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Mathematics: Post your doubts here!

DragonCub

DragonCub

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Aarjit said:
Having difficulty deciphering it? Must be a Mac OSX snag. Apologies...;)
Click to expand...
It can be decoded by copying it to the reply zone and choose a normal font for them. :p
The deciphered text: :D
  1. P(X > 30.0) = 0.1480
    or, P [z > (30.0 - μ)/σ] = 0.1480
    or, P [z < - (30.0 - μ)/σ] = ϕ [(μ - 30.0)/σ] = 0.1480....... [We've inverted the '>' to '<' here, see why z is negative?]

    > obtain an equation involving μ and σ .... [1]

    Similarly,

    P(X > 20.9) = 0.6228
    or, P [z > (20.9 - μ)/σ] = 0.6228
    or, P [z < - (20.9 - μ)/σ] = ϕ [(μ - 20.9)/σ] = 0.6228 ......     [you can always rearrange the inequality to get rid of the negative sign]

    >     obtain another equation involving μ and σ .... [2]
Solve [1] and [2] simultaneously !
Click to expand...
 
Aarjit

Aarjit

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shan5674 said:
But the 30.0 does not give the negative z value as for the marking scheme only the 29.0 value gives the negative but from ur working both z values for 30 and 29 gets a negative sign? :S
Click to expand...

That's because z switches back to negative since you do not have ϕ(z) = 0.1480 in the table. That's why you need to solve for ϕ(-z) = 1 - 0.1480 = 0.8520.
 
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shan5674

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Aarjit said:
That's because z switches back to negative since you do not have ϕ(z) = 0.1480 in the table. That's why you need to solve for ϕ(-z) = 1 - 0.1480 = 0.8520.
Click to expand...

I still dont get it :/ do you mind working it on a piece of paper showing the steps and explaining them and uploading it? :/ only if its not too much and your not too busy :(
 
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farrukh

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Q.2 The position of 9 trees which are to be planted along the sides of a road is five on the north side and four on the south side.
(a) find the number of ways in which this can be done if the trees are all of different species.



(b) If the trees in (a) are planted at random, find the probability that two particular trees are next to each other on the same side of the road.



(c) if there are 3 cupressus, 4 prunus and 2 magnolias, find the number of different ways in which these could be planted assuming that the trees of the same species are identical. (d) If the tress in (c) are planted at random, find the probability that the 2 magnolias are on the opposite sides of the road.
 
RGBM211

RGBM211

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I still dont get the digit questions what yu hav to do when yu have digits repeated,digits not repeated,digits starting with even no./odd no. :(

sorry for not posting a particular question if sum one can explain what we normally do in such digit questions like use permutaion nd all :/
 
umarFM

umarFM

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RGBM211 said:
I still dont get the digit questions what yu hav to do when yu have digits repeated,digits not repeated,digits starting with even no./odd no. :(

sorry for not posting a particular question if sum one can explain what we normally do in such digit questions like use permutaion nd all :/
Click to expand...

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 digits are nt to be repeated, then
fr frst digit we can choose frm 5 digits... lets say we choose nmber 3... nw fr the 2nd digit we are left 1,2,4 &5...so we hv 4 possibilities... so total arrangements are 5x4x3x2x1= 5!
 
RGBM211

RGBM211

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umarFM said:
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 digits are nt to be repeated, then
fr frst digit we can choose frm 5 digits... lets say we choose nmber 3... nw fr the 2nd digit we are left 1,2,4 &5...so we hv 4 possibilities... so total arrangements are 5x4x3x2x1= 5!
Click to expand...

so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
 
umarFM

umarFM

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RGBM211 said:
so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
Click to expand...

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 first digit.... we have to choose frm either 2 or 4...so 2 possibilities fr 1st digit.....
fr the last digit we choose either frm 1,3 or 5...so 3 possibilites fr last digit...
so the remainin 3 digits left.. we are left 3 nmbrs...
as they are nt to be repeated, so we hv 3 possibilites fr 2nd digit, 2 possibilities fr 3rd digit and 1 possibility fr 4th digit...
hence total arrangements are 2x3x3x2x1 = 36....

hope that wrkd.. :)
 
M

mukki

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RGBM211 said:
so when digits are repeated its always smthing^no.

and mmm in question like wher they say it starts with an even no. and ends with odd no. /etc
Click to expand...
Example we have numbers 1 2 3 4 5 6 and we have to maka a 4 digit number out of which. This 4 digit number should be even.
Make 4 boxes and write 2,4,6 below the last box because and even number should be in the last box for the 4 digit number to be even. now we have to arrange one from 2,4,6 in the last box hence 3P1 and since there are three boxes left to be arranged 5P3 from remaining 5 numbers . hence this leads to 3P1 *5P3
 
Aarjit

Aarjit

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smartangel said:
Q7..part ii onwards.. PLEASE HELP! its permutations n combinations question!
Click to expand...

[Q7] Nine cards, each of a different colour, are to be arranged in a line. The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card? [3]

It's easier to solve by reversing the question, here's how you do it:

The total number of arrangements which do have the pink (P) card next to the green card (G):

[G P] 8! ....................................................[Here we are treating G and P as a single card attached together]
But remember that this can be [P G] 8! too !
n(P next to G) = 2! x 8! = 80640

now, finding the arrangements do not have the pink card next to the green card;

n(P away from G) = total # arrangements - n(P next to G) = 9! - 80640 = 282240

Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.

(iii) How many different arrangements in total of 3 cards are possible? [2]

arranging 3 cards from of 9;
= 9P3 = 504

(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card? [2]

Consider the restriction here: there MUST be a pink card (P) in the 3 cards, the remaining two can be randomly selected from the remaining 8 cards and arranged in the _ spaces;

n (P _ _ ) = 3P1 x 8P2...............................[Remember that P can be placed in a total of 3 ways P _ _ or _ P _ or _ _ P]
..............= 168

(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green card?

Again, It's easier to solve by reversing the question like we did in (ii);

The total number of arrangements which do have the pink (P) card next to the green card (G):

[P G] 7P1................but this can also be [GP] 7P1 or 7P1 [PorG]!

n (P next to G) = 2! x 2! x 7P1 = 28

or, n (P away from G) = total # arrangements [see (iii)] - n(P next to G) = 504 - 28 = 476

Q.E.D.
 
Aarjit

Aarjit

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smartangel said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf
Q2..how to find upper n lower quartile.my ans doesnt match the ms..
Click to expand...

Draw an ORDERED stem and leaf diagram from the discrete data [n = 21 (odd number)] and use the following formulae:

Q1 = [(n+1)/4]th item = 5.5th item
now move to the 5th leaf in your stem and leaf plot. Q1 is the average of the 5th and 6th leaf.

Q3 = [3(n+1)/4]th item = 16.5th item
now move to the 16th leaf in your stem and leaf plot. Q3 is the average of the 16th and 17th leaf.

The answers will definitely concur with the m.s.!
 
Aarjit

Aarjit

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smartangel said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf
Q6 please help..
Click to expand...

Q6. Permutations and Combinations

(i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy and Monica if each receives an odd number of mugs. [3]

The possible distributions between Lucy (L) and Monica (M) are:

...L9M1,...L7M3,...L5M5,..L3M7,..L1M9

= 10C9 + 10C7 + 10C5 + 10C3 +10C1
= 10 + 120 + 252 + 120 + 10 = 512

(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a different design. Find in how many ways these 9 mugs can be arranged in a row if the china mugs are all separated from each other. [3]

Alright, let's try this out in a slightly creative way; ;)

Plastic Mugs ~ \_/
China Mugs ~ (_)3

If the 3 China mugs are all separated from each other, they must fit in any of the alternating 7 _ !

_ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _

The (_)3 have 7 _ to fit in and the \_/ can arrange in 6! ways between themselves;

n = 7P3 x 6! = 151200

(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs. These 14 mugs are placed in a row. Find how many different arrangements of the colors are possible if the red mugs are kept together. [3]

If the red mugs are kept together, the remaining 11 mugs can be arranged randomly;

\_/ \_/ \_/ x 11 !...................... [Here, we are considering the \_/ to be 'attached' so they're always kept together]

= 3! x 11 ! x 2!......................[since there are 2 possible arrangements: \_/ \_/ \_/ x 11 ! or 11! x \_/ \_/ \_/ and the \_/ can arrange in 3! ways between themselves!]

but wait! we still have to consider the 7 and 4 identical yellow and blue mugs!

= 3! x 11! x 2! = 3960......Q.E.D
........7! x 4!
 
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leosco1995

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Help please:

Find the number of different ways in which the 9 letters of the word GREENGAGE can be arranged if exactly two of the Gs are next to each other.

The answer is 5040.. but I'm confused because there's 3 Gs and I don't know how to find the # of ways where 2 Gs are separated from the other one.
 
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