Universecolors
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Idk how to solve 6 i and iican someone please help , attached image
is below
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Universecolors
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(Original post by Universecolors)
Idk how to solve 6 i, can someone please help , attached image
is below
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Answer is 648 and 104 respectively
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zattyzatzat
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6ai) X Y Z. these three set of variables which are all different numbers holds all the different possible combinations
The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1
7 _ _
8 _ _
9 _ _


edit: typo
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Universecolors
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(Original post by zattyzatzat)
6ai) X Y Z. these three set of variables which are all different numbers holds all the different possible combinations
The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist. Do the same type of working for the even 7 and 9 but remember the first number is already an odd number so the combination is 4C1
7 _ _
8 _ _
9 _ _
For a i , why does variable Y have 9 options , and not 8 , since one is already used in X ? Btw thanks
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(Original post by Universecolors)
For a i , why does variable Y have 9 options , and not 8 , since one is already used in X ? Btw thanks
X excludes the number zero, whereas Z and Y can accept it.
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(Original post by zattyzatzat)
6ai) X Y Z. these three set of variables which are all different numbers holds all the different possible combinations
The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1
7 _ _
8 _ _
9 _ _


edit: typo
For 6 ii I don’t really get what u do for the middle value , can u please explain it more , im sorry : (
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zattyzatzat
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(Original post by Universecolors)
For 6 ii I don’t really get what u do for the middle value , can u please explain it more , im sorry : (
8 _ _. Since it must be odd the last digit must be either 1,3,5,7,9. This gives a combination of 5C1. The middle value now can accept numbers 0, 2, 4 , 6, and 4 odd numbers since one odd number is used at the end. Remember no repetition. So between 800 and 899 there are 5C1 * 8C1 = 40 different combinations.

Also don't worry it took me some time to solve this question. I absolutely hate permutation and combination questions in statistics
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Universecolors
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(Original post by zattyzatzat)
8 _ _. Since it must be odd the last digit must be either 1,3,5,7,9. This gives a combination of 5C1. The middle value now can accept numbers 0, 2, 4 , 6, and 4 odd numbers since one odd number is used at the end. Remember no repetition. So between 800 and 899 there are 5C1 * 8C1 = 40 different combinations.

Also don't worry it took me some time to solve this question. I absolutely hate permutation and combination questions in statistics
Thank yiuuu so muchhhh , can you please tell me how you would approach this question too 7 ii? If it’s not a bother
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(Original post by Universecolors)
Thank yiuuu so muchhhh , can you please tell me how you would approach this question too 7 ii? If it’s not a bother
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the 3 different possible arrangements of W.

_ W _ _ _ _ _ _ W _
W _ _ _ _ _ _ W _ _
_ _ W _ _ _ _ _ _ W

There are 8 spaces so there are 8! different ways to arrange them however there the letter L is repeated 3 times. So for 1 arrangement of W there are 8!/3! different possible arrangements. Multiply by 3 to get the total amount of arrangements. Good luck this wednesday? (Am having my s1 exam aswell )
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Universecolors
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(Original post by zattyzatzat)
the 3 different possible arrangements of W.

_ W _ _ _ _ _ _ W _
W _ _ _ _ _ _ W _ _
_ _ W _ _ _ _ _ _ W

There are 8 spaces so there are 8! different ways to arrange them however there the letter L is repeated 3 times. So for 1 arrangement of W there are 8!/3! different possible arrangements. Multiply by 3 to get the total amount of arrangements. Good luck this wednesday? (Am having my s1 exam aswell )
How do u feel about the exam ? Wait u did p1 too right ? How was it for u , btw thanks for the answer I get it now
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(Original post by Universecolors)
How do u feel about the exam ? Wait u did p1 too right ? How was it for u , btw thanks for the answer I get it now
A bit scared of the permutation and combination questions and the binomials distribution questions. P1 was fine, but I realized after the exam i made a silly mistake on the trigonometry question so I lose like 5 marks there :P
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(Original post by zattyzatzat)
A bit scared of the permutation and combination questions and the binomials distribution questions. P1 was fine, but I realized after the exam i made a silly mistake on the trigonometry question so I lose like 5 marks there :P
P1 was really bad for me , which variant were u
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(Original post by Universecolors)
P1 was really bad for me , which variant were u
13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year
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(Original post by zattyzatzat)
13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year
I was variant 12 : ( it was very difficult
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(Original post by zattyzatzat)
13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year
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Can you help me with that question please
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(Original post by Universecolors)
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Can you help me with that question please
i can't see the question
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Universecolors
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(Original post by zattyzatzat)
i can't see the question
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Please do help ; (
It’s from 2018 feb
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i think am missing some information, can i see part 1?
(Original post by Universecolors)
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Please do help ; (
It’s from 2018 feb
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Universecolors
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(Original post by zattyzatzat)
i think am missing some information, can i see part 1?
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This is i , I have already solved it
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(Original post by Universecolors)
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This is i , I have already solved it
since it's asking for weights 1.5 sd from the mean, we need to calculate x. which is 400 + 1.5*5.62 (your S.d) and 400 - 1.5*5.62. Using these values of x (408.43 and 391.57) find the probability between these two values and then multiply by 500 to get the number of packets. I think this is correct? can you message the markscheme when you're done
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