

[matttayl]

for 28c i dived how far he could see out to sea somehting like 4.5k by how high his eyes where. like 1.6 that equalled 2.8 then i times that by 16 or however high it was and bingo there is your answer...... i hope, i did do in no joke like 5seconds

the answer was 17.8

using the equation h = k d^2

k = 4.5^2/1.6

 

[BlueCatBulletCD]

hopefully for you

Well if I've shown all the working out to the answer (which I did) they are looking for I'll get the marks.

 

[BlueCatBulletCD]

the answer was 17.8

using the equation h = k d^2

k = 4.5^2/1.6

You mean 1.6/4.5^2

 

[MzBiiBii]

you know what i hate, coming on BOD and reading everyones answers. Then seeing i got questions wrong. It just makes me more depressed..

lol me to... but i can't seem to stay away

 

[matttayl]

You mean 1.6/4.5^2

yes my mistake

 

[I Study Hard]

also. the last question was in my opinion the hardest. i said juan was correct, because there were 8 differences between the theoretical probability and the first experiment, where as there were only 4 differences between the theoretically probable answer and experiment two... what did everyone else get, and what were your explanations?

I said that since the chance of rolling any one face on the dice was 1/6 that if the dice were rolled 18 times then theoretically each number should have come up 3 times, and therefor he was wrong, because experiment 1 was closer to the expected result.

 

[eagle09]

fuckkk i didnt fully simplify 25a . - still get one mark if i wrote -2x - 9 ?

you cant simplify dat any further

 

[mystiques4]

I said that since the chance of rolling any one face on the dice was 1/6 that if the dice were rolled 18 times then theoretically each number should have come up 3 times, and therefor he was wrong, because experiment 1 was closer to the expected result.

The question was difference between two dice, not what the value of the dice was.

 

[emma211]

I said that since the chance of rolling any one face on the dice was 1/6 that if the dice were rolled 18 times then theoretically each number should have come up 3 times, and therefor he was wrong, because experiment 1 was closer to the expected result.

I think the question was talking about the difference between the numbers rolled on the dice e.g. if 4 and 5 were rolled the difference is 1

i think you had to draw one of those tables with like 1-6 across the top and 1-6 down the side. then fill in what the difference's would be in the middle.

if that's right then exp 2 was closest to the theoretical probability

 

[matttayl]

I said that since the chance of rolling any one face on the dice was 1/6 that if the dice were rolled 18 times then theoretically each number should have come up 3 times, and therefor he was wrong, because experiment 1 was closer to the expected result.

thats what i said, because from the sample space there were 36 possible outcomes and of which the 6 different differences were possible 6 times. eg 6/36 - 1/6. they all had equal chances of occuring

 

[Question?]

I said that since the chance of rolling any one face on the dice was 1/6 that if the dice were rolled 18 times then theoretically each number should have come up 3 times, and therefor he was wrong, because experiment 1 was closer to the expected result.

DUDDE sorry but ur rong

 

[BlueCatBulletCD]

There was a 6/36 chance of getting a difference of 0, 10/36 for 1, 8/36 for 2, 6/36 for 3, 4/36 for 4, 2/36 for 5. You just multiply these chances with 18 to get the difference they should theoretically show. So experiment two was more correct..

 

[tegan-leanne]

I think the question was talking about the difference between the numbers rolled on the dice e.g. if 4 and 5 were rolled the difference is 1

i think you had to draw one of those tables with like 1-6 across the top and 1-6 down the side. then fill in what the difference's would be in the middle.

if that's right then exp 2 was closest to the theoretical probability

that's how i did it. i didn't have time to write soem examples of the theoretical probability though. hopefully i would still get 3 marks for that though because i have an answer and the other parts to the question.I don't think i did very will in the test. byebye uni! :[

 

[Question?]

There was a 6/36 chance of getting a difference of 0, 10/36 for 1, 8/36 for 2, 6/36 for 3, 4/36 for 4, 2/36 for 5. You just multiply these chances with 18 to get the difference they should theoretically show. So experiment two was more correct..

that is exactly what i done

 

[emma211]

that's how i did it. i didn't have time to write soem examples of the theoretical probability though. hopefully i would still get 3 marks for that though because i have an answer and the other parts to the question.I don't think i did very will in the test. byebye uni! :[

haha yer don't worry about it, you should be sweet.

 

[cmanassa]

QUESTION 27c

3 raffles - 100 tickets each

MARY: 2/100 = 1/50 = 0.02 <B>(1 mark)</B>

JANE:
Propability of winning both ticks is 1/100 x 1/100 = 1/10,000 <B>(1 mark)</B>
Probabilty of winning first draw but not the second = 1/100 x 99/100 = 99/10,000
Propability of winning second but not first = 99/100 x 1/100 = 99/10,000

CONCLUSION...
Propability of winning AT LEAST 1 (is the sum of the above 3 calculations)
1/10 000 plus 99/10 000 plus 99/10 000 = 199/10 000 = 0.0199 <B>(1 mark)</B>

Hence, mary has a slightly beter chance... <B>(1 mark)</B>

 

[Question?]

QUESTION 27c

3 raffles - 100 tickets each

MARY: 2/100 = 1/50 = 0.02 <B>(1 mark)</B>

JANE:
Propability of winning both ticks is 1/100 x 1/100 = 1/10,000 <B>(1 mark)</B>
Probabilty of winning first draw but not the second = 1/100 x 99/100 = 99/10,000
Propability of winning second but not first = 99/100 x 1/100 = 99/10,000

CONCLUSION...
Propability of winning AT LEAST 1 (is the sum of the above 3 calculations)
1/10 000 plus 99/10 000 plus 99/10 000 = 199/10 000 = 0.0199 <B>(1 mark)</B>

Hence, mary has a slightly beter chance... <B>(1 mark)</B>

FUCKKK YESSS. are you right tho. please say yes lmao