Posted by White Group Mathematics on November 28, 2016 at 4:40 AM | comments (1) |
Hi Mr Koh it's XXX. I have a question for you:
Why did the teacher use cross product instead of dot product? Thank you.
The way the respective dot and cross product results arise are attributed to the orientation of the known direction vector. In the context you cited, the known direction vector, ie the normal to the plane is vertically placed, hence things are phrased as such.
On the other hand, if you consider the typical projection of a given vector onto a line, the known direction vector d is seen running horizontally instead of vertically. Therefore, the dot and cross product results are switched.
I have crafted two separate diagrams (attached below) for your reference, hopefully that will clarify matters.
Best Regards,
Mr Koh
Posted by White Group Mathematics on September 28, 2016 at 4:50 AM | comments (1) |
Hi Mr Koh,
I chanced upon your website online and I'd really like to engage you as my private tutor. I'm currently a JC2 student studying in SAJC. I know this is very last minute, but I really do need the help; in summary, I've tried really hard to work on my Math, however I really don't see any improvements in my grades and I feel like I need someone to tutor me over this last month.
I'm currently living at MacPherson, near Taiseng MRT station and I'm willing to meet outside my home if it is more convenient for you. I will understand if you are completely packed and cannot accommodate any more students. Apologies for this late message. Please get back to me soon-thank you and have a nice day!
Student X
Hi,
Thank you for your message. Firstly, if you work with me, bear in mind I cannot produce short term miracles and more than a fair bit of effort has to be invested on your part to secure a reasonably positive outcome at the A Levels. Secondly, while commuting to your residence isn't an issue, I am pretty swarmed right now, so please undersand it is only fair I accord priority to my existing students schedule-wise.
Thirdly, kindly note that a fee premium will be charged for extremely last minute engagements. If you still sincerely wish to set things in motion, give me a holler and we will take it from there. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on August 26, 2016 at 2:25 AM | comments (0) |
Hi Mr Koh,
I can't seem to get around this problem, could you please help?
Student X
Gradient of the line = -4/3 by rewriting its equation in the form y=mx+c. As such, it is as good as saying that for every 1 unit one moves in the x direction, it also has to move -4/3 units in the y direction. This general movement is hence represented by i - 4/3j.
Since OA is parallel to the line, then we say vector OA= k (i - 4/3j) , where k is a real valued constant. Given the magnitude of vector OA is 20 units, we can further surmise that k √ [ 1² + (-4/3)² ] = 20, solving this gives k=12 or -12. Thus vector OA= 12 (i - 4/3j) =12i -16j or -12 (i - 4/3j) = -12i+16j (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on February 13, 2016 at 9:05 PM | comments (0) |
I am having considerable difficulty grappling with this OCR problem. Could you kindly assist?
Student X
Worked it out for you, hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on February 2, 2016 at 6:00 AM | comments (0) |
I was given some basic calculus revision questions to do for an undergrad module, but since there's nothing new in them, so I decided to give these questions an A-level tag. I've been working on them for a while now, but I'm stuck at this. Any help would be appreciated; I'm not particularly bothered if you want to post the answers as it's not an assessed homework, and I'm starting to get tired of these.
Student X
Please find the integral worked out as follows. Hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on January 24, 2016 at 11:50 AM | comments (0) |
Sales of cooking oil bought in a shop during a week follow a Poisson distribution with mean 100. How many units should be kept in stock to be at least 99% certain that supply will be able to meet demand? Use normal approximation to Poisson distribution, if appropriate.
I have already approximated the distribution to Y ~ N (100, 100). Then, I have tried to use P(Y>y) = 0.99 . That didn't work so I'm trying to use P( Y = y) = 0.99. However, with this, I'm stuck with P [ (y-100.5)/10 < Z < (y - 99.5)/10] = 0.99
The correct answer is 123. Please help.
Student X
You kinda got your understanding of stuff mixed up. Firstly, recognize that it is the demand that varies, not the supply. It is the amount demanded that follows a Poisson distribution, which is subsequently approximated to that of a normal distribution.
The part about defining the original underlying distribution modelling demand as Y ~ N (100, 100) is correct.
If you let k be the number of units you wish to store as stock, then the required inequality satisfying the requirements of the question should be P(Y ≤ k) ≥ 0.99, which after adjusting for continuity correct would yield P(Y ≤ k+0.5) ≥ 0.99
As such, k+0.5 ≥ InvNorm of a normal distribution curve with mean and variance both =100 *
ie after standardization, solve (k+0.5-100)/ 10 ≥ InvNorm ( 0.99, 0, 1)
Hope this clarifies. Peace.
* The graphic calculator should help you compute the inequality in k here; unless you are using basic tables, then standardization is needed.
Best Regards,
Mr Koh
Posted by White Group Mathematics on November 4, 2015 at 10:10 AM | comments (1) |
Hi Mr Koh,
I am unsure how about how the solution for part (a) is formulated. Could you please help?
Student X
I have drafted a detailed set of explanations-please see as attached below. Hope it helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on August 4, 2015 at 9:40 PM | comments (0) |
Hi Mr Koh,
How do I do this question- In an Argand diagram, the points A, B, C and D represent the complex numbers a, -2+5i, c and 3/2 - 1/2 i respectively. Given that ABCD is a rectangle described in the clockwise sense with BC=2AB, find a and c.
I tried drawing B and D on the same diagram but they don't even have a right angle between them, so how can it form a rectangle? And I can't place points A and C on the diagram because they said ABCD is read clockwise but it is also mentioned BC=2AB....so I am lost. Haha. Thanks Mr Koh!
Student X
Based on the construct described, the rectangle's length is twice its breadth. I would recommend drawing a generic upright rectangle instead of trying to place the points specifically in Argand space. Do note that vectors are involved in the solving process. I have drafted a solution template for you:
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on July 19, 2015 at 10:40 PM | comments (0) |
Hi,
Here in my country, when you buy lottery tickets, you try to guess the six (6) combination of numbers out of 42 numbers. Unlike in Sweepstakes, the order of numbers does not really matter.
http://www.philippinepcsolotto.com/6-42-lotto-result-summary
Some say that the probability of winning the jackpot is simply 6/42 = 14.29%. That is high!
Since I was in high school, that formula seemed to make sense for me. After all, what is the probability of guessing one of the six combinations? 1/42. That is the probability of guessing each number in the combination of six. That is the result when you add the probability for each number.
(1/42) X 6 = 6/42
1/42 + 1/42 + 1/42 + 1/42 + 1/42 + 1/42 = 6/42
But I've come across another possibility today. Without repetitions of numbers, you can create 5,245,786 combinations of six digits from 1 to 42. What is the probability that you will guess the right combination out of more than 5 million combinations? 1 out of 5,245,786. That is 1.906 X 10 raised to negative 7.
I'm confused now because both solutions seem to make sense, but this second idea makes more sense. Each one is a computation based on a different point of view.
I hope you can enlighten me on this one.
Thanks,
Student X
The first manner of calculating the required probability which you cited is flawed. If say, each of these 6 combinations can be recycled, the probability would be computed as (1/42)^6 = 1/5489031744.
However, I am assuming they cannot be recycled, as such there will be 42 possible combinations for the first number, 41 for the second, 40 for the third, so on and so forth. That said, the probability would be computed as (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37)= 1/3776965920.
You may therefore question: why isn't this equivalent to 1/5245786, which is considerably larger in fact? The reason is (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37) actually examines the instance when sequencing of numbers matters, as opposed to 1/( 42 C 6) = 1/5245786 which doesn't pay any regard to the ordering of numbers. Clearly, the former would involve much greater stakes and thus even lower likelihood of winning.
Hope this clarifies. Peace.
Best Regards,
Mr Koh
Posted by White Group Mathematics on July 14, 2015 at 10:25 PM | comments (0) |
Hi Mr Koh,
I encountered this question. It says: if a = (a • b) b, where a and b are vectors, what are the possible angles between a and b? Also, what is the magnitude of vector b?
Student X
a = (a • b) b actually implies a is parallel to b. Note that a • b is in fact a scalar quantity, such that a=kb, where we can set a • b = k . In view of this, the only possible angles between a and b would be 0 and 180 degrees.
To discover |b|, we shall proceed to take modulus of both sides of the original equation a = (a • b) b ; this gives us |a| = |a • b| |b| = [ |a| |b| ] |b| = |a| |b|^2 . Cancelling |a| on both sides leads to |b|^2 =1, and thus |b|=1 (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh