Bertie

I'm not a statistician as such so a proper one, (a proper statistician that is ), might take issue, however, so far as I'm aware, the S.E. that's being calculated here is the S.E. for the parent population, not the sample.  We don't have a standard deviation for the sample.

BTW, this is the method used in opinion poles for elections and so on.

This is a non-linear equation, and requires a numerical method for its solution, Newton-Raphson for example.

A simple sketch graph (y = log(x) and y = -0.064x), shows that there is a single root and that its value will be slightly less than 1.

Rather than use a numerical method, you could simply use one of those clever graph plotters.

The root turns out to be 0.87856 5dp.

The Standard Error for the proportion of weight gains in the whole population will be

$$\displaystyle \sqrt{\frac{0.65 \times 0.35}{30}}=0.087\quad \text{ 3 dp},$$

which means that the 99.7% confidence limits will be

$$\displaystyle 0.65 \pm 3\times 0.087 = 0.65 \pm 0.26.$$

D is the mid-point of AC, AB = BC = a and CE is the perpendicular from C to AB and is of length h.

$$\displaystyle \frac{h}{a}=\sin(\angle EBC) = \sin(180 - 2\angle BAC)=\sin(2\angle BAC)\\ \\ =2\sin(\angle BAC)\cos(\angle BAC)\\ \\=2\times \frac{30}{a}\times \frac{\sqrt{a^{2}-30^{2}}}{a}$$

If h is to equal 20, then, multiplying by a squared and cancelling,

$$\displaystyle a=3\sqrt{a^{2}-900},$$

from which, (square both sides),

$$a=\sqrt{8100/8}\approx 31.8198.$$

The base angle can then be calculated. Repeat the calculation for BD = 20, and h = 30.

To save on typing, let e^(i.theta) = z.

Then, using the binomial expansion,

$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$

so,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

You already have the method Melody, you used it in your earlier post.

To illustrate what's going on here, consider the,  integrand cos(x).sin^3(x).

We can integrate that in, at least, two different ways.

We can integrate it directly as sin^4(x)/4  or we can use the trig identity sin^2 = 1 - cos^2 to get

cos(x)(1-cos^2(x))sin(x) = cos(x)sin(x) - cos^3(x).sin(x) which integrates to -cos^2(x)/2 + cos^4(x)/4.

The problem we are discussing is dealt with in exactly the same way except that the functions are cosec and cot.

Come back if you need further details.

I didn't want to spoil your, or someone elses, enjoyment. The answer is in a private message, publish it if you wish.

Both are correct.

Simply make use of the identity 1+cot^2 = cosec^2 to go from the first form to the second. The -1/4 gets lumped in with the constant of integration.

Picture 52 locations divided into 4 groups of 13.

The first Ace can go anywhere, the second Ace has 39 of 51 possible locations in which to go, the third has 26 of a possible 50 and the fourth 13 of a possible 49.

That works out to about 10.5498%, you should be able to work out the exact fractional value.

It's taken a while, but, 6789 = 5.