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Write it down. 4000-4999. 4P3Using 2, 3 , 4, 5, 6
!) How many integers greater than 4000 can be formed, if there is no repetition?
2) How many integers which are multiples of 5 can be formed if there is no repetition?
Thank You
Write it down. 4000-4999. 4P3
5000-5999: 4P3
6000-6999: 4P3
total : 72?
i dont exactly understand the quartiles part.. what do u mean by deleting them from the list? :-Sdude if the total frequency is odd like 21 .. to find the median u add 1 to it and divide by 2. 21+1/2 so its the 11th value.
for a total that is even for example if its 22 u take both the 11th and 12th values from the data and half them.
here is a nice tip for the quartiles. if the median is odd find it and delete it from the list. now calculate the quartiles of the numbers before the median which should be q1 and of the numbers after the median which should be q3. if the median is an even number then take it with your calculation like if the median is 22 take the 11th value and values before it to calculate q1 and take the 12th value and values after it to calculate q3.
Write it down. 4000-4999. 4P3
5000-5999: 4P3
6000-6999: 4P3
total : 72?
Lower boundary, 20-12 = 8, upper 20+12 = 32.5 The weights of letters posted by a certain business are normally distributed with mean 20 g. It is
found that the weights of 94% of the letters are within 12 g of the mean.
(i) Find the standard deviation of the weights of the letters. [3]
This is from Oct/Nov 2011, paper 61. I only have a doubt with how the question is phrased. It says 94% are WITHIN 12 g of the mean. So if we say that 94% of the weights lie between, per say, 26 g and 14 g, shouldn't that also be right? In the marking scheme they use 32 g. Shouldn't they stick to saying that 94% of the weights are below a value which is 12 g above the mean instead of writing within?
Thanks.
then won't that go on??no there will be more combinations... such as 20000 or above. 30000 and above and so on
u know that 0.028 = 0.85r -.82qcan someone please explain how to solve question 6 (ii)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_63.pdf
it doesn't ever come as 4 and 2
no there will be more combinations... such as 20000 or above. 30000 and above and so on
if theres a 5 digit number. and without repetition. it will be 5! as all those 5 numbers can be either way. 5! + 72 = 192I got that as well but the answer in the book is 192
How is that possible?
Pals_1010
I got it.
if theres a 5 digit number. and without repetition. it will be 5! as all those 5 numbers can be either way. 5! + 72 = 192
yes it can b either a 4 digit number or 5 digit numer..I got that as well but the answer in the book is 192
How is that possible?
i mean that circle the median and dont use it in your calculation its for un grouped data. look if the median is odd make a mark on each side of the median and calculate the MEDIAN for all the values before the median which will result as q1 and calculate the MEDIAN for all the values after the median which will result in q3. TRY IT OUT BUT PLZ DONT INCLUDE THE MEDIAN IF IT IS FROM AN ODD NUMBER.i dont exactly understand the quartiles part.. what do u mean by deleting them from the list? :-S
i dont get itt.. :'(3 diff colors:
Red,red,not red, not red = 2C2 * (5 * 2C1) x 4!/2! since 4 pegs and 2 are same
Since the rest 5 colors have 2 pegs each.
Total= (5*2C1) * 4!/2! * 6
= 720
found it.thanks xDyes the cambridge s1 book by steve dobbs
can you plz do part iii as well! thanks
kind of sigmoid curveIn stats 1 when drawing the cumulative graph do u draw a curve or one with straight lines?
oh... thanks a lot!i mean that circle the median and dont use it in your calculation its for un grouped data. look if the median is odd make a mark on each side of the median and calculate the MEDIAN for all the values before the median which will result as q1 and calculate the MEDIAN for all the values after the median which will result in q3. TRY IT OUT BUT PLZ DONT INCLUDE THE MEDIAN IF IT IS FROM AN ODD NUMBER.
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