• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
1,394
Reaction score
12,123
Points
523
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.
 
Messages
23
Reaction score
18
Points
3
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.
 
Messages
1,394
Reaction score
12,123
Points
523
Integrate this to get the volume of revolution:

pi * y ^2 <---- y here is the equation of the curve.
 
Messages
2,538
Reaction score
17,571
Points
523
Can someone help with i) c)
i have uploaded the tree diagram for the qs
 

Attachments

  • 13.png
    13.png
    121.4 KB · Views: 7
  • 12.png
    12.png
    152.9 KB · Views: 5
Messages
340
Reaction score
339
Points
73
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
 
Messages
185
Reaction score
37
Points
38
help
how ms find the ANSWERS
only help required in parta for oct nov 2009 p61
 

Attachments

  • 123.JPG
    123.JPG
    76.6 KB · Views: 12
  • 1234.JPG
    1234.JPG
    52.7 KB · Views: 10
Last edited:
Messages
456
Reaction score
280
Points
73
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.
 
Messages
2,538
Reaction score
17,571
Points
523
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 :)
 
Messages
185
Reaction score
37
Points
38
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
 
Messages
64
Reaction score
18
Points
18
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.
 
Messages
340
Reaction score
339
Points
73
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.
We are given:
Σ(x-110) = -47
Σ(x-110)^2 = 5460
n=30
let y= x-110
Σy=-47.2 and Σy^2=5460

ȳ=-47.2/30 =-1.57
since y=x-1110
ȳ= x̅-110
x̅=ȳ+110
=-1.57 +110
=108.43

For SD calculate the sd of y
√Σy^2 /n -(ȳ)^2 (you know the normal formula we use to find the standard deviation, I can't seem to type it correctly)
√(5460/30)-(-1.57)^2
=13.4
SD of x =SD of y
so SD of x is 13.4 as well
 
Top