Section II (1 Viewer)

[Bootawa_Boy]

I put it would be true if there was more girls than boys because the graph shows that boys spend longer on the internet than girls.

I put is the mean 5 and upper quartile range 6 where both in girls and boys

 

[Macdwg]

Yeah, it's more girls than boys. It's gotta be, if there is to be the same number of girls outside their interquartile range as there are boys close to their median.

Yeah but you're only talking about the amount of hours between 5-6 which means that there would have to be the same amount in total if 25% for each exist in that range. Thats the only way that you could conclude the same amount go onto the internet for 5-6 hours. It's not asking for all hours, its asking for between 5-6.

 

[Intermission]

Yes, but there are less in the extremities of the distributions.

If 5-6 hours is outside the interquartile range for girls, and inside it for boys, then for there to be the same amount of kids in those hours, there's got to be more girls surveyed than boys.

 

[Bond/Catalyst]

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

Sorry to burst your bubble peeps. They have the exact same chance.first chick equals 2/100 chance fullstop of winning the prize in her barrel. Thats her probability of winning.second chick has a 1/100 chance of winning in the second barrel and 1/100 in the third barrel. Therefore the second chick's probablity is.1/100 + 1/100 = 2/100therefore second chick = first chick.U need to plus them together as they are separate events. If they were multi stage events then taht means u multiply them. Second chicks chances of winning in each barrel are 1/100 and are unaffetced by any other outcome, so u add the two togethr.Pretty sure its right.

 

[rowdyroddy]

What did everyone get for Simpson's Rule question?

cant remember answer but was quite easy, just use the given measurements to work out the substitutes for the simpson rule formula, i think it was 20+35 for the left side, and then 30+10 for the right side then 15 because 55 all up but only 40 was there. and the middle is same thing. volume is just x.60

 

[rowdyroddy]

Sorry to burst your bubble peeps. They have the exact same chance.first chick equals 2/100 chance fullstop of winning the prize in her barrel. Thats her probability of winning.second chick has a 1/100 chance of winning in the second barrel and 1/100 in the third barrel. Therefore the second chick's probablity is.1/100 + 1/100 = 2/100therefore second chick = first chick.U need to plus them together as they are separate events. If they were multi stage events then taht means u multiply them. Second chicks chances of winning in each barrel are 1/100 and are unaffetced by any other outcome, so u add the two togethr.Pretty sure its right.

actually the answer was they both have a probability of 0 because chicks never win anything.

 

[Bond/Catalyst]

actually the answer was they both have a probability of 0 because chicks never win anything.

lol

 

[lookleft]

Sorry to burst your bubble peeps. They have the exact same chance.first chick equals 2/100 chance fullstop of winning the prize in her barrel. Thats her probability of winning.second chick has a 1/100 chance of winning in the second barrel and 1/100 in the third barrel. Therefore the second chick's probablity is.1/100 + 1/100 = 2/100therefore second chick = first chick.U need to plus them together as they are separate events. If they were multi stage events then taht means u multiply them. Second chicks chances of winning in each barrel are 1/100 and are unaffetced by any other outcome, so u add the two togethr.Pretty sure its right.

Negative. First chick has 2/100 chance of winning AT LEAST ONE PRIZE The at least bit is the key to to the qn. Therefore, the second chick has the probability of winning both, winning first one and not the second, and winning the second and losing the first one. Like this: (1/100 x 1/100) + (99/100 x 1/100) + ( 1/100 x 99/100) = 199/10,000 That's what I got. So first chick has a 2/100, or a 200/10,000 chance of winning at least one prize. The second chick has a 199/10,000 chance of winning at least one prize.

 

[noturningback]

cant remember answer but was quite easy, just use the given measurements to work out the substitutes for the simpson rule formula, i think it was 20+35 for the left side, and then 30+10 for the right side then 15 because 55 all up but only 40 was there. and the middle is same thing. volume is just x.60

Wasnt the lake in the middle of the page?
So you had to find the total area of rectangle

then
find the area of grass at the bottom and top using simpsons rule

then
total area minus grass area = lake in middle surface area

the lake was in the middle right?

 

[bgriffith]

What was the answer to 25d ii)???

 

[Nivek5]

What was the answer to 25d ii)???

83.85%
68% + ((95-68) /2)+((99.7-95)/2) i beleive

 

[2pmhottest]

Guys for Question 27 C GUYSMary has the better chance of winning let me explain this to youMary has 2/100 tickets which is 1/50 which is 2 percent. Jane has 1/100 + 1/100 so she also has 2 percent chanceHOWEVER the question states determine the better chacne of winning AT LEAST one PRIZEso there for mary still has a 2/100 chance of winning at least one prize and also 98/100 to LOSEJANE HOWEVER has 1/100 to win 1st raffle and 99/100 to lose the 2nd raffle is the same 1/100 to win and 99/100 to loseBUT JANE U MUST CALCULATE HER CHANCE OF WINNING AT LEAST 1 prize so it is1/100 x 99/100 + 1/100 x 99/100 WHICH EQUAL to 1.98% of her winningso therefore MARY HAS a 2 % chance of winning while jane has a 1.98 chance of WINNING AT LEAST ONE PRIZEand i am 99% sure this is the correct answer let me know what u guys think

 

[ajay12]

yeah 27 C) i had them as equal chance? hmm.. what you guys do for 25 A)?

5 - 2(x + 7) = 5x + 35 -2x -14 = 3x + 21 .:. x = 7 Fucked up Question 24 (i) bullshitted my answer for (ii). Fucked up Question 25 (b) had something like 2 x 10^-5 or something. Fucked up Question 25 (c) (ii) did Simpsons rule 508m x 0.6m, but didn't times that by 1000, only dividing by 4, lol. RE: Q26 (i) Do you just minus the Upper Quartile from the Lower Quartile? Probably not. Question 27 (c) Did anyone get 1/50 for both Jane and Mary? I heard people saying they got 1/10,000 for Jane, but I'm pretty sure you don't 1/100 x 1/100, instead 1/100 + 1/100. Fucked up Question 28 (a:i) And didn't know what a sample space... Other than that, I aced it.

 

[sir-study-alot]

what did every one get for 23 d ?

 

[hazza9]

I think I got around 500 (maybe 508 or 540ish?) for the Simpson's Rule or something. Don't know exactly don't have the paper n can't remember.What did other people get?

 

[2pmhottest]

man i have no idea how to put gaps into the sentences anyway LOOK AT MY SOLUTION FOR QUESTION 27 C !!!

 

[rowdyroddy]

83.85%
68% + ((95-68) /2)+((99.7-95)/2) i beleive

we have a winnar

 

[eagle09]

well some otha dude said that . you dissing me brother . oi where did you go after school ?

jussss went to go find som motivation ayyyy. haha. nahh i had ta get gas from faulco servooo. did ya c dat car crash?

 

[sir-study-alot]

i agree with the winner

 

[sir-study-alot]

yeah bro , they crashed into the rail . oi that post has got me like 40 rep. points lol . and im sooo pissed off i got 15/22 for mutliple choice ..buh i think i got 80% for the ewhole paper