Mathematics: Post your doubts here!

[Dug]

wait could you elaborate please? why did you take 6^3?

Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.

 

[ZainH]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

The last question's last part , I'm getting one of the signs wrong for some reason. They changed the equation when taking the inverse? :\
If someone could do it with steps included I'd love them forever .

 

[Scafalon40]

Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.

Sorry but I still don't understand
I've never encountered a question like this before

 

[Minato112]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

The last question's last part , I'm getting one of the signs wrong for some reason. They changed the equation when taking the inverse? :\
If someone could do it with steps included I'd love them forever .

h(x) = 6x−x^2
= - (x^2 - 6x)
= - [ (x-3)^2 - 9]
= 9 - (x-3)^2

To find h*inverse*
Let y = h(x)
y = 9 - (x-3)^2
(x-3)^2 = 9 - y
x - 3 = (+or -) *root* 9 - y
x = (+or -) [*root* 9 - y ] + 3

Since the domain is positive, we take the positive root.
Therefore h*inverse* = [*root* 9 - x ] + 3

 

[ZainH]

h(x) = 6x−x^2
= - (x^2 - 6x)
= - [ (x-3)^2 - 9]
= 9 - (x-3)^2

To find h*inverse*
Let y = h(x)
y = 9 - (x-3)^2
(x-3)^2 = 9 - y
x - 3 = (+or -) *root* 9 - y
x = (+or -) [*root* 9 - y ] + 3

Since the domain is positive, we take the positive root.
Therefore h*inverse* = [*root* 9 - x ] + 3

The part I highlighted, how come you took 9 to the other side but changed the sign of 'y' ?

 

[Dug]

Sorry but I still don't understand
I've never encountered a question like this before

Read this

 

[Minato112]

The part I highlighted, how come you took 9 to the other side but changed the sign of 'y' ?

I didnt move the 9. I switched - (x-3)^2 with y. Got it now?

 

[ZainH]

I didnt move the 9. I switched - (x-3)^2 with y. Got it now?

Not exactly, why are you equating the equation while finding the inverse?
This is what I did.

To find H inverse
Let y = h(x)

h(x)= 9-(x-3)^2
h(x)-9= (x-3)^2
*root*h(x)-9= x-3
[*root*h(x)-9]+3 = x

H inverse = [*root*h(x)-9]+3 where as yours and the MS's is "9-x" instead of "x-9"


EDIT: I undestand what you did, but why is my method wrong?

 

[princess787]

can anyone help me in 9709_w12_qp_31 question 3 and 10 ii) ! please!

 

[Minato112]

Not exactly, why are you equating the equation while finding the inverse?
This is what I did.

To find H inverse
Let y = h(x)

h(x)= 9-(x-3)^2
h(x)-9= (x-3)^2
*root*h(x)-9= x-3
[*root*h(x)-9]+3 = x

H inverse = [*root*h(x)-9]+3 where as yours and the MS's is "9-x" instead of "x-9"


EDIT: I undestand what you did, but why is my method wrong?

"To find H inverse
Let y = h(x)

h(x)= 9-(x-3)^2
h(x)-9= - (x-3)^2

The '-'... U omitted it! Correction in red

 

[princess787]

can anyone help me in 9709_w12_qp_31 question 10 ii) plaease!!! please!

 

[princess787]

some body!!!! please help poor me!! i'm sitting on this question since morning! can anyone help me in 9709_w12_qp_31 question 10 ii

 

[Kumkum]

can someone pls help me with question 5 and 6bii) nd 6c
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_63.pdf
thnx in advance

 

[princess787]

HELP MEEEE!!!

 

[Sanis]

Hello for statistics p6 i am having difficulities for finding the variance .. it's equation is 1/n sigma (x-random number)^2 - mean^2 ... however the mean to be calculated differ sometimes i dnt know why, i need to find something called the coded mean not the ordinary mean .. I need help to know when should I have to place the coded mean in the mean to be deducted, rather than the ordinary .. take number 1 of this exam as example http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf .... http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_ms_6.pdf

Thank You

 

[ZainH]

"To find H inverse
Let y = h(x)

h(x)= 9-(x-3)^2
h(x)-9= - (x-3)^2

The '-'... U omitted it! Correction in red

Ah thanks

 

[yubakkk]

pLz help this question. its from book.

For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.​

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Q3 last part i know the answer but i want to asl how we got that

 

[Minato112]

Ah thanks

No prob!

 

[aleezay]

can someone pls help me with question 5 and 6bii) nd 6c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_63.pdf
thnx in advance

Q5.(i) Apply continuity correction. If u are rounding off a number to 84, it must be in the range 83.5<X<84.5. Now calculate the probability of the result ( by converting the data to normal distribution by subtracting mean from each value and dividing by standard deviation)

(ii) Simply calculate P(X>87)- no need to apply continuity correction since there is no mention of rounding off numbers. U will get 0.3282
now convert the data to binomial distribution: X - B(5,0.3282) and calculate the result for P(X=0) + P(X=1) (at most 1 out of 5 trials- I'm assuming you'd be knowing how to evaluate this expression)

(iii) Convert data to normal distribution
Fi{(k-82)/under root 126} - Fi(5/under root 126) = 0.3
The rest is all simple algebra.
Please let me know if you still have a confusion.