Mathematics: Post your doubts here!

[Esme]

Its better to make it in horizontal ..

ok thanks.

 

[Esme]

its better you draw it horizontally as done in the marking scheme because in class and in books i have always seen it horizontally

alright thank you !

 

[Diamona151294]

um last minute problems.....:/

Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

 

[Mustehssun Iqbal]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

same questions here...

 

[JulyMei]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

first attachment is Q3

second attachment is Q7

hope it helps
gonna have my math paper 6 later.. wish me luck!

 

[hassam]

now ,......how to know class boundaries here???
question says less than 45....less than 50.............bt ms writes this Class as 45-50 widout paying attention to the damn inequality signs............now why the hell were they there/....if we dun have to use them.....m cnfused.....is there any rule to detrmine class widths

 

[smartangel]

P(k-125/4.2 <Z< 128-125/4.2) =.7465
Fi(3/4.2) - fi(k-125/4.2) = .7465
fi(k-125/4.2) = .0158
since its lower than .5,
1-fi(125-K/4.2) = .0158
.9842 = fi(125-K/4.2)
Using normal table:
2.15 = 125-k/4.2
9.03 = 125-K
K = 115.97 = 116

how'd the k - 125 change to 125 - k???

 

[smzimran]

Assalamu alaikum,
Q.3 part 1 pls...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_61.pdf

Total possible selections are 8C4 = 70.

If there are 1 girl and 3 boys in the team,
Selections = 5C1 * 3C3 = 5
P(1G) = 5/70 = 1/14

If there are 2 girl and 2 boys,
Selections = 5C2 * 3C2 = 30
P(2G) = 30/70 = 3/7

If there are 3 girls and 1 boy,
Selections = 5C3 * 3C1 = 30
P(3G) = 30/70 = 3/7

If there are 4 girls and 0 boys,
Selections = 5C4 * 3C0 = 5
P(4G) = 5/70 = 1/14

 

[Jaf]

now ,......how to know class boundaries here???
question says less than 45....less than 50.............bt ms writes this Class as 45-50 widout paying attention to the damn inequality signs............now why the hell were they there/....if we dun have to use them.....m cnfused.....is there any rule to detrmine class widths

Apparently, the mark scheme is wrong.

 

[Mattman]

So...how did the paper 6 go? Went somewhat badly..

 

[IsaacNewton]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

Question 3(i)
36 Total Values.

S = Sigma sign.

S(x-45) = -148
Open the brackets:
Sx-S45 = -148

Sx = sum of all the different values of x.
s45 = Sum of all the 45's. Since there are 36 total values, it means sum of 36 45's. So, S45 = 36(45) = 1620

Sx = -148 + 1620
Sx= 1472

Mean = Sx / 36 = 1472 / 36 = 40.9

S(x-45)^2 = 3089
Expand Bracket
S(x^2 -90x + 45^2) = 3089 Quadratic Expansion
Open brackets

Sx^2 -90Sx+ S2052 = 3089 90Sx = 90(1472) S2052 = 36(2052)

So Sx^2 = 3089 - 73872 +132480
Sx^2=61697

S.D^2 = Variance.

Var = (Sx^2 / 36 ) - (mean)^2
Var= (61697/36)- (40.9)^2
Var = 40.99
S.D = 6.4


(ii)

A value "29" was added to our Data. So our total values are now 37

This means that the sum of all the values now, is Sx + 29
So our New Sx = 1472 + 29 = 1501

We will also calculate our new Sx^2 in the same way.
Our new Sx^2 = 61697 + 29^2 = 62538

use the Variance Formula

(Sx^2 / 37) - (Sx/37)^2
(62538/37)-(1501/37)^2

Variance = 44.5
S.D = 6.67

 

[IsaacNewton]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

Question 7 (iii)

70 days at random, how many would have temperature more than 3 times the mean.

So first we calculate the probability, that a single day has, that its minimum temperature is more than 3 times the mean.


So, p(X > 3u )

p(Z > (3u - u)/2u )

p(Z > 2u/2u)

p( Z > 1 )
1 - P(Z<1)

after looking from the table:
1 - 0.8413
0.1587

So the probability of a single day having a minimum temperature more than 3 times the mean:
p = 0.1587

n, number of days = 70.

Expected value, = np, = 70(0.1587) = 11.109,
Therefore 11 days.

 

[IsaacNewton]

So...how did the paper 6 go? Went somewhat badly..

When did you give Paper 6??
Which variant, and what came in it ?
I still have the paper in a few hours.

 

[smzimran]

When did you give Paper 6??
Which variant, and what came in it ?
I still have the paper in a few hours.

Its wrong dude, don't ask what is to come!

 

[hassam]

So...how did the paper 6 go? Went somewhat badly..

r u kidding....cos paper 62 is taken most early....and there are still 5-6 hours in that

 

[hassam]

Its wrong dude, don't ask what is to come!

well Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other?
ms does 8!/3! -2*7!/3!....i cnt understand why he is multiply by 2

 

[hassam]

?????

 

[smzimran]

leadingguy , Mustehssun Iqbal , Khan_971 , hassam , user , RGBM211 , alphabravocharlie
and
@Everyone:
Some important points....

1. If a box and whisker plot is more stretched towards the lower values, it is said to be positively skewed.

2. If a box and whisker plot is more stretched towards the higher values, it is said to be negatively skewed.

3. An outlier is a value that is more than 1.5 times the interquartile range.
(a) A high outlier is given by = Q3 + 1.5 (Q3 - Q1)
(b) A low outlier is given by = Q1 - 1.5 (Q3 - Q1)
*Q1 and Q3 are lower quartile and upper quartile respectively.

4. Remember to include +/- 0.5 in class boundaries when drawing histograms, although they are not necessary for cumulative frequency curves.

5. Conditional probabilty; when asked probability of B given that A is occuring, the probability is given by
P (B | A) = P (A and B) / P (A)
This can be rewritten as
P (A and B) = P (A) * P (B | A)

6. Independent events are events that have no effect on one another. For two independent events A and B ,
P (A and B) = P (A) * P (B)

7. If A1 , A2 , ... , An are n independent events, then
P (A1 and A2 and ... and An) = P (A1) * P (A2) * ... * P (An)

8. Two events which have no outcomes in common are called mutually exclusive events. The addition law of mutually exclusive events is given by
If A1 , A2 , ....... , An are n mutually exclusive events, then
P (A1 or A2 or .... or An) = P (A1) + P (A2) + ... + P (An)

 

[Diamona151294]

first attachment is Q3

second attachment is Q7

hope it helps
gonna have my math paper 6 later.. wish me luck!

Thanku soooooooooooooo much!!!!!! gdluk for ur paper!!
oh um 1 thing, can u exp why we multiply the probability by 70?

 

[smzimran]

well Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other?
ms does 8!/3! -2*7!/3!....i cnt understand why he is multiply by 2

Im not getting that also,
Which year ?