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Mathematics: Post your doubts here!

Saad the Paki

Saad the Paki

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Part (ii) I used probability as 0.8 however the mark scheme has used probability of 0.9... can somebody explain why? This is M/J 2005 p6
20160304_171756-1.jpg
 
Rizwan Javed

Rizwan Javed

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Saad the Paki said:
Part (ii) I used probability as 0.8 however the mark scheme has used probability of 0.9... can somebody explain why? This is M/J 2005 p6
View attachment 59563
Click to expand...
P(1.9 - b < X < 1.9 + b) = 0.80

P(X < 1.9 + b) - P(X < 1.9 - b) = 0.80

Standardizing X, using Z = X - μ / σ

P( Z < 1.9 + b - 1.9 / 0.15 ) - P(Z < 1.9 - b -1.9) = 0.80

P(Z< b/0.15) - P(Z <-b/0.15) = 0.80
Ф(b/0.15) - Ф(-b/0.15) = 0.80
2Ф (b/0.15) - 1 = 0.80
2Ф (b/0.15) = 1.80
Ф (b/0.15) = 0.90

Using Normal Distribution tables,

b/0.15 = 1.281
b = 0.192 (3sf)

You can now use this value of b to find the safety limits. :)
 
Saad the Paki

Saad the Paki

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Rizwan Javed said:
P(1.9 - b < X < 1.9 + b) = 0.80

P(X < 1.9 + b) - P(X < 1.9 - b) = 0.80

Standardizing X, using Z = X - μ / σ

P( Z < 1.9 + b - 1.9 / 0.15 ) - P(Z < 1.9 - b -1.9) = 0.80

P(Z< b/0.15) - P(Z <-b/0.15) = 0.80
Ф(b/0.15) - Ф(-b/0.15) = 0.80
2Ф (b/0.15) - 1 = 0.80
2Ф (b/0.15) = 1.80
Ф (b/0.15) = 0.90

Using Normal Distribution tables,

b/0.15 = 1.281
b = 0.192 (3sf)

You can now use this value of b to find the safety limits. :)
Click to expand...
Damn mashallah! Thanks
 
qwertypoiu

qwertypoiu

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Saad the Paki said:
Part (ii) for this please :).View attachment 59564
Click to expand...
Untitled.png

I think the diagram above will help.
You can find the pink area using sector area formula.
You can find yellow area by first finding area of sector CAB, then subtracting the right angle triangle CAB. (multiply two since there are two yellow parts)
You can find shaded area by first finding area of circle CBED, then subtracting yellow part then subtracting pink part.

Sorry I didn't do the actual calculation :D
 
Rizwan Javed

Rizwan Javed

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Saad the Paki said:
Part (ii) for this please :).View attachment 59564
Click to expand...
Find the angle CBD.
θ = 2 sin^-1 (1/√2)
= pi/2
Then,
find the area of the sector, CBD.
A = 1/2 (2√r)^2 * pi/2
= (pi r^2 )/2

Now, you know the area of sector CBD, so you can find the area of the region, CADF. (I have taken F a point on the arc to the left of E)
Area of region CADF = (pi r^2 )/2 - (1/2 * 2r * r)
= (pi r^2) /2 - r^2

Find the area of the hemisphere of the smaller circle and subtract the area of the region CADF from it.
Hence, area of shaded region will be:

A = (pi r^2) /2 - ((pi r^2) /2 - r^2)
= r^2 Ans.
 
Saad the Paki

Saad the Paki

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qwertypoiu said:
View attachment 59565

I think the diagram above will help.
You can find the pink area using sector area formula.
You can find yellow area by first finding area of sector CAB, then subtracting the right angle triangle CAB. (multiply two since there are two yellow parts)
You can find shaded area by first finding area of circle CBED, then subtracting yellow part then subtracting pink part.

Sorry I didn't do the actual calculation :D
Click to expand...
Wow thanks. Did u make that?
 
Saad the Paki

Saad the Paki

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Rizwan Javed said:
Find the angle CBD.
θ = 2 sin^-1 (1/√2)
= pi/2
Then,
find the area of the sector, CBD.
A = 1/2 (2√r)^2 * pi/2
= (pi r^2 )/2

Now, you know the area of sector CBD, so you can find the area of the region, CADF. (I have taken F a point on the arc to the left of E)
Area of region CADF = (pi r^2 )/2 - (1/2 * 2r * r)
= (pi r^2) /2 - r^2

Find the area of the hemisphere of the smaller circle and subtract the area of the region CADF from it.
Hence, area of shaded region will be:

A = (pi r^2) /2 - ((pi r^2) /2 - r^2)
= r^2 Ans.
Click to expand...
Thanks I got most of it except the CADF part. What is (1/2*2r*r)? Is it a triangle? And can you be more specific about the location of point F. Appreciate it
 
The Sarcastic Retard

The Sarcastic Retard

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Raj21999 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s14_qp_11.pdf

Please help me with question 10 (iii)
Also, in this question they say "set of values for which each expression is valid." Do they mean find domain of the function or the range?
I'm really confused.Thanks in advance.
Click to expand...
f(x) = 3x - 2
replace x with y and f(x) with x
x = 3y - 2
make y subject.
y = 1/3(x + 2)
f^-1(x) = 1/3(x + 2)
Range of f(x) = domain of f^-1(x)
range of f(x) :domain of f^-1(x) = -5 <= x <= 1
f(x) = 4/(5-x)
x = 4/(5 - y)
y = 5 - (4/x)
f^-1(x) = 5 - (4/x)
domain of f^-1(x) : range of f(x) = 1 < x <= 4
 
Rizwan Javed

Rizwan Javed

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Copy Cat said:
View attachment 59575
(ii) & (iii) part
Click to expand...
There are 3! ways to arrange the 3 business men in the first row.

The two couples can only sit on the side, which has two seats together (with no aisle between!). So select two pairs of such seats from the 3 and arrange the two couple in those two pairs of seats. This can be done in 3P2 Ways. Now arrange both the couples in their adjacent seats. So the total arrangements for arranging the couples would be: 3P2 * 2! * 2!

There are now only 5 windows seats remaining. So arrange the students in these 5 window seats. The arrangements would be 5!

Hence, the total no. of arrangements for placing these 12 passengers in the seats would be:
3! * 3P2 * 2! * 2! * 5! = 17280 arrangements.
 
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