Mathematics: Post your doubts here!

[muhammadali233]

View attachment 58941

I did the same paper today cheers!Here is the solution,just make x the subject and square it,integrate it,solve the limits,multiply with pie you have your answer,here is the working excuse the writing please.

 

[Eugene99]

I did the same paper today cheers!Here is the solution,just make x the subject and square it,integrate it,solve the limits,multiply with pie you have your answer,here is the working excuse the writing please.
View attachment 58942

*high-five* Thank you!

Just make x the subject of formulae and integrate like other normal questions. (because it says the region is shaded about the y-axis NOT x axis)

thank you!

 

[zeejay]

Hey guys, Can anyone of u solve Q14 and Q15 exercise 5A mechanics 1 with explanation?.......Thanks in advance ✊

 

[zeejay]

Hey guys, Can anyone of u solve Q14 and Q15 exercise 5A mechanics 1 with explanation?.......Thanks in advance ✊

 

[Anum96]

Hey guys, Can anyone of u solve Q14 and Q15 exercise 5A mechanics 1 with explanation?.......Thanks in advance ✊

Please post the question

 

[zeejay]

I can't upload it using my cell

 

[Rizwan Javed]

I can't upload it using my cell

are you talking about these?

 

[zeejay]

are you talking about these?

Yeah...Q14 and 15

 

[Eugene99]

sir i need help in s1..
Kevin hosts the TV programme Thank Your Lucky Stars. During the show he picks members of the large studio audience at random and asks them what star sign they were born under. (There are 12 star signs in all and you may assume that the probabilities that a randomly chosen person will be born under each star sign are equal)
-> Show that the probability that the first three people picked were all born under different star signs is approximately 0.764

The probability of the first person being chosen is unique is:
12/12 since no other member has been chosen yet
The probability for the 2nd person would be:
11/12 since one has already been chosen
and for the 3rd it is:
10/12 (2 are already chosen)
Multiplying all three: (12/12) x (11/12) x (10/12) = 0.764

 

[Eugene99]

What is the probability that at least two of the first five people picked were born under same star sign

First, we find the probability that the first five people picked were born under the same sign, which goes the same like we did previously:
(12/12) x (11/12) x (10/12) x (9/12) x (8/12) = 0.382
Now, to find the probability that at least two of the first five were born under the same sign:
1 - 0.382
=0.618
(The inverse of the probability that no two of them were born under the same sign gives the probability that at least two share a birth sign)

 

[Eugene99]

Thanks but i didnt understand this part: (The inverse of the probability that no two of them were born under the same sign gives the probability that at least two share a birth sign) eagerly waiting for your reply sir!

just meant to explain why we subtracted 0.38 from 1...You know, that rule P= 1 - (not P)

 

[nehaoscar]

s13-63


For part iii , can I not use the graph to find the mean ... ?? like the point 0.5*144 = at 72 cf i will look for the corresponding value of the weight on the x -axis for the mean ... will this be correct or no??
Anum96

 

[Eugene99]

s13-63
View attachment 58951
View attachment 58952
For part iii , can I not use the graph to find the mean ... ?? like the point 0.5*144 = at 72 cf i will look for the corresponding value of the weight on the x -axis for the mean ... will this be correct or no??

No! You can't use the graph to solve (iii) because the question asks to draw the graph in (i). If you had been asked to draw a graph before you attempted any part of the question, you could've used it to solve (iii) but since drawing the graph itself is a part of the question, you can't use it for another part what you attempted there. And also since the mark scheme shows that the mean must be worked through , this is how it must be!

 

[Eugene99]

yes i know that but i am kind of confused i guess

Okay...I guess I got things mixed up. Look, we first calculated P which is: 'first five people picked were all born under different signs' (I wrote 'under same sign' in my answer which was a mistake, I'm sorry). So, we calculated P(all born under different signs) to be 0.382.
If they are not "all born under different sign", then at least two of them are born under the same sign. The complementary event includes the possibility that two are under the same sign, or three, or four, of even all five.

Hope you get it now

 

[nehaoscar]

No! You can't use the graph to solve (iii) because the question asks to draw the graph in (i). If you had been asked to draw a graph before you attempted any part of the question, you could've used it to solve (iii) but since drawing the graph itself is a part of the question, you can't use it for another part what you attempted there. And also since the mark scheme shows that the mean must be worked through , this is how it must be!

But then will using the graph give the same answer??

 

[Eugene99]

But then will using the graph give the same answer??

you can try...depends upon whether your method is correct or not

 

[zeejay]

Anyone pls help me out with M1 exercise 5A Q15 and Q14

 

[Anum96]

Anyone pls help me out with M1 exercise 5A Q15 and Q14

Q14:
Friction Force = µR
µR = ma
(0.8)(75x10) = 75a
a = 8 m/s² ( You might get 7.84 if you use 9.8 instead of 10)

Second part:
Deceleration/acceleration = 7.84 m/s^2

V^2 = 2as + u^2
0^2 = 2(-7.84)(s) + (6)^2
s = 2.29 ~ 2.3m Ans.

Q15:
Mass = 6 kg
µ = 0.4
g = 9.8 (or 10)

Frictional Force = µR
R = mg
Ff = ma
----> m*a = µ*m*g
m will cancel out.

a = u*g
a = 0.4*9.8 = 3.92m/s^2 Ans.

 

[DarkEclipse]

I need help!

It is claimed that a certain 6-sided die is biased so that it is more likely to show a six than if it was fair.
In order to test this claim at the 10% significance level, the die is thrown 10 times and the number of sixes is noted.
(i) Given that the die shows a six on 3 of the 10 throws, carry out the test.

On another occasion the same test is carried out again.
(ii) Find the probability of a Type I error.

I got part (i) right (0.225), but I can't understand how I'm supposed to attempt part two. The answer to (ii) is 0.0697.

 

[Anum96]

I need help!

It is claimed that a certain 6-sided die is biased so that it is more likely to show a six than if it was fair.
In order to test this claim at the 10% significance level, the die is thrown 10 times and the number of sixes is noted.
(i) Given that the die shows a six on 3 of the 10 throws, carry out the test.

On another occasion the same test is carried out again.
(ii) Find the probability of a Type I error.

I got part (i) right (0.225), but I can't understand how I'm supposed to attempt part two. The answer to (ii) is 0.0697.

Part i is related to part ii here I guess.
So if you let the number of sixes thrown be X
You need to find out the P(X>=4)
Probabiilty of success is 1/6
probability of failure is 5/6
Just use the binomial formula and you'll get the answer.