Author Topic: VCE Methods Question Thread! (Read 5648513 times) Tweet Share

darklight · « **Reply #1830 on:** April 20, 2013, 05:38:51 pm »

The weekly error (in seconds) of a brand of watch is known to be normally distributed. Only those watches with an error of less than 5 seconds are acceptable.

Find the mean and standard deviation of error if 3% of watches are rejected for losing time and 3% are rejected for gaining time.

b^3 · « **Reply #1831 on:** April 20, 2013, 06:03:01 pm »

Let $X$ be the time that a watch loses in a week (negative time means it's running fast).
$\begin{alignedat}{1}\text{Pr}\left(X\leq-5\right) & =0.03 \\ \text{Pr}\left(X\leq5\right) & =1-0.03 \\ & =0.97 \end{alignedat}$
Transform it to the standard normal distribution.
Then
$\begin{alignedat}{1}z & =\frac{x-\mu}{\sigma} \\ \text{Pr}\left(X\leq-5\right) & =\text{Pr}\left(Z\leq\frac{-5-\mu}{\sigma}\right) \\ \text{Pr}\left(Z\leq\frac{-5-\mu}{\sigma}\right) & =0.03 \\ \text{Pr}\left(X\leq5\right) & =\text{Pr}\left(X\leq\frac{5-\mu}{\sigma}\right) \\ \text{Pr}\left(X\leq\frac{5-\mu}{\sigma}\right) & =0.97 \end{alignedat}$
Using the Inverse Normal Function on the calculator, [Menu] [5] [5] [3] with an area of 0.03 and 0.97 (Ti-nspire calc).
$\begin{alignedat}{1}\frac{-5-\mu}{\sigma} & =-1.8808 \\ \sigma & =\frac{-5-\mu}{-1.8808} \\ & =\frac{\mu+5}{1.8808}....[1] \\ \frac{5-\mu}{\sigma} & =1.8808....[2] \\ \text{Substitute }[2]\text{ into }[1] \\ \left(-5-\mu\right)\div\left(\frac{\mu+5}{1.8808}\right) & =1.8808 \\ \frac{-5-\mu}{\mu+5} & =1 \\ -5-\mu & =\mu+5 \\ 2\mu & =0 \\ \therefore\mu & =0 \\ \sigma & =\frac{0+5}{1.8808} \\ \therefore\sigma & =2.6584 \end{alignedat}$

Smiley_ · « **Reply #1832 on:** April 21, 2013, 11:16:42 am »

Find the coordinates of the points of intersection for the following:

x2 + y2 = 178
x + y = 16

i keep getting the answer wrong

Jeggz · « **Reply #1833 on:** April 21, 2013, 11:25:44 am »

Quote from: fishandchips on April 21, 2013, 11:16:42 am

Find the coordinates of the points of intersection for the following:

x2 + y2 = 178
x + y = 16

i keep getting the answer wrong

If we transpose the second equation we get $y=16-x$
We can then substitute this into the 1st equation as such : $x^2 + (16-x)^2 = 178$

$x^2 + (256 - 32x + x^2) = 178$
$2x^2 -32x + 256 = 178$
$2x^2 - 32x + 78=0$
$2(x^2 - 16x+39) = 0$
$(x-3)(x-13) = 0$
Hence the intersection points are when $x=13$ and $x=3$ . But since it asks for co-ordinates you would have to substitute the x-value in one of the equations to the find the corresponding y-value.

$13+y=16$ , hence $y=3$
$3+y=16$ , hence $y=13$ .

The co-ordinates of intersection would be $(13,3)$ and $(3,13)$

Henreezy · « **Reply #1834 on:** April 22, 2013, 07:54:58 pm »

Hey! can somebody help me with this simultaneous question?

-2x + ky = k+2
(k+1)x-6y=-10
where k is a real constant, find the values for which k gives a
>unique solution
>infinitely many solns
>no solution

Alwin · « **Reply #1835 on:** April 22, 2013, 07:56:16 pm »

Quote from: Henreezy on April 22, 2013, 07:54:58 pm

Hey! can somebody help me with this simultaneous question?

-2x + ky = k+2
(k+1)x-6y=-10
where k is a real constant, find the values for which k gives a
>unique solution
>infinitely many solns
>no solution

Hint: think of matrices and determinants

Henreezy · « **Reply #1836 on:** April 22, 2013, 08:04:29 pm »

Quote from: Alwin on April 22, 2013, 07:56:16 pm

Hint: think of matrices and determinants

Hm, I've set up my matrices but I'm really unclear on how to solve using matrices (it's not heavily emphasised in our textbook)

Alwin · « **Reply #1837 on:** April 22, 2013, 08:10:09 pm »

Quote from: Henreezy on April 22, 2013, 08:04:29 pm

Hm, I've set up my matrices but I'm really unclear on how to solve using matrices (it's not heavily emphasised in our textbook)

Okay, so you have the 2x2 matrix
$\begin{bmatrix} -2 & k \\ (k+1) & -6\end{bmatrix} $

You find the determinant and put it equal to 0. then sub in the values of k you get to see if no solution or infinitely many.

b^3 · « **Reply #1838 on:** April 22, 2013, 08:12:50 pm »

There are two approaches to this question, the rearranging method and the matrix method.

Firstly the rearranging method.

For the system of equations to have a unique solution, the two lines must not be parallel, that is their gradients have to be different.
So we can rearrange the equations into $y=mx+c$ form.
$y=\frac{2}{k}x+\frac{k+2}{k}$
$y=\frac{k+1}{6}x+\frac{10}{6}$

So to have different gradients
$\begin{alignedat}{1}\frac{2}{k} & \neq\frac{k+1}{6} \\ 12 & \neq k^{2}+k \\ k^{2}+k-12 & \neq0 \\ \left(k+4\right)\left(k-3\right) & \neq0 \\ \therefore k & \neq-4,\:3 \\ k & \in R\backslash\left\{ -4,3\right\} \end{alignedat}$

For there to be infinitely many solutions, we need to have both lines being the same line, that is same gradient and same y intercept.
We've already done the work for the same gradient, it's the same as before, except using an equals sign. So that is for the gradients to be the same we have $k=-4,\:3$
Now for the same y intercept we
$\begin{alignedat}{1}\frac{k+2}{k} & =\frac{10}{6} \\ 6k+12 & =10k \\ 12 & =4k \\ \therefore k & =3 \end{alignedat}$
Now we need to satisfy both of these conditions at the same time, that is we have to take the intersection of the ${-4,\:3}$ and ${3}$ which is $3$ .

For there to be no solutions, we need the two lines to be parallel but not the same line, that is we need to have the same gradients, but different y intercepts. We've already done the working for both of these, for the same gradients we have $k=-4,\:3$ . For different y intercepts, we have the same method as solving above, except with a not equals to sign, which gives $k \neq 3$ . Again we take the intersection of these two sets, and we arrive at $k=-4$

Now lets look at the second method, using matricies.
For there to be a solution to the $AX=b$ (where $A$ is the matrix of the coefficients, $X$ is a column matrix of $x$ and $y$ and $b$ is a column matrix of the RHS of the equation), we need A to be invertible, so that we would arrive at $X=A^{-1}b$ . For A to be invertible, the determinate of A must be non zero. So we can find A and let it not equal zero.
$\begin{alignedat}{1}A & =\begin{bmatrix}-2 & k \\ k+1 & -6 \end{bmatrix} \\ \det\left(A\right) & =\left(-2\times-6\right)-k\left(k+1\right) \\ & =-k^{2}-k+12 \\ det\left(A\right) & \neq0 \\ -k^{2}-k+12 & \neq0 \\ k^{2}+k-12 & \neq0 \\ \left(k+4\right)\left(k-3\right) & \neq0 \\ \therefore k & \neq-4,\:3 \end{alignedat}$

Now if there are infinite solutions or no solutions, then the determinate of A must be zero, as that will mean we won't have a unique solution to the system AX=b.
So again, we've done the working above, just with the wrong sign, so when $\det (A)=0$ , we will have $k=-4,\:3$ . To distinguish between the two cases, we substitute the value of $k$ back into the equation and see if we get the same line, or two different lines.
For $k=-4$
$\begin{alignedat}{1}-2x-4y & =-4+2 \\ x+2y & =1....[1] \\ \left(-2+1\right)x-6y & =-10 \\ -3x-6y & =-10....[2] \end{alignedat}$
That is we have two different lines, so for $-4$ we have no solutions.

For $k=3$
$\begin{alignedat}{1}-2x+3y & =5....[1] \\ 4x-6y & =-10 \\ -2x+y & =5....[2] \end{alignedat}$
So we do have the same line, hence for $k=3$ we have infinite solutions.

Anyways, hope that helps

Henreezy · « **Reply #1839 on:** April 22, 2013, 08:17:16 pm »

Thanks b^3 + alwin :^)

I just wanted to get in some more matrix practice because even if it didn't pop up on the SAC it could show up on the exam and I'd be screwed. So thank you a lot for that :^D

Alwin · « **Reply #1840 on:** April 22, 2013, 08:24:39 pm »

Quote from: b^3 on April 22, 2013, 08:12:50 pm

There are two approaches to this question, the rearranging method and the matrix method.

Firstly the rearranging method.

$...$

Anyways, hope that helps

Okay, I'm still a newbie, got my account 5 days ago. But I already have a goal... BEAT b^3 in a post haha

Sorry b^3.. but you're just so bloody fast and detailed!

Wish me luck guys, needa try somehow out-ninja him..

shadows · « **Reply #1841 on:** April 22, 2013, 08:31:22 pm »

Doing some revision and I just have a few questions

Which values of m does the equation mx^2 -2mx + 3 = 0 have one solution for x

Determinant = 4m^2 -12m = 0

That gives 2 answers... 3. and 0...

But it says one solution o.o? Answer says 3. How do I disregard 0?

Show that (k+1)x^2 -2x-k =0 has a solution for all values of k?

Show that ax^2 -(a+b)x +b=0 has a solution for all values of a and b?

Alwin · « **Reply #1842 on:** April 22, 2013, 08:40:35 pm »

Quote from: shadows on April 22, 2013, 08:31:22 pm

Doing some revision and I just have a few questions

Which values of m does the equation mx^2 -2mx + 3 = 0 have one solution for x

Determinant = 4m^2 -12m = 0

That gives 2 answers... 3. and 0...

But it says one solution o.o? Answer says 3. How do I disregard 0?

Show that (k+1)x^2 -2x-k =0 has a solution for all values of k?

Show that ax^2 -(a+b)x +b=0 has a solution for all values of a and b?

If you sub $m=0$ in, the equation becomes $3=0$ with is clearly inconsistent (wrong). Hence, you disregard m=0

$(k+1)x^2 -2x-k =0$
$discriminant=(-2k)^2 - 4 (k+1)(-k) = 4k^2 + 4k +4$
since $4k^2 + 4k +4$ clearly $>0$ , there is always a solution no matter the value of k

$ax^2 -(a+b)x +b=0$
$discriminant=(-(a+b))^2 - 4 (a)(b) = a^2+2ab+b^2 - 4ab = (a-b)^2$
since $(a-b)^2$ is always $>=0$ then there is always a solution

Hope that helps!

EDIT: Silly typo, that's what I get for rushing things. Thanks for the pick up guys !

shadows · « **Reply #1843 on:** April 22, 2013, 08:45:23 pm »

You guys are so quick in replying!

Thank you so much!

Phy124 · « **Reply #1844 on:** April 22, 2013, 09:13:28 pm »

Quote from: Alwin on April 22, 2013, 08:40:35 pm

If you sub $m=0$ in, the equation becomes $3=0$ with is clearly inconsistent (wrong). Hence, you disregard m=0

$(k+1)x^2 -2x-k =0$
$determinant =(-2k)^2 - 4 (k+1)(-k) = 4k^2 + 4k +4$
since $4k^2 + 4k +4$ clearly $>0$ , there is always a solution no matter the value of k

$ax^2 -(a+b)x +b=0$
$determinant =(-(a+b))^2 - 4 (a)(b) = a^2+2ab+b^2 - 4ab = (a-b)^2$
since $(a-b)^2$ is always $>=0$ then there is always a solution

Hope that helps!

I think you're a little confused with your terms;

$\text{Determinant} = ad - bc$ (relates to square matrices)

$\text{Discriminant} = \Delta = b^2 - 4ac$ (relates to the roots of a polynomial)

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darklight

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