Mathematics: Post your doubts here!

[mukki]

I assume that there cant be 0 students in a school. either way. I take at least 1 in a school. so therfore to find the midpoint (100+1)/2
try doing this way for all the columns.

well if its < then -.5 but if its <equal to then add .5

 

[mominzahid]

can someone please explain thiss.. i cant figure out how to solve the 2nd part.

 

[mukki]

can someone please explain thiss.. i cant figure out how to solve the 2nd part.

since the time for the standard journey is between two times use P(x1<X<x2) which becomes P(-z<Z<z) for such types of qs. so ul get 2fi(z)-1=.34

 

[mukki]

since the time for the standard journey is between two times use P(x1<X<x2) which becomes P(-z<Z<z) for such types of qs. so ul get 2fi(z)-1=.34

Sorry bro really sleepy crappy drawing cant sleep though

 

[RGBM211]

How many numbers are there between 1245 and 5421 inclusive which contain each of the digit 1,2,4 and 5 once and once only? Ans:- 24

can someone explain ?

 

[arlery]

Ermm how can I solve Q 3 (b) in http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

 

[mukki]

Ermm how can I solve Q 3 (b) in http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

 

[Khan_971]

Hmm.. I just looked through some old notes I had, it said:

P(A or B) = P(A) + P(B) - P(A and B)

..which probably applies here because P(F) comes twice I think. Not sure, but I hope questions like these don't come in the paper.

p63 is really a difficult one. but GT has 38+ as A :O

 

[Khan_971]

How many numbers are there between 1245 and 5421 inclusive which contain each of the digit 1,2,4 and 5 once and once only? Ans:- 24

can someone explain ?

since there are 4 numbers. and each order matters. 4!= 24
Since there is no number greater than 5412, this is easy.

 

[RGBM211]

since there are 4 numbers. and each order matters. 4!= 24
Since there is no number greater than 5412, this is easy.

lol i got it i was thinking of smthing else...
anyways another question

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

 

[Khan_971]

lol i got it i was thinking of smthing else...
anyways another question

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

i'll solve both parts simultaneously
the simplest way is to write possibilities alphabetically
.1 EEOP . . 12 . . [ 4!/2! ]
.2 EEOS . . 12
.3 EEPS . . 12
.4 EESS . . . 6 . . [ 4! / (2!*2!) ]
.5 EOPS . . 24 . . [ 4! ]
.6 EOSS . . 12
.7 EPSS . . 12
.8 ESSS . . . 4 . . [ 4! / 3! ]
.9 OPSS . .12
10 OSSS . . 4
11 PSSS . . 4
12 SSSS . . 1

12 combos , 115 arrangements
----------------- --------------------------

note:
------
of course, there is a shorter way,
but not without pitfalls unless very clear-headed

TYPE ................. CASES .......... PERMS....... TOTAL
all different ........ ......1 .................... 4! ............. 24
2 singles, 2 same ... 3C2*2 = 6 ...... 4!/2! ........... 72
1 single, 3 same ..... 3C1 = 3 ......... 4!/3! ............12
2 each .................... 1 ................. 4!/(2!2!) ......... 6
all same .................. 1 ....................1 ............... 1
............................... 12 ............................... 115

 

[RGBM211]

Part a asks for 4 different letters???

yep as yu cn see there is P,O,S,E

 

[Khan_971]

yep as yu cn see there is P,O,S,E

Done. Look up.

 

[Khan_971]

yep as yu cn see there is P,O,S,E

The 12s, 4s are the no.of arrangements possible of those 4 letters.

 

[RGBM211]

Done. Look up.

thanks alot brother

 

[arlery]

Thank you!

 

[Mustehssun Iqbal]

Assalamu alaikum.
Q) Two fair dice are thrown simultaneously. Find the probability that ;
b ) the total score is at least 8...

 

[usmiunique]

Assalamu alaikum.
Q) Two fair dice are thrown simultaneously. Find the probability that ;
b ) the total score is at least 8...

draw up a possibility diagram.. it will be like this
Since the sum of the scores is required,

_|1 2 3 4 5 6
1 |2 3 4 5 6 7
2|3 4 5 6 7 8
3|4 5 6 7 8 9
4|5 6 7 8 9 10
5|6 7 8 9 10 11
6|7 8 9 10 11 12
from the diagram, it can be seen that there are 15 outcomes greater than or equal to 8 so, the probablity is 15/36
hope u understood!

 

[usmiunique]

ASA,
i wanted to know when to take the mean of the data when calculating the quartiles.
for eg. in M/J/2010 paper 61 stats,
in question 2, we get the following steam-and-leaf diagram.
0|2 5 6 8 8 (5)
1|2 4 6 7 7 9 (6)
2|1 2 3 3 3 5 6 7 (8)
3|1 5 (2)
now in calculating the median, we will divide the total number of data by 2 which in this case is
21/2=10.5 so, which value will we take? 10 th value or 11thvalue or the mean of the 10th and 11th value.
also in calculating the lower quartile, we get 21/4=5.25 here also, do we take the 5th value or the 6th value or the mean of the 5t and 6th value?
again in calculating the upper quartile, we get 21*0.75=15.75 here again do we take the 15th value or the 16th value or the mean of the 15th and 16th value?
sometimes in the mark scheme i have seen them taking the mean and sometimes the value immediately next to it.. im confused as to what should i do?
WILL appreciate any help!​

 

[hassam]

The 1st interval is about 1-100. therfore mid class value = (100+1)/2
2nd interval is 101-150, therfore (101+150)/2 = 125.5
And so on..

why its 1-100 and not 0-100?