Mathematics: Post your doubts here!

[Rizwan Javed]

In the first part why isn't 4C2*9C2*2C2 the right way to get to the answer since we have to choose from 4 9 and 2 types of tress?
And part iiii as well please explain.

What are the answers? I'll post the solutions if my answers will be correct.

 

[Rizwan Javed]

In the first part why isn't 4C2*9C2*2C2 the right way to get to the answer since we have to choose from 4 9 and 2 types of tress?
And part iiii as well please explain.

In part (i) it says that there must be ATLEAST 2 trees of each type. Concentrate on the word ATLEAST. It means that this is the minimum requirement of each tree and there can more than two trees of each type.
The available quantity of trees is : 4 Hibiscus tree, 9 Jacaranda trees, and 2 oleandars.

Now the possible selections in which there are atleast 2 trees of each type are:

2 hibiscus, 8 jacaranda, and 2 oleandars <---- the ways for doing this are : 4C2 * 9C8 * 2C2
3 hibiscus, 7 jacaranda, and 2 oleanders <---- the ways for doing this are : 4C3 * 9C7 * 2C2
4 hibiscus, 6 jacaranda, and 2 oleanders <---- the ways for doing this are : 4C4 * 9C6 * 2C2

Now sum the above, to find the total no. of ways, which are : 54 + 144 + 84 = 282 ways.

In part (iii), it says that there should be any Hibiscus trees together.

Let the hibiscus trees be represented by ' * 's
and the other two types to be represented by ' X 's.

Now the possible positions for the trees that no hibiscus trees are together are :
* X * X * X * X * X * X * X * X *

Now arrange the trees in these positions.

The no. of arrangements of the 9 trees (jacaranda and oleander) in places marked by X are : 8!

There are 4 hibiscus trees, so select 4 positions marked by * from the shown 9 positions. This will be done in 9C4 ways.
Now the positions have been selected, so arrange the 4 hibiscus trees in these 4 selected positions: 4!

Now combine all this stuff to find the total no. of possible arrangements: 8! * 9C4 * 4! = 121927680 arrangenments.

 

[leenz98]

In part (i) it says that there must be ATLEAST 2 trees of each type. Concentrate on the word ATLEAST. It means that this is the minimum requirement of each tree and there can more than two trees of each type.
The available quantity of trees is : 4 Hibiscus tree, 9 Jacaranda trees, and 2 oleandars.

Now the possible selections in which there are atleast 2 trees of each type are:

2 hibiscus, 8 jacaranda, and 2 oleandars <---- the ways for doing this are : 4C2 * 9C8 * 2C2
3 hibiscus, 7 jacaranda, and 2 oleanders <---- the ways for doing this are : 4C3 * 9C7 * 2C2
4 hibiscus, 6 jacaranda, and 2 oleanders <---- the ways for doing this are : 4C4 * 9C6 * 2C2

Now sum the above, to find the total no. of ways, which are : 54 + 144 + 84 = 282 ways.

In part (iii), it says that there should be any Hibiscus trees together.

Let the hibiscus trees be represented by ' * 's
and the other two types to be represented by ' X 's.

Now the possible positions for the trees that no hibiscus trees are together are :
* X * X * X * X * X * X * X * X *

Now arrange the trees in these positions.

The no. of arrangements of the 9 trees (jacaranda and oleander) in places marked by X are : 8!

There are 4 hibiscus trees, so select 4 positions marked by * from the shown 9 positions. This will be done in 9C4 ways.
Now the positions have been selected, so arrange the 4 hibiscus trees in these 4 selected positions: 4!

Now combine all this stuff to find the total no. of possible arrangements: 8! * 9C4 * 4! = 121927680 arrangenments.

Oh I really got this so there's still some hope. Thanks a bunch. Actually I was skipping the 9C4 thing in the second part because I wasn't considering the fact that those 4 positions could be any among the 9.

 

[tiki-taka]

Is the markscheme ans wrong for 10 (i)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_ms_12.pdf

 

[Saad the Paki]

Is the markscheme ans wrong for 10 (i)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s15_ms_12.pdf
View attachment 59695

Ummm nope, if u use b^2-4ac you'll get (-6)^2-4 (2)(5-p) = 36-8 (5-p)<0 then solve.

 

[tiki-taka]

Ummm nope, if u use b^2-4ac you'll get (-6)^2-4 (2)(5-p) = 36-8 (5-p)<0 then solve.

Q10 i not Q11 thanks

 

[Rizwan Javed]

Q10 i not Q11 thanks

It's also correct in the ms.

 

[Rizwan Javed]

Integrate this to get the volume of revolution:

pi * y ^2 <---- y here is the equation of the curve.

 

[Saad the Paki]

Q10 i not Q11 thanks

Oops sorry

 

[usman]

Some books for A level Maths:

https://www.mediafire.com/folder/tsqgdwqtwqf44/A_Level_Maths

 

[The Sarcastic Retard]

Q3

 

[Lola_sweet]

Can someone help with i) c)
i have uploaded the tree diagram for the qs

 

[Eugene99]

Can someone help with i) c)
i have uploaded the tree diagram for the qs

At least two of the witnesses select accused person means that we can either have all three correct (cuz it says AT LEAST) or chose two correct and one wrong:
you can see from the tree diagram it can be done in four ways:
correct x correct x correct (all three correct)
wrong x correct x correct (first wrong and the next two correct)
correct x wrong x correct (second wrong and the rest correct)
correct x correct x wrong (third wrong)

multiplying the probabilities now
(1/12) x (1/12) x (1/12) = 0.00058
(11/12) x (1/12) x (1/12) = 11/1728
(1/12) x (11/12) x (1/12) = 11/1728
(1/12) x (1/12) x (11/12) = 11/1728

Now add them all together:
0.00058 +11/1728 + 11/1728 + 11/1728
= 0.0196 approx 0.02

 

[mohmed ahmed soliman]

help
how ms find the ANSWERS
only help required in parta for oct nov 2009 p61

 

[Saad the Paki]

help
how ms find the ANSWERS
only help required in parta for oct nov 2009 p61

You have four digits to fill. The first one HAS to be the number 5, otherwise you won't get a number between 5000 and 6000.
i) 3 digits left, 5 numbers to choose from. And since none can be repeated, you can choose from 5 numbers for the 2nd digit, from 4 for the 4th and from 3 for the 5th.
So the number of possibilities = 5*4*3 (aka 5P3) = 60

ii) If digits can be repeated, it means you can choose from all 6 numbers for the 3 digits left.
So, no. of possibilities: 6^3 = 216

This is awesomaholic101 's explanation so credit goes to her.

 

[Lola_sweet]

At least two of the witnesses select accused person means that we can either have all three correct (cuz it says AT LEAST) or chose two correct and one wrong:
you can see from the tree diagram it can be done in four ways:
correct x correct x correct (all three correct)
wrong x correct x correct (first wrong and the next two correct)
correct x wrong x correct (second wrong and the rest correct)
correct x correct x wrong (third wrong)

multiplying the probabilities now
(1/12) x (1/12) x (1/12) = 0.00058
(11/12) x (1/12) x (1/12) = 11/1728
(1/12) x (11/12) x (1/12) = 11/1728
(1/12) x (1/12) x (11/12) = 11/1728

Now add them all together:
0.00058 +11/1728 + 11/1728 + 11/1728
= 0.0196 approx 0.02

Thank you so much

 

[mohmed ahmed soliman]

help in part iv

 

[mohmed ahmed soliman]

help

 

[mohmed ahmed soliman]

You have four digits to fill. The first one HAS to be the number 5, otherwise you won't get a number between 5000 and 6000.
i) 3 digits left, 5 numbers to choose from. And since none can be repeated, you can choose from 5 numbers for the 2nd digit, from 4 for the 4th and from 3 for the 5th.
So the number of possibilities = 5*4*3 (aka 5P3) = 60

ii) If digits can be repeated, it means you can choose from all 6 numbers for the 3 digits left.
So, no. of possibilities: 6^3 = 216

This is awesomaholic101 's explanation so credit goes to her.

thanks but part ii is still unclear why 6 to power 3 is this a new rule

 

[DXDEVIL130]

Can anyone help me with this question?
In a spot check of the speeds x km/hr of 30 cars on a motorway, the data were summarised by Σ(x-110) = -47.2 and Σ(x-110)^2 = 5460. Calculate the mean and standard deviation of these speeds.