Mathematics: Post your doubts here!

[areeba240]

In quadrilateral ABCD, AB produced is perpendicular to DC produced. If Angle A is 44 and angle C is 148, calculate angle D and B

 

[areeba240]

send me any sort of doubts........ill b eager to help u out

then please solve the question which i have asked.

 

[Manisha98]

then please solve the question which i have asked.

sorry fr d late reply ..........welll here u go


Draw the quadrilateral ABCD and let AB produced meet DC produced at E

Given,

angle A = 44 degrees

angle C = 148 degrees


The triangle AED is a right triangle with right angle at E (since AB is perpendicular to DC) thus

angle D = 180 - (90 + angle A) = 180 - (90 + 44) = 46 degrees

angle D = 46 degrees


The sum of the angles in the quadrilateral is 360 thus

angle B = 360 - (angle A + angle C + angle D) = 360 - (44 + 148 + 46)

angle B = 122 degrees


I hope u have undestood =)

 

[zahra azam]

guys can anybody help me with math 0580/42/o/n/12
question 8 part a(ii)
husnain99
yasoob_shah

 

[Manisha98]

s

guys can anybody help me with math 0580/42/o/n/12
question 8 part a(ii)

send d paper

 

[zahra azam]

s

send d paper

Ok

 

[zahra azam]

https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230QUVLN0ZkTklsUnc

its in the 15th position

 

[husnain99]

X axis per 2.5 say line oper extend kro .. Us curve ko line touch kray gi .. Wahan say line full lay jana ..
Then 2 point lenay hain for gradient ..

 

[zahra azam]

X axis per 2.5 say line oper extend kro .. Us curve ko line touch kray gi .. Wahan say line full lay jana ..
Then 2 point lenay hain for gradient ..

2 Points konsey leney hein

 

[husnain99]

2 Points konsey leney hein

Any two points from that line , answer would be same

To make sure ans is correct , 2 baar 2 different points lo...

 

[areeba240]

sorry fr d late reply ..........welll
I hope u have undestood =)
Yes I understood, thanks

 

[yasoob_shah]

https://drive.google.com/folderview...sp=drive_web&tid=0BzumkDfi9230QUVLN0ZkTklsUnc

its in the 15th position

isme tw itne saare khul rahe hain...konsa wala hai?

 

[NOneed2speedd]

Can someone explain me Q.7 last part (b)
and last past of Q.9 (e)

 

[Mr.Physics]

Ans 7 b) Mean mass of 20 oranges is 70g so total mass of 20 oranges will be 70 x 20 = 1400 g

Mean mass of 19 oranges is 70.5g so total mass of 19 oranges will be 70.5 x 20 = 1339.5 g

The mass of the orange which is eaten = 1400 - 1339.5 = 60.5 g


Ans 9 e) This question is bit hard !! But here we go

The probability of weather is fine is 3/4

The probability of weather is not fine will be 1 - 3/4

The probability of weather is not fine for 1 day in 5 days will 1 - (3/4)^5 = 781 / 1024

Note ( ^ means raise to power)

 

[NOneed2speedd]

Ans 7 b) Mean mass of 20 oranges is 70g so total mass of 20 oranges will be 70 x 20 = 1400 g

Mean mass of 19 oranges is 70.5g so total mass of 19 oranges will be 70.5 x 20 = 1339.5 g

The mass of the orange which is eaten = 1400 - 1339.5 = 60.5 g


Ans 9 e) This question is bit hard !! But here we go

The probability of weather is fine is 3/4

The probability of weather is not fine will be 1 - 3/4

The probability of weather is not fine for 1 day in 5 days will 1 - (3/4)^5 = 781 / 1024

Note ( ^ means raise to power)

Wonderful. ... Thanks bro

 

[Mr.Physics]

Wonderful. ... Thanks bro

No prob !! feel free to ask !!

 

[lina1999]

Need HELP
maths problem November 2014 question 10(c)
https://docs.google.com/file/d/0B108wb6vqF_8TDZydUdObjVFXzg/edit?pli=1

 

[masterex567]

Need HELP
maths problem November 2014 question 10(c)
https://docs.google.com/file/d/0B108wb6vqF_8TDZydUdObjVFXzg/edit?pli=1

Oh this question! Just cracked it yesterday, but its incredibly long!

Alright first we gather the possibilities. So to get to the 100th place, these are the possible ways:

1. Roll a 3.
2. Roll a 2 on first throw and then a 4 on second throw.
3. Doesn't roll a 2 or 3 on first throw and then rolls a 3 on second throw.

Now we gather the probabilities for each of these cases.
To roll a 3, he can either roll a 1 and 2 on die, or a 2 and 1 on die. Which is (1/6 * 1/6) + (1/6 * 1/6).
Therefore to roll a 3, the probability is 2/36.

Next, for roll a 2 on first throw and 4 on second.
To roll a 2, he needs to roll 1 and 1 on die. Therefore it is (1/6 * 1/6). which is 1/36.
Now to roll a 4 there are numerous possibilities. He can roll a 1 and 3, or a 3 and 1, or a 2 and 2. Calculate the probabilities for this as follows:
(1/6*1/6) + (1/6*1/6) + (1/6*1/6).
Therefore probability for rolling a 4 is 3/36.

Next, we find probability for not rolling a 2 or 3. Which is the same as 1 - (probability of rolling 2 or of rolling 3).
1 - (1/36 + 2/36). This is (1 - 3/36) or 33/36. same thing. And now probabibility of rolling a 3 on second throw will be 2/36 as calculated before.


FINALLY. We calculate all these.
So probabibility of getting a 100 is:
Probability of rolling a 3, OR, not rolling 2 or 3 on first throw, AND rolling 3 on second, OR probability of rolling a 2 on first throw AND 4 on second. Remember OR is adding. AND is multiplying.
[2/36] + [1-(3/36) * 2/36] + [1/36 * 3/36]

And that gives the final answer of 47/432. If you check the calculation above is similiar to one in marking scheme!

If you need any clarification on this, let me know.
Cheers

 

[masterex567]

I have one question on the marking of these papers, do they give all marks if answer is correct or the correct method should also be there?
What if the method is different from the one in marking scheme but answer is correct, is full credit awarded then?

 

[reem maqlad]

How Do I Factorize ?