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Yes you are.. for your first q, let me give you the main idea. If you still have trouble solving it, ill post the whole workinghttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .
Q6ii) Firs you need to find the probability of lengths more than one standard deviation from mean. To get this idea, imagine the STANDARD NORMAL DISTRIBUTION curve in which 0 is the mean. When it says more than one standard deviation from mean, it means more than +1 and -1 from the mean 0. See the attached image for better clarification.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .
For q7) Yes you have got the idea.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .
Can someone explain this?Can anyone explain why they gave Z= 1.882 and p=0.97? Q5) I)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
Honestly you are so helpful . Thank a tonne for the help you are providing me May you pass with bright colours in all your exams ! Good Luck !part 3, sorry it took so much time
oh and i forgot to mention ( though its obvious)
the denominators cancel out a part of the numerators so only the 8P8 's are left
Can anyone explain why they gave Z= 1.882 and p=0.97? Q5) I)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
part (i)
12P8
=19958400
part (ii)
so frances and mary do not sit together,
just find out the number of ways in which they DO sit together and subtract from the total
if they sit together then 11P7 x 2P2 - - - - (11P7 because since they sit together you count them as one unit, 2P2 because they can arrange themselves in their group, sit left or right)
11P7 x 2P2
=3326400
this is when they do sit together, so subtracting from total ( part (i))
19958400 - 3326400
=16632000
part (iii)
i'll just upload the working for it in a bit
"Jiyad Ahsan, post: 564620, member: 31829"]part 3, sorry it took so much time
oh and i forgot to mention ( though its obvious)
the denominators cancel out a part of the numerators so only the 8P8 's are left
Thanks. Q1 Is it okay to write the probabilities in simplified form? For eg P(0)= 7/24 instead of 210/720?All credit to JiyadAhsan ....
Thanks. Q1 Is it okay to write the probabilities in simplified form? For eg P(0)= 7/34 instead of 210/720?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf
So 7/24 is accepted even though it is not written in the marking scheme?YOU NEED TO WRITE in standard form![/quote)
So 7/24 is accepted even though it is not written in the marking scheme?
So 7/24 is accepted even though it is not written in the marking scheme?
is this right?
jub < hota hai to direct table read krte hain
jub ≤ hota hai tou 1 se minus krte hain
jub > hota hai tou 1 se minus krte hain
jub ≥ hota hai tou direct lete hain
agar jo hum nikal rhe hain wo negative ho
tou 1 mai se minus krskte hain
Please?Somebody help me with 3b, please.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
So 7/24 is accepted even though it is not written in the marking scheme?
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