Mathematics: Post your doubts here!

[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.

 

[nehaoscar]

Parts ii and iii please

 

[Anum96]

View attachment 58958
Parts ii and iii please

For part ii. Rearrange the equation:
sina - ((1.5-cosa)/a)
Put 1 and calculate. Then put 2 and calculate. You will notice a change in the signs of your answer. Second one will be positive and so just write a statement that since the answer is negative when a is 1 and positive when a is greater than 1 then a is by calculation greater than 1

For part iii. Use a1 as 1 and keep calculating like other iterations. Stop when you start getting the same answer on ur calculator. That will be your answer.

 

[nehaoscar]

Parts ii and iii please ... I'm stuck on ii so I can't go onto iii either and the ms is confusing me how to plot it....

Why do we need a bisector for part ii ...
Please can you solve and explain this question?
Anum96

 

[nehaoscar]

For part ii. Rearrange the equation:
sina - ((1.5-cosa)/a)
Put 1 and calculate. Then put 2 and calculate. You will notice a change in the signs of your answer. Second one will be positive and so just write a statement that since the answer is negative when a is 1 and positive when a is greater than 1 then a is by calculation greater than 1

For part iii. Use a1 as 1 and keep calculating like other iterations. Stop when you start getting the same answer on ur calculator. That will be your answer.

For part ii I'm getting negative for a as 1 and as 2
when a = 1
sin(1) - ((1.5-cos(1))/(1)) = -0.118
when a = 2
sin(2) - ((1.5-cos(2))/(2)) = -0.0487
there is no sign change :/
This is in radian mode btw and even in degree mode there is no sign change :S

 

[Anum96]

For part ii I'm getting negative for a as 1 and as 2
when a = 1
sin(1) - ((1.5-cos(1))/(1)) = -0.118
when a = 2
sin(2) - ((1.5-cos(2))/(2)) = -0.0487
there is no sign change :/
This is in radian mode btw and even in degree mode there is no sign change :S

There should beee. Lol, Wait!

 

[DarkEclipse]

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.

Thanks!

 

[DESTROYER1198]

can anyone please solve q5 ii in this year i would really appreciate any help http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s14_ms_31.pdf
Thanks

 

[nehaoscar]

There should beee. Lol, Wait!

You got it yet??

 

[Anum96]

You got it yet??

I cant recall any other method. Maybe ***amd*** can help.

 

[nehaoscar]

I cant recall any other method. Maybe ***amd*** can help.

Well this is what the ms says:

Oh got it! in the question a has limits within 0<a<pi/2 so we can't take 2
there is sign change at 1.5

 

[Anum96]

Well this is what the ms says:
View attachment 58977
Oh got it! in the question a has limits within 0<a<pi/2 so we can't take 2
there is sign change at 1.5

Oh there u have it.

 

[nehaoscar]

part ii please

 

[nehaoscar]

how to integrate this! Please someone do it... the ms is too vague on this question
w15 - 32

 

[Anum96]

View attachment 58982
part ii please

Just use your answer to part i. If your answer lies between 2 and 3 use 2 as x1 and just keep solving till you get the same answer thrice.

 

[Anum96]

View attachment 58983
how to integrate this! Please someone do it... the ms is too vague on this question
w15 - 32

This was too easy. Just use the double angle formula.

 

[Anum96]

View attachment 58983
how to integrate this! Please someone do it... the ms is too vague on this question
w15 - 32

replace sin^2(2theta) with (1-cos4theta)/2 and then solve.

 

[nehaoscar]

Just use your answer to part i. If your answer lies between 2 and 3 use 2 as x1 and just keep solving till you get the same answer thrice.

Thats part iii where it asks to use iterative formula to find value of a
How to prove that it's converging to a ?