Recent content by chwala

It is not given.

I do not seem to understand this question nor the Ms. I understand the box-whisker plot and its relationship to quartiles of which ##Q_2## is the Median. This question is from an As level- statistics past paper question.

Just went through this...steps pretty clear. I refreshed on Riemann integrals { sum of rectangles approximate area under curves}. My question is on the highlighted part in Red. The approximation of area under curve may be smaller or larger than the actual value. Thus the inequality may be ##<##...

...still not getting it. You mind giving some reason or thought. Thks.

O level question; i used similarity would appreciate an easier approach for 2 marks. The ms solution (approach) is not clear to me. Here it is; My approach; using similarity Any insight welcome its a 2 mark question- cannot seem to find easier way though i suspect reflection.

My interest is on the highlighted (In Red). Otherwise the other steps are clear. We have on that part of the problem, ##(-p\sin t -q\cos t)-12(p\cos t -q \sin t)+36p\sin t +36q\cos t = 37 \sin t + 0 \cos t## Ah I just realized we are solving a simultaneous equation for ##p## and ##q## ...

Correct. Cheers guys.

Substitution for ##dx## using change of variables... It should be fine. Let me check that again.

Ok bass (boss). :wink: Gday @Mark44 :cool:

For part (b) i was able to use equations to determine the eigenvectors; For example for ##λ =6## ##12x +5y -11z=0## ##8x-4z=0## ##32x+10y-26z=0## to give me the eigen vector, ##\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}## and so on. My question is to get matrix P does the arrangement of...

Thanks, i was confusing... the correct identity is ##\cos^{2} 2θ = \dfrac{1}{2}(1+\cos 4θ) ## anyway i realized that I will still need it in doing my work ; ##\int_0^{\frac{π}{6}} [cos ^4θ] dθ= \int_0^{\frac{π}{6}} \left[\dfrac{1}{4} + \dfrac{\cos 2θ}{2} + \dfrac{\cos^{2} 2θ}{4}\right] dθ##...

that is ##\cos^4θ = \dfrac{1}{2}(1+\cos 4θ) ?##

I have ##1-x^2 = 1- \sin^2 θ = \cos^2 θ## and ## dx =cos θ dθ## ##\int_0^{0.5} (1-x^2)^{1.5} dx = \int_0^{\frac{π}{6}} [cos ^2θ]^\frac{3}{2} dθ = \int_0^{\frac{π}{6}} [cos ^4θ] dθ## Suggestions on next step.

yeah typo error. Cheers man.

Thanks man!!! i was blind. ##z=x+y## ##\dfrac{dz}{dx} = 1 + \dfrac{dy}{dx}## ##\dfrac{dz}{dx} -1 = \dfrac{dy}{dx}##