Mathematics: Post your doubts here!

[i_try9621]

just a quick question , is it accelerating or decelerating ?

 

[i_try9621]

Doubt in 6(ii) and 6(iii) . It's from http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_42.pdf

 

[Konstantino Nikolas]

View attachment 59950 just a quick question , is it accelerating or decelerating ?

Decelerating, since the slope of graph gets less steep with time. So speed decreases.

 

[Konstantino Nikolas]

View attachment 59951 Doubt in 6(ii) and 6(iii) . It's from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_42.pdf

 

[i_try9621]

View attachment 59955

In 6(ii) why is the whole graph shaded if the time limit is 20 < t < 26?

 

[Konstantino Nikolas]

In 6(ii) why is the whole graph shaded if the time limit is 20 < t < 26?

Because, as stated in the question, 's' represents the displacement of P from O (the starting position) at any time 't'. Here, our 't' lies anywhere in between 20 and 26, the limits included.

 

[i_try9621]

Doubt in 4(iv)

 

[i_try9621]

Because, as stated in the question, 's' represents the displacement of P from O (the starting position) at any time 't'. Here, our 't' lies anywhere in between 20 and 26, the limits included.

ohhh! thanks for that

 

[Anum96]

http://studyguide.pk/Past Papers/CI... AS Level/9709 -Mathematics/9709_w07_qp_6.pdf
Q7 last part

 

[The Sarcastic Retard]

Please anybody give full solution to this question

Yeah! I solved this few pages back : Here

 

[Rizwan Javed]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9709 -Mathematics/9709_w07_qp_6.pdf
Q7 last part

There are three possible no. of times a red paper clip is taken:
1) 0 times red is taken : white from box A and White from box B
2) 1 time red is taken out: Red from box A and White from Box B OR White from Box A and Red from Box B
3) 2 times red is taken out: Red from box A and Red from Box B

So first consider the 1st situation in which no red is taken out.
P(white from A and White from B) = 1/6 * 3/10 = 3/60

Now the second situation:
P(red from A and white from B OR white from A and red from B) = 5/6 * 2/10 + 1/6 * 7/10 = 17/60

Finally, the third situation:
P( red from A and red from B) = 5/6 * 8/10 = 40/60

So the probability distriburion of X is :

x | P(X = x)
0 | 3/60
1 | 17/60
2 | 40/60

 

[Saad the Paki]

View attachment 59945
View attachment 59946
help in I

First we need to find the probability that a carton contains more than 1002 ml. This will be 225/900 = 0.25. Now we need to find the value of Z by working backwards from the table. Since we don't have a probability of 0.25 in the table we can find Z from (1_0.25) = 0.75. The value of Z corresponding to a probability of 0.75 is Z=0.674.
Now standardise (X >1002) ; (Z > (1002-u)/8 ) = 0.674 ; and solve

 

[Saad the Paki]

View attachment 59943 View attachment 59944
help in II

At least one damaged tape means (P>=1) which is the same as (1-P (0)).
1-[ nC0*(0.2)^0*(0.8)^n ] >=0.85
1-(0.8)^n >=0.85
Therfore solve to get n>=8.5 so the smallest value will be n=9 (we need an integer value)

 

[Saad the Paki]

Correct answer legit obtained ?? in MS what does this mean

Means no fortuitous answers.
Like for example they say "Show the probability is X=0.25"
And you simply wrote 0.5*0.5=0.25, even though no probability of 0.5 is given in the question.
Your answer is correct but your method is wrong

 

[Anum96]

There are three possible no. of times a red paper clip is taken:
1) 0 times red is taken : white from box A and White from box B
2) 1 time red is taken out: Red from box A and White from Box B OR White from Box A and Red from Box B
3) 2 times red is taken out: Red from box A and Red from Box B

So first consider the 1st situation in which no red is taken out.
P(white from A and White from B) = 1/6 * 3/10 = 3/60

Now the second situation:
P(red from A and white from B OR white from A and red from B) = 5/6 * 2/10 + 1/6 * 7/10 = 17/60

Finally, the third situation:
P( red from A and red from B) = 5/6 * 8/10 = 40/60

So the probability distriburion of X is :

x | P(X = x)
0 | 3/60
1 | 17/60
2 | 40/60

I got it. Thank you so much

 

[Anum96]

http://studyguide.pk/Past Papers/CI...AS Level/9709 -Mathematics/9709_w09_qp_62.pdf
Q4 (a) ii

 

[Rizwan Javed]

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9709 -Mathematics/9709_w09_qp_62.pdf
Q4 (a) ii

Divide this question into parts. First find all the 3 digit numbers, then find all the 4 digit numbers.

In the 3 digit numbers, the number can either start with 5 or 6 since, all the numbers must be greater than 500. If 5 is the starting number, then the ending number can be either 1 or 3 since number must be odd. If it starts with 6, then it can end with either 1, 3 or 5. The middle number can then be selected in 2C1 ways for both of the cases.
So the total number of possible 3 digit numbers are:
2 * 2C1 + 3 * 2C1 = 10

Now consider the 4 digit numbers. They can start with any number. So deal them separately:
1) 4-digit number starts with 1, so it can end with either 3, or 5. The middle 2 digits can be arranged in 2! or 2 ways. So possible arrangements: 2 * 2!
2) 4-digit number starts with 3, so it can end with either 1, or 5. The middle 2 digits can be arranged in 2! or 2 ways. So possible arrangements: 2 * 2!
3) 4-digit number starts with 5, so it can end with either 1, or 3. The middle 2 digits can be arranged in 2! or 2 ways. So possible arrangements: 2 * 2!
4) 4-digit number starts with 6, so it can end with either 1, 3, or 5. The middle 2 digits can be arranged in 2! or 2 ways. So possible arrangements: 3 * 2!

Add ^ them up to find the TOTAL possible numbers:
10 + 4 + 4 +4 + 6 = 28 numbers

 

[Eugene99]

so for the (i) it's (3C1) (5C1 x 2C1) (3C1) = (3)(5x2) (3) =90
and for (ii) 3C1 (5C1 2C1) 3C0 + 3C1 (5C0 2C0) 3C1 + 3C0 (5C1 2C1) 3C1
and so becomes 69
but I've no idea how we got this! somebody help please

 

[Rizwan Javed]

View attachment 59961

so for the (i) it's (3C1) (5C1 x 2C1) (3C1) = (3)(5x2) (3) =90
and for (ii) 3C1 (5C1 2C1) 3C0 + 3C1 (5C0 2C0) 3C1 + 3C0 (5C1 2C1) 3C1
and so becomes 69
but I've no idea how we got this! somebody help please

There are 3 courses, and he can take one meal from each course. So just make selections:
Select 1 meal from the 3 meals in starter course: 3C1
Select 1 meak from 5 meals in main course. Also note that each meal is served with either new potatoes OR french fries. So you need to select any from these 2 as well. So the number of selections will become: 5C1 * 2C1
Finally select 1 meal from the Dessert course: 3C1

The total no. of possible selections can then be calculated by multiplying all the above possible selections, since these events are occuring in side by side: 3C1 * 5C1 * 2C1 * 3C1 = 90

(ii) Similarly in part (ii) you are to adapt a similar method. But now you're restricted to choosing from only courses at a time.
So you can select either Starter and Main couse, Starter and Dessert Or Main Course and Dessert.
The no. of possible selections will remain the same.

Starter + Main Course: 3C1 * 5C1 * 2C1
Starter + Dessert: 3C1 * 3C1
Main course + Dessert : 5C1 * 2C1 * 3C1

^ Sum them up to find the total possible selections. The answer will be the one you have mentioned that is 69.

 

[Eugene99]

There are 3 courses, and he can take one meal from each course. So just make selections:
Select 1 meal from the 3 meals in starter course: 3C1
Select 1 meak from 5 meals in main course. Also note that each meal is served with either new potatoes OR french fries. So you need to select any from these 2 as well. So the number of selections will become: 5C1 * 2C1
Finally select 1 meal from the Dessert course: 3C1

The total no. of possible selections can then be calculated by multiplying all the above possible selections, since these events are occuring in side by side: 3C1 * 5C1 * 2C1 * 3C1 = 90

(ii) Similarly in part (ii) you are to adapt a similar method. But now you're restricted to choosing from only courses at a time.
So you can select either Starter and Main couse, Starter and Dessert Or Main Course and Dessert.
The no. of possible selections will remain the same.

Starter + Main Course: 3C1 * 5C1 * 2C1
Starter + Dessert: 3C1 * 3C1
Main course + Dessert : 5C1 * 2C1 * 3C1

^ Sum them up to find the total possible selections. The answer will be the one you have mentioned that is 69.

GOt that!! just needed a li'l bit of activation energy to start the topic!