Mathematics: Post your doubts here!

[whitecorp]

Hey guys, I'm having some trouble with this A2 statistics question. I can't quite figure it out, please help :{

The heights of a population of male students are distributed normally with mean 178cm and standard deviation 5cm. The heights of a population of female students are distributed normally with mean 168cm and standard deviation 4cm. Find the probability that a randomly chosen female is taller than a randomly chosen male.

Let X be the random variable denoting the height of a male student. Then X~N( 178, 5^2 =25)

Let Y be the random variable denoting the height of a female student. Then Y~N( 168, 4^2 =16)

Also, based on the above, we have Y-X~N( -10, 25+16=41)

P( Y>X) =P(Y-X>0)=0.0592 (shown)

Hope this helps. Peace.

 

[whitecorp]

express cos^2(2x) in terms of cos4x..... pls help

This simply requires a tweaking of the double angel formula, where cos 2x = 2 cos^2 (x) -1
We rewrite things as cos 4x =2 cos^2 (2x) -1 ; making cos^2 (2x) the subject, we therefore have

cos^2 (2x) = 1/2* (cos4x +1) (shown)

Hope this helps. Peace.

 

[salvatore]

Please help me solve the above question. Seems easy.. but I'm confused.
Thanks

 

[Beca1206]

Let X be the random variable denoting the height of a male student. Then X~N( 178, 5^2 =25)

Let Y be the random variable denoting the height of a female student. Then Y~N( 168, 4^2 =16)

Also, based on the above, we have Y-X~N( -10, 25+16=41)

P( Y>X) =P(Y-X>0)=0.0592 (shown)

Hope this helps. Peace.

THANK YOU SO MUCH

 

[whitecorp]

Aoa wr wb
Can someone please help me with part iii) of this qn?

please provide detailed step-by-step explanation, i didn't understand the ms here:/


View attachment 18982


whitecorp and everyone else. thanx a million.

There you go:





Hope this helps. Peace.

 

[whitecorp]

THANK YOU SO MUCH

No problem. Peace.

 

[Alice123]

This simply requires a tweaking of the double angel formula, where cos 2x = 2 cos^2 (x) -1
We rewrite things as cos 4x =2 cos^2 (2x) -1 ; making cos^2 (2x) the subject, we therefore have

cos^2 (2x) = 1/2* (cos4x +1) (shown)

Hope this helps. Peace.

Show that (cosx+3sinx)^2 equals 5-4cos2x+3sin2x.........quick.....thanks in advance

 

[whitecorp]

Please help me solve the above question. Seems easy.. but I'm confused.
Thanks

Here are the full workings:



Hope this helps. Peace.

 

[whitecorp]

Show that (cosx+3sinx)^2 equals 5-4cos2x+3sin2x.........quick.....thanks in advance

(cosx+3sinx)^2 = cos ^2 (x) + 6 sinx cosx + 9 sin ^2 (x)

= 0.5* [cos(2x)+1] + 3 (2 sinx cosx) + 9 * 0.5* [1-cos(2x)] (use the identities cos (2x) = 2 cos ^2 (x) -1 = 1- 2sin ^2 (x) for transformation)

= 0.5 cos (2x) + 0.5 +3 sin (2x) + 4.5 -4.5 cos(2x)

= 5-4cos2x+3sin2x (shown)

Hope this helps. Peace.

 

[Soldier313]

There you go:

Hope this helps. Peace.

Thanx a million !!

 

[Alice123]

thank

(cosx+3sinx)^2 = cos ^2 (x) + 6 sinx cosx + 9 sin ^2 (x)

= 0.5* [cos(2x)+1] + 3 (2 sinx cosx) + 9 * 0.5* [1-cos(2x)] (use the identities cos (2x) = 2 cos ^2 (x) -1 = 1- 2sin ^2 (x) for transformation)

= 0.5 cos (2x) + 0.5 +3 sin (2x) + 4.5 -4.5 cos(2x)

= 5-4cos2x+3sin2x (shown)

Hope this helps. Peace.

thanks a lot

 

[Alice123]

∫ (3/2-x)+(4x/4+x^2) dx
help

 

[salvatore]

Here are the full workings:



Hope this helps. Peace.

Thanks a lot man..

 

[Silent Hunter]

Do we have to memorise all the trigonometric formula etc? or not?
what are specific things to be learned for P2 and P3 ?

 

[whitecorp]

thank
thanks a lot

No problem. Peace.

 

[whitecorp]

Thanks a lot man..

Welcome. Peace.

 

[whitecorp]

∫ (3/2-x)+(4x/4+x^2) dx
help

I am assuming (3/2-x) = 3/(2-x) and (4x/4+x^2) = 4x/ (4+x^2)

Then ∫ (3/2-x)+(4x/4+x^2) dx

= ∫ -(-3/2-x)+2*(2x/4+x^2) dx (shape the both parts of the integral in the format ∫ f '(x)/ f(x) dx, so that this can be reduced to ln |f(x)| )

= -3 ln |2-x| + 2 ln |4+x^2 | +C (shown)

Hope this helps. Peace.

 

[Silent Hunter]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf


Its question number 6

The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,
where a is a positive constant and 0 < t < pie/2
(i) Express dydx in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.

part 1 done.How to do part (ii) ?

whitecorp or anybody ? Thank You

 

[Navi Don]

how we can go back from probabillity to its value in terms of phi from normal distribution table?

 

[salvatore]

Find the definite integral of √(5x + 4) from x = 0 to x= 1

This question seems easy but I'm not able to solve it.. please help!
Thanks!