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

 

[Manisha98]

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

 

[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