Maths, Addmaths and Statistics: Post your doubts here!

[Nabeel.Ahmad]

You dont need to do that either. just sit back, eat, drink, party and the A* will come flying to you -_-

Well, yeah..
But AT LEAST i'll have to work during the exam days. I mean, that's necessary.

 

[Winter]

Hi guys.. I wanted to know what books i should buy for maths 4024 o level for good practise and preparation .. Plz reply quick

 

[Jenn]

Here is a probablity question that has confused me, kindly help me in this. ASAP.

There are 23 boys and 35 girls in the school hall. After x boys and x+4 girls left the hall, the probablity of selecting a boy becomes 2/5. Find the value of x.

Thanks in advance.

 

[Jenn]

Here is a probablity question that has confused me, kindly help me in this. ASAP.

There are 23 boys and 35 girls in the school hall. After x boys and x+4 girls left the hall, the probablity of selecting a boy becomes 2/5. Find the value of x.

Thanks in advance.

 

[Zuhsid]

Here is a probablity question that has confused me, kindly help me in this. ASAP.

There are 23 boys and 35 girls in the school hall. After x boys and x+4 girls left the hall, the probablity of selecting a boy becomes 2/5. Find the value of x.

Thanks in advance.

2/5= (23-x)/(58-2x-4)
2/5= (23-x)/(54-2x)
2(54-2x) = 5(23-x)
108-4x = 115-5x
5x-4x = 115-108
x= 7

According to me, this should be it....but you can take some elses opinion!!

Hope it helps!!

 

[Jenn]

T

2/5= (23-x)/(58-2x-4)
2/5= (23-x)/(54-2x)
2(54-2x) = 5(23-x)
108-4x = 115-5x
5x-4x = 115-108
x= 7

According to me, this should be it....but you can take some elses opinion!!

Hope it helps!!

Thank you so much. Helped a lot.

 

[Zuhsid]

T
Thank you so much. Helped a lot.

Always welcome to help!!

 

[Ram97]

Guys I need help
http://papers.xtremepapers.com/CIE/...(Calculator Version) (4024)/4024_w08_qp_1.pdf
Q.12 part b

Please help

 

[Zuhsid]

Guys I need help
http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w08_qp_1.pdf
Q.12 part b

Please help

f(x) = (4x + 3)/2x

As f(x) = y
& x = f^-1( y)
Therefor;
y = (4x+3)/2x
2xy = 4x+3
4x - 2xy = -3
x(4 - 2y) = -3
x = -3/(4 - 2y)

Therefor;
f^-1( y) = -3/(4 - 2y)
f^-1(x) = -3/(4-2x)


Hope it helps!!

 

[Ram97]

f(x) = (4x + 3)/2x

As f(x) = y
& x = f^-1( y)
Therefor;
y = (4x+3)/2x
2xy = 4x+3
4x - 2xy = -3
x(4 - 2y) = -3
x = -3/(4 - 2y)

Therefor;
f^-1( y) = -3/(4 - 2y)
f^-1(x) = -3/(4-2x)


Hope it helps!!

Thanks a ton man!!!
I see where I was going wrong

 

[TheLeagueofShadows]

If the probabilities that Jane, Tom, and Mary will be chosen chairperson of the board are 0.5, 0.3 and
0.2 respectively, what is the probability that the chairperson will be either Jane or Mary?
Please explain too.

 

[TheLeagueofShadows]

Suppose a box contains 3 defective light bulbs and 12 good bulbs. Two bulbs are chosen from the box
without replacement. To find the probability that one of the bulbs drawn is good and one is defective,
what expression would you use?
a.
12 3
15 14
+
b.
12 3 3 2
15 15 15 15
⋅ + ⋅
c.
12 3
15 14
⋅
d.
12 3 3 12
15 14 15 14
⋅ + ⋅
e. none of these

 

[Jenn]

If the probabilities that Jane, Tom, and Mary will be chosen chairperson of the board are 0.5, 0.3 and
0.2 respectively, what is the probability that the chairperson will be either Jane or Mary?
Please explain too.

0.5+0.2/0.5+0.3+0.2

I think we do it like this. After solving it, we get the answer.
It's either Jane OR Mary, so we take both of them and add them, divide it by the total.

 

[Nabeel.Ahmad]

If the probabilities that Jane, Tom, and Mary will be chosen chairperson of the board are 0.5, 0.3 and
0.2 respectively, what is the probability that the chairperson will be either Jane or Mary?
Please explain too.

The answer is 0.7.
Whenever they use OR, we have to add the two individual probabilities.
Since, it says either Jane OR Mary..
The answer will be calculated by adding the two probabilities.

P(Either Jane or Mary) = P(Jane) + P(Mary)
= 0.5 + 0.2
= 0.7

 

[Nabeel.Ahmad]

Suppose a box contains 3 defective light bulbs and 12 good bulbs. Two bulbs are chosen from the box
without replacement. To find the probability that one of the bulbs drawn is good and one is defective,
what expression would you use?
a.
12 3
15 14
+
b.
12 3 3 2
15 15 15 15
⋅ + ⋅
c.
12 3
15 14
⋅
d.
12 3 3 12
15 14 15 14
⋅ + ⋅
e. none of these

P(One Defective and One Good) = P(Defective, Good) + P(Good, Defective)
= (3/15 x 12/14) + (12/15 x 3/14)
= 72/210
=12/35

 

[333333*]

In circle properties questions , do we are supposed to write answer along with reasons or its okay to write only answer .

 

[Saad Mughal]

In circle properties questions , do we are supposed to write answer along with reasons or its okay to write only answer .

With reasons, always.

 

[333333*]

With reasons, always.

if we dont mention reasons do they deduct marks ????

 

[Saad Mughal]

if we dont mention reasons do they deduct marks ????

When they say 'prove/show' congruence/similarity/ratios then yes, they do deduct marks.
But when they are asking for calculations of ratios only then they don't deduct marks.

 

[haj.arsh]

Five competitors in a quiz are isolated from each other. They are asked, in a random order, the
same question, until one of them gives the correct answer.
At this point the questioning stops. Two of them know the correct answer, and three do not.
(i) State the maximum number of times the question would have to be asked.
PLZ HELP ASAP.