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51+1/4 for LQby which formula u got this result ? why its not LQ=520 and UQ = 690 ?
and (51+1)*3/4 UQ
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51+1/4 for LQby which formula u got this result ? why its not LQ=520 and UQ = 690 ?
What is the difference between separated and not together ?? no differenceYou are finding the probability where all are separated but we have to find where all are not together. This can be done: All possible - All vowels together
For together: 8! × 4/2!2!=40320
Subtract this from the total ways.
7C2 or 7C1 cannot be used as 2 or 3 E's are already chosen so remaing are 4 not 7.
When 2 are chosen, it will be 4C2 for remaining are 4C1 when all 3 E's are chosen.
0 will be when it will be answered in first attempt which is 1/2.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_06.pdf
Q6 part two how to get each probability with brief explanation
Thanks got it was confused bcz didnot think of the prob of being answered0 will be when it will be answered in first attempt which is 1/2.
2 when are 2 unanswered and 1 answered so (1/2)^2*1/2.
3 when are 3 unanswered and 1 answered so (1/2)^3*1/2.
4 when are 4 unanswered and 0 answered so (1/2)^4.
NOT ALL next means there can also be 2 or 3 together, whereas you are finding where all are completely separated.What is the difference between separated and not together ?? no difference
noo they are choosing 4 letters from all and 2 are already there from the four ??
I am not convinced with what u said can u explain more ???
Thankss
We don't, weight is on the x-axis and frequency density on the y-axis.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
Q4 iv) can anyone PLEASE explain why we will use frequencies on xaxis
ohhh okaaaii ,Thank youu aloootNOT ALL next means there can also be 2 or 3 together, whereas you are finding where all are completely separated.
When 2 E's are chosen, 2 are already there. Now we have to choose rest of 2 from the 4 which are not E's
When 3 E's are chosen, only 1 has to be chosen from the rest of 4.
We divide when items are identical but here it is mentioned that they are different.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_06.pdf
Q3 part one should we divide the total by the no. of repetitions bcz in the ms they have not can u tell why did they not divide
yeaWe don't, weight is on the x-axis and frequency density on the y-axis.
Its completely Ok. Glad to helpohhh okaaaii ,Thank youu alooot
I think your mad of me sorry
Do you have any idea of the query I posted above?ohhh okaaaii ,Thank youu alooot
I think your mad of me sorry
No what is it ??Do you have any idea of the query I posted above?
okkaii i will try it nowNo what is it ??
You are finding the probability where all are separated but we have to find where all are not together. This can be done: All possible - All vowels together
For together: 8! × 4/2!2!=40320
Subtract this from the total ways.
7C2 or 7C1 cannot be used as 2 or 3 E's are already chosen so remaing are 4 not 7.
When 2 are chosen, it will be 4C2 for remaining are 4C1 when all 3 E's are chosen.
why do you wanna do thathttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf
Q6 (ii) What if plastic mugs were separated instead of the china mugs? Anyone? ZaqZainab
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