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  1. umarFM

    Statistics Paper 62 -- How was it?!

    n = 22
  2. umarFM

    Statistics Paper 62 -- How was it?!

    coz more than 2 and less than 12 was required...
  3. umarFM

    Statistics Paper 62 -- How was it?!

    marks cant say anything...probably 1 or 2
  4. umarFM

    Statistics Paper 62 -- How was it?!

    fr the second question.. dnt remember the anser... but P(good) = .65 i guess.. we hv to find out the probability for =0,1,2&12.....add these together and subtract it from 1...
  5. umarFM

    Statistics Paper 62 -- How was it?!

    Q5 i) 11C3 ii) (8C4 x 3C2) + (8C5 x 3C1) + (8C6) iii) 9C6 + 9C4
  6. umarFM

    Mathematics: Post your doubts here!

    provided u r doing the question in binomial before...
  7. umarFM

    Mathematics: Post your doubts here!

    http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w11_qp_61.pdf Q1 look at this question.... u hv fixed nmber of trial.... probability of success and failure....all the data fr binomial...u need to find the probability for more...
  8. umarFM

    Mathematics: Post your doubts here!

    do u hv any???
  9. umarFM

    Mathematics: Post your doubts here!

    when lets say, u are given...a fixed nmber of trial eg a dice is thrown 200 times.... use a suitable approximation to find the probability that more than 60 times it landed on a 3..... or when u hv all the binomial related data ie; number of trials....probability of success and failure...and u...
  10. umarFM

    Mathematics: Post your doubts here!

    original value value is used when the question is only of normal.... we use continuity correction when we convert from binomial to normal distribution....
  11. umarFM

    Mathematics: Post your doubts here!

    Probability(Z>(1002-u)/8)=225/900 let (1002-u)/8) = a 1 - phi(a)=0.25 phi(a)=0.75 phi(a)=phi(0.674) a=0.674 (1002-u)/8)=0.674 u=997
  12. umarFM

    Maths GT for P3 and M1

    p3 probably 57
  13. umarFM

    Mathematics: Post your doubts here!

    when digits are repeated..its (number of digits given)^spaces given... spaces means how many digit nmber is required..... taking the same example again.... 1,2,3,4&5...nd making a 5 digit nmber.... if it has to start with an even and end with an odd...nd digits are nt to be repeated so for the...
  14. umarFM

    Mathematics: Post your doubts here!

    lets say we hv nmbers... 1,2,3,4 nd 5.... no of arrangements of a 5 digit nmber if digits can be repeated.... in first digit we can choose frm 5 digits, so 5 possibilites..as the digits can be repeated hence we again hv 5 possibilities fr 2nd digit... so total arrangements are 5x5x5x5x5=5^5 if...
  15. umarFM

    Economics, Accounting & Business: Post your doubts here!

    anser is C...... opening net book value - disposals + additions - closing book value = depreciation 245000 - 16000 + 92000 - 268000 = 53000....
  16. umarFM

    Economics, Accounting & Business: Post your doubts here!

    anser is C..... in income and expenditure account. u need to show the subscription for the year 2009 (regardless of in which year it is being recieved) which can be caluclated by adding 2nd row (300+2100+400=2800).... fr the reciepts and payments account, u need to show the subscription amount...
  17. umarFM

    Mathematics: Post your doubts here!

    Help me out ! Nov 09,Q5 (a) Find how many numbers between 5000 and 6000 can be formed from the digits 1,2,3,4,5 and 6 i) if no digits are repeated now u hv to make a 4 digit nmbr between 5000 nd 6000..... so we hv 4 spaces... - - - -.... the first digit must be 5, so we can have only 1 digit...
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