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Mathematics: Post your doubts here!

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1 hour 45 min isnt enough! If the papers were 2hours i wouldnt be so damn depressed right now lol.
 
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The 94% probability is the area in the MIDDLE, which means there is a "tail" on either side, of 3% probability. When you look for z value you need to count from the left side, or negative infinity. So here the probability is 0.94 + 0.03 = 0.97
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whenever they use the word USING SUITABLE APPROXIMATION in the question
ok but there is a problem in nov 11 variant 2 or 3 i think they say random sample and in the mark scheme they used the continuity correction !!
 
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ok but there is a problem in nov 11 variant 2 or 3 i think they say random sample and in the mark scheme they used the continuity correction !!
i cant check it right now busy with my chem exam tomorrow when i am free i will try and figure it out and explain
 
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[number without pink next to green] = [total number] - [number with pink next to green]
total number, as calculated in (iii), is 9P3 = 504
number with pink next to green can be done this way:
- find the number of choices that has both the pink and green cards, which is just, with the pink and green removed, choosing 1 extra from 7.
1 × 1 × 7C1 = 7
- find the number of arrangements of the three chosen cards that have the pink next to green.
2! × 2! = 4
So overall with pink next to green there are 7 × 4 = 28 arrangements
504 - 28 = 476 arrangements without pink next to green.
 
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[number without pink next to green] = [total number] - [number with pink next to green]
total number, as calculated in (iii), is 9P3 = 504
number with pink next to green can be done this way:
- find the number of choices that has both the pink and green cards, which is just, with the pink and green removed, choosing 1 extra from 7.
1 × 1 × 7C1 = 7
- find the number of arrangements of the three chosen cards that have the pink next to green.
2! × 2! = 4
So overall with pink next to green there are 7 × 4 = 28 arrangements
504 - 28 = 476 arrangements without pink next to green.
thanks
 
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please tell me some rules regarding reading reverse table in normal distribution.i get so confused..
 
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please any help when we use the continuity correction in solving the normal distribution problems, for example in june 11 paper 62 there is a question using normal approximation and in the mark scheme they used the continuity correction !!!!
 
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please any help when we use the continuity correction in solving the normal distribution problems, for example in june 11 paper 62 there is a question using normal approximation and in the mark scheme they used the continuity correction !!!!
Continuity correction is used when we approximate a discrete variable using a continuous variable. Like using X~N to approximate X~B, or X~N approximate X~Po
 
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i really don't understand it well, i mean the question terms in the past papers
when i know from these terms that i have to solve using continuity correction??
 
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can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)

- Each of the digits 1, 1,, 2,3, 3, 4, 6,is written on a separarte card. The seven cards are then laid out in a row to form a 7-digit number. How many of these 7-digit numbers is divisible by 4?( ans:300)

-The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
a) how many possible selections are there of four letters?( ans :12)
b)how many arrangements are there of four letters? ( ans:115)

thanx in advance
 
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