Search results

Part ii asks for the probability that he will choose less than 7, which implies that he does not actually choose 7, but 6,5... ie P(x<7) Otherwise your approach to the question is correct, but rather than 1 - P(x>7) you would use 1 - P(x\geq7).

let |z-(2+i)|=|z-2| then let z=x+iy which gives |(x-2)+i(y-1)|=|(x-2)+iy| \\\sqrt{(x-2)^2+(y-1)^2}=\sqrt{(x-2)^2+y^2}\\\\(y-1)^2=y^2\\\\-2y+1=0\\\\y=\frac{1}{2} ie Im(z)=\frac{1}{2} now this gives us a line for which |z-(2+i)|=|z-2| to find which side of the line gives |z-(2+i)|\leq|z-2| we...

The only factor you forgot to take into account was that there were 6 spots that the teachers could sit in (as seen in the diagram), but only 4 teachers. So you just have to multiply what you already had 5!\times4! by {6}\choose4 to account for the additional arrangements. Which gives the same...

First substitute a point in, it would be good to choose a point that is different for all of the graphs but subbing in \frac{\pi}{2} is alright so \sin\left(\frac{\pi}{2}\right)=1 hence \cos^{-1}(-1)=\pi thus we can take out options A and B as they do not have the point. now option C is just...

we know that \\proj_\mathbf{b}\mathbf{a}=\frac{\mathbf{a}\cdot{\mathbf{b}}}{\mathbf{b}\cdot{\mathbf{b}}}\mathbf{b}\\\\=\frac{\mathbf{a}\cdot{\mathbf{b}}}{|\mathbf{b}|^2}|\mathbf{b}|\times\mathbf{\hat{b}}\\\\=\frac{\mathbf{a}\cdot{\mathbf{b}}}{|\mathbf{b}|}\mathbf{\hat{b}} As we can see here the...

Since all the means are the same for each option we can just ignore it. The formula for variance is given in the formula sheet Var(X)=np(1-p) hence the standard deviation is \sigma=\sqrt{np(1-p)} but this is the sample proportion so we actually have to divide the standard deviation by the...

for question 16) a) use IBP I=\int_0^1{x^p(1-x)^qdx let u=(1-x)^q\implies{}du=-q(1-x)^{q-1}dx let dv=x^p\ dx\implies{}v=\frac{x^{p+1}}{p+1} hence the integral can be written as...

use substitution first of s=\sqrt{x} \\ds=\frac{dx}{2\sqrt{x}}\\\\2sds=dx then \int{\sin(\sqrt{2x})dx}=2\int{s\sin(\sqrt{2}s)ds} then use integration by parts to get the answer

I think the question is trying to ask you to prove that at least one of the expressions given have to be less than or equal to one-quarter, no matter what real numbers are used.

Since this is a multi-part question, we probably need to use one of the previous parts. From part ii we get that 1=\frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}+\frac{1}{n!} now we need to show that the sum of n! is equal to the sum of n distinct divisors, so we should try to make n! the...

It was an assumption that if 4|z^2| was equivalent to the the imaginary component of z^4 then you could say z^4=a+4i|z^2| for some real number a. I just made the number a equal zero cause I wanted the smallest value for |z^4|. Looking at it now, this is probably wrong. Cause there is no...

I have a slight feeling that this could have something to do with complex numbers, cause if you let z=x+iy x^2+y^2=|z|^2, \Re(z^2)=x^2-y^2 and \frac{1}{2}\Im(z^2)=xy so the equation can be |z|^2=\frac{1}{2}\Im(z^2)\Re(z^2) or |z|^2=\frac{1}{4}\Im(z^4) I'm not sure where to go on from there...

I think you could add all the labels into one frame, then set the fill to tk.X for each label. I believe this will make the labels will expand horizontally until it reaches the edges of the frame.

I think for the question you can think of a regular cos graph for the flux, then you shift it by pi cause the normal of the armature starts perpendicular to the mangetic field which gives a -sin graph. This means that the flux is decreasing at a decreasing rate intially as t increases. Hence...

I would recommend using the substitution u=\tan^{-1}(x^3) I=\int{\frac{6x^2}{1+x^6}\times\tan^{-1}(x^3)dx \\let\ u=\tan^{-1}(x^3)\\\\du=\frac{3x^2}{1+x^6}dx \\I=2\int{u}du\\\\I=u^2+c\\\\I=(\tan^{-1}(x^3))^2+c I can kinda see where you got the answer \tan^{-2}(x^3), but you must know that...

multiply top and bottom by e^{-x} \\\int{\frac{e^{-x}}{e^{-x}\sqrt{e^{2x}-1}}}dx\\\\=\int{\frac{e^{-x}}{\sqrt{1-e^{-2x}}}dx then you can integrate it either using the substitution u=e^{-x} or just directly by inverse trig integration.

Suppose that f(x)=\cos^{-1}x and g(x)=\sin^{-1}\sqrt{1-x^2} by part a their derivatives are the same hence f'(x)=g'(x) integrating both sides \int{f'(x)}dx=\int{g'(x)}dx\implies{}f(x)=g(x)+c now let x = 0 f(0)=\frac{\pi}{2} g(0)=\frac{\pi}{2} therefore we can say that c=0. hence...

for part i differentiate the polynomial x^3 + px + q = 0\implies3x^2+p=0 then we find the roots of the derivative \\3x^2+p=0\\\\x^2=\frac{-p}{3}\\\\x=\pm\sqrt{\frac{-p}{3}} one of these roots will also have to be a root of the original function, so lets just sub it into the polynomial as it...

Basically what I am doing is making all the fractions have the same denominator which is 2k(k-1) So for example \frac{1}{2} would become \frac{k(k-1)}{2k(k-1)} by multiplying numerator and denominator by k(k-1). another example say for \frac{1}{30} the working would be...

\\LHS=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{2(k-1)(2k-3)}+\frac{1}{2k(2k-1)}\\\\=\frac{k(2k-1)}{2k(2k-1)}+\frac{\frac{1}{6}k(2k-1)}{2k(2k-1)}+\frac{\frac{1}{15}k(2k-1)}{2k...