Mathematics: Post your doubts here!

[The Sarcastic Retard]

Need help on this question ! I couldn't get the answer after many times of trials,

2010 O/N paper 33 question 8 ii) b)

b)
Let theta = @
@ = 41.41, -41.41
@/2 - 52.24 = -41.41
@ = 21.7

 

[Saad the Paki]

Part (ii). The ms says graph has line of symmetry at x=3 therefore A=6. What?
The completed square form is 2(x-3)^2-5

 

[sj0007]

Part (ii). The ms says graph has line of symmetry at x=3 therefore A=6. What?View attachment 59662
The completed square form is 2(x-3)^2-5

U know how the turning point obtained from this form is the line of symmetry?
So imagine this: The curve passes through zero, reaches its minimum point at x=3, now when we say that its the line of symmetry, we mean that its the exact half and so there will be an exact reflection on the other side........... So the point at which it will cross the x axis again will be 6.........
Something like this:

 

[Saad the Paki]

U know how the turning point obtained from this form is the line of symmetry?
So imagine this: The curve passes through zero, reaches its minimum point at x=3, now when we say that its the line of symmetry, we mean that its the exact half and so there will be an exact reflection on the other side........... So the point at which it will cross the x axis again will be 6.........
Something like this:

View attachment 59663

Ooh kay. Thank you!

 

[Saad the Paki]

Q5 (a) both parts

 

[awesomaholic101]

View attachment 59665
Q5 (a) both parts
Why can't it simply be 5! for (i) ?

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

 

[Saad the Paki]

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

Thanks

 

[Saad the Paki]

Q3 part (ii) also please

 

[Anum96]

μ = 100 min
σ = 7 min


X -> journey times of a certain car in the given journey.
Let the lower limit of times to be termed as standard to be μ - a and upper limit to be μ + a
P(μ - a < X < μ + a)
standardizing X, using Z = X - μ /σ

P( -a / 7 < Z < a / 7) = 0.34
P(Z < a / 7) - P(Z < -a / 7) = 0.34
ϕ (a/7) - ϕ(-a/7) = 0.34
2 ϕ(a/7) - 1 = 0.34
ϕ(a/7) = 1.34/2

Use normal distribution tables,

a = 3.08

So the upper limit of time will be 100+3.08 = 103.08 ~ 103.1 min
lowe limit of time wll be 100-3.08 =96.92 ~ 96.9 min

Credit goes to : Rizwan Javed

 

[Saad the Paki]

μ = 100 min
σ = 7 min


X -> journey times of a certain car in the given journey.
Let the lower limit of times to be termed as standard to be μ - a and upper limit to be μ + a
P(μ - a < X < μ + a)
standardizing X, using Z = X - μ /σ

P( -a / 7 < Z < a / 7) = 0.34
P(Z < a / 7) - P(Z < -a / 7) = 0.34
ϕ (a/7) - ϕ(-a/7) = 0.34
2 ϕ(a/7) - 1 = 0.34
ϕ(a/7) = 1.34/2

Use normal distribution tables,

a = 3.08

So the upper limit of time will be 100+3.08 = 103.08 ~ 103.1 min
lowe limit of time wll be 100-3.08 =96.92 ~ 96.9 min

Credit goes to : Rizwan Javed

Wow this is great! Thank you (both of you)

 

[sj0007]

Ooh kay. Thank you!

 

[leenz98]

Someone please explain Q5 part iii. Since X=0 why won't it be 12C0*(0.20)^0 * (0.8)^12
Also any trips and tricks on how to solve the permutations and combination questions like question 6

 

[leenz98]

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.

 

[leenz98]

For this question p= 5/12 and q= 7/12. Can anyone explain what's wrong with how I am doing it.
1-(12C0*(5/12)^0 * (7/12)^12 + 12C1* (5/12)^1 * (7/12)^11)

 

[Saad the Paki]

For this question p= 5/12 and q= 7/12. Can anyone explain what's wrong with how I am doing it.
1-(12C0*(5/12)^0 * (7/12)^12 + 12C1* (5/12)^1 * (7/12)^11)

It won't be 12C0 or 12C1 it will be 3C0 and 3C1 as the newspaper is delivered to 3 houses not 12

 

[Saad the Paki]

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.

I suck at permutations and combinations as well so sorry can't help u there

 

[leenz98]

I suck at permutations and combinations as well so sorry can't help u there

Still appreciate the fact that you replied.
We literally studied the whole s1 in like 15 days and I can't make heads or toes of it now.

 

[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.