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Mathematics: Post your doubts here!

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can someone tell me when continuity correction has to be done for calculation mean/variance for a histogram data in s1?
When the given data is not continuous, you to correct it using upper and lower bounds to plot histogram and calculate the mean, varaiance, SD, or anything asked.
 
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http://onlineexamhelp.com/past-pape.../9709-mathematics-a-as-level-past-papers-2014

Hi. can someone please take a look at s14_qp_13 question 7 (i) ?
I seem to never get it right.

Hey, in 7) were given vectors OA, OB and OC.
We have to show that angle BAC is = cosinverse (1/3)
This angle is at the vertex A, draw a quick sketch it'll make things clear.
This angle is between vectors AB and AC also, a quick sketch will help you.
We know that Costheta= A.B/|A|.|B|
Where theta is the angle between the two vectors, and modulus is their lengths.
Now find AB, we know that AB= OB - OA= (0,-1,7) -(2,1,3) = (-2,-2,4)
Find AC= OC - OA= (2,4,7)- (2,1,3)= (0,3,4)
Now substitute in the equation, Costheta= (-2,-2,4).(0,3,4)/root((-2)^2+(-2)^2+(4)^2).root((0)^2+(3)^2+(4)^2)
Make theta the subject, Costheta = 10/(root24).(5)
Theta = cosinverse(1/root6)
I think there's something wrong here, ms says OB - OA= (4,-2,4) but how ? I think there's a typo in the first number of the given OB :/ or I might have a silly mistake who knows
 
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Hey, in 7) were given vectors OA, OB and OC.
We have to show that angle BAC is = cosinverse (1/3)
This angle is at the vertex A, draw a quick sketch it'll make things clear.
This angle is between vectors AB and AC also, a quick sketch will help you.
We know that Costheta= A.B/|A|.|B|
Where theta is the angle between the two vectors, and modulus is their lengths.
Now find AB, we know that AB= OB - OA= (0,-1,7) -(2,1,3) = (-2,-2,4)
Find AC= OC - OA= (2,4,7)- (2,1,3)= (0,3,4)
Now substitute in the equation, Costheta= (-2,-2,4).(0,3,4)/root((-2)^2+(-2)^2+(4)^2).root((0)^2+(3)^2+(4)^2)
Make theta the subject, Costheta = 10/(root24).(5)
Theta = cosinverse(1/root6)
I think there's something wrong here, ms says OB - OA= (4,-2,4) but how ? I think there's a typo in the first number of the given OB :/ or I might have a silly mistake who knows
hi! Could you help me with further statistics? Ques 7ii please https://thol.sunway.edu.my/examdbase/alv/math/p7/math_p7_n02.pdf Thanks.
And this question too, A fair coin is tossed 5 times and the number of heads is recorded.
The number of heads is doubled and denoted by the random variable Y. State the mean and variance of Y.
 
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Hey, in 7) were given vectors OA, OB and OC.
We have to show that angle BAC is = cosinverse (1/3)
This angle is at the vertex A, draw a quick sketch it'll make things clear.
This angle is between vectors AB and AC also, a quick sketch will help you.
We know that Costheta= A.B/|A|.|B|
Where theta is the angle between the two vectors, and modulus is their lengths.
Now find AB, we know that AB= OB - OA= (0,-1,7) -(2,1,3) = (-2,-2,4)
Find AC= OC - OA= (2,4,7)- (2,1,3)= (0,3,4)
Now substitute in the equation, Costheta= (-2,-2,4).(0,3,4)/root((-2)^2+(-2)^2+(4)^2).root((0)^2+(3)^2+(4)^2)
Make theta the subject, Costheta = 10/(root24).(5)
Theta = cosinverse(1/root6)
I think there's something wrong here, ms says OB - OA= (4,-2,4) but how ? I think there's a typo in the first number of the given OB :/ or I might have a silly mistake who knows

Thank you very much!!
 
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3 The times taken by students to get up in the morning can be modelled by a normal distribution with
mean 26.4 minutes and standard deviation 3.7 minutes.
(i) For a random sample of 350 students, find the number who would be expected to take longer
than 20 minutes to get up in the morning. [3]
(ii) ‘Very slow’ students are students whose time to get up is more than 1.645 standard deviations
above the mean. Find the probability that fewer than 3 students from a random sample of 8
students are ‘very slow’.

please explain part 2,i didnt get what does question mean???
 
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Hey, in 7) were given vectors OA, OB and OC.
We have to show that angle BAC is = cosinverse (1/3)
This angle is at the vertex A, draw a quick sketch it'll make things clear.
This angle is between vectors AB and AC also, a quick sketch will help you.
We know that Costheta= A.B/|A|.|B|
Where theta is the angle between the two vectors, and modulus is their lengths.
Now find AB, we know that AB= OB - OA= (0,-1,7) -(2,1,3) = (-2,-2,4)
Find AC= OC - OA= (2,4,7)- (2,1,3)= (0,3,4)
Now substitute in the equation, Costheta= (-2,-2,4).(0,3,4)/root((-2)^2+(-2)^2+(4)^2).root((0)^2+(3)^2+(4)^2)
Make theta the subject, Costheta = 10/(root24).(5)
Theta = cosinverse(1/root6)
I think there's something wrong here, ms says OB - OA= (4,-2,4) but how ? I think there's a typo in the first number of the given OB :/ or I might have a silly mistake who knows
I think its a typo cuz 0 and 2 can never give an i vector of 4,i also did it and got the ans as cos-1 (1/(square root 6).There has to be a typo in the question.
 
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dx=2u du
Substitute x and dx to get ∫2ucosu du. Integrate by parts.
2u(sin u) - ∫sin u(2)
2u sinu + 2 cos u
Limits will be 0 and p as when x is 0 u is 0 and when x=p^2 then u=p. Area of R is equal to 1 so equation will be equated to 1.
(2p sinp + 2 cosp) - (0 + 2)=1
2p sinp +2 cosp -2=1
Make sinp the subject to get sinp= (3-2 cosp)/2p
 
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View attachment 52293
in my ans i get +2 instead of -1 cn smone point out my mistake??


if you're writing 4x^2 +8x+3 in the form of a(x+b)^2+c. I suggest you fully expand the form of -->a(x+b)^2+c and compare it to your quadratic equation.
example, the full expansion of a(x+b)^2+c is ---> ax^2 + 2abx+ab^2+c.
comparing it to your quadratic equation would be like:
  • 4x^2 = ax^2, cancel the x^2 on both sides gives you a=4.
  • 8x= 2abx (substitute a here) , 8x= 2 (4)bx, cancel x on both sides gives you b=1.
  • 3=ab^2+c (substitute both a and b here) , 3= (4)(1)^2 +c, gives you c= -1
hence overall substitute all values you found in the form of a(x+b)^2+c gives you 4( x+1)^2 -1
I hope this helps.
 
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