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Re: Nairaland Mathematics Clinic by Mrshape: 1:26am On May 01, 2020 |
Martinez39s:Lol. Mathematician don't give up hahaha |
Re: Nairaland Mathematics Clinic by naturalwaves: 1:33am On May 01, 2020 |
Martinez39s: From the probability gotten, it is possible to then estimate the number but the problem jackpot has with the solution is that she expects a figure (100-n) and that is not possible given the circumstances of the question. |
Re: Nairaland Mathematics Clinic by naturalwaves: 1:41am On May 01, 2020 |
jackpot: Hahahaha. Not exactly. I remember how long it took me to find a birthday mate and I had interacted with at least 150 people asking for their birthdays before I saw one. It can be 0 too. Another angle you should look at it from is that the 100 referred to in the question may be a sub group. These sort of questions require a large sample size. The 100 may be a broken-down version from 100,000 because for every 100,000,you will get 3 with an unshared birthday. |
Re: Nairaland Mathematics Clinic by Adedap25(m): 5:31am On May 01, 2020 |
Pls lets create a mathematics clinic group chat...... Like online class Tanx very much |
Re: Nairaland Mathematics Clinic by jackpot(f): 8:53am On May 01, 2020 |
Adedap25:my friend, go and pay for the online class fee your school billed you Your teachers must chop whether COVID-19 or not |
Re: Nairaland Mathematics Clinic by jackpot(f): 9:13am On May 01, 2020 |
naturalwaves:You're seeing it from your perspective (your own birthday matching with others). From your perspective, you'll have to do 150 comparisons of birthdays in your case of trying to see your birthday mate. The chances of seeing other matching birthday mates increase greatly if you start comparing among all the 150 persons (plus you=151), for then you'll have to do 151Combination2=11325 comparisons naturalwaves:I think that with a sound knowledge of statistics especially all these joint probability distribution/density function (PDF), the solution can be readily obtained. It might just be a case of solving one or two multiple integrals as solved in joint PDFs |
Re: Nairaland Mathematics Clinic by naturalwaves: 2:05pm On May 01, 2020 |
jackpot: I'm interested in the source of this question o Jackpot.....and I know that whatever is finally solved will definitely not give you the (100-n) you want . It will tend towards 0.00003%. Please,keep me updated. |
Re: Nairaland Mathematics Clinic by love2017(m): 4:06pm On May 01, 2020 |
Help me to derive a single explicit formula for solving quartic equation. |
Re: Nairaland Mathematics Clinic by duchaB(m): 3:02pm On May 02, 2020 |
DeeCherry: Hey DeeCherry.. Howdy? |
Re: Nairaland Mathematics Clinic by Strech(m): 3:24pm On May 02, 2020 |
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Re: Nairaland Mathematics Clinic by DeeCherry(f): 3:30pm On May 02, 2020 |
duchaB: I'm good |
Re: Nairaland Mathematics Clinic by duchaB(m): 5:12pm On May 02, 2020 |
DeeCherry: Sorry, I sent you a PM.. Don't know if u have seen it. I have St important to talk to you about if u don't mind. Thanks. |
Re: Nairaland Mathematics Clinic by DeeCherry(f): 7:34pm On May 02, 2020 |
duchaB:Didn't get any |
Re: Nairaland Mathematics Clinic by jackpot(f): 12:14am On May 04, 2020 |
naturalwaves:source: myself really. Question was motivated by the birthday paradox. Any update on the little experiment concerning analyzing the last 100 players? |
Re: Nairaland Mathematics Clinic by naturalwaves: 1:14am On May 04, 2020 |
jackpot:Hahahaha. Chai. I will check out the PDF tomorrow in my free time. I will also try some other experiments. |
Re: Nairaland Mathematics Clinic by jackpot(f): 2:13pm On May 07, 2020 |
Hello house kindly help to solve this. CC: martinez39s, dejt4u, Richiez, mathefaro, naturalwaves, Mrshape
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Re: Nairaland Mathematics Clinic by Mrshape: 2:22pm On May 07, 2020 |
jackpot:I think game theory will apply here |
Re: Nairaland Mathematics Clinic by HajiTooMuch: 7:34am On May 11, 2020 |
Good day everyone. I need someone who understands Stochastic Programming (farming problem) well. I have a problem with modelling a problem statement and I need insight. I've modeled it actually just need someone to help me check where I'm wrong. Kindly help me if you know it or know someone who knows it. I'm ready to pay if need be |
Re: Nairaland Mathematics Clinic by phemolisti(m): 7:32pm On May 13, 2020 |
Please help me to evaluate these questions using logarithms table only question 23 and 25
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Re: Nairaland Mathematics Clinic by ndyfaby: 1:24am On May 14, 2020 |
ONLINE COURSES 2020 FOR INTERNATIONALS STUDENTS interested students can apply |
Re: Nairaland Mathematics Clinic by jackpot(f): 7:43pm On May 17, 2020 |
FUN QUESTION There are two packs filled with identical cards as follows: Pack 1 contains 4 red cards and 3 yellow cards; Pack 2 contains 3 red cards and 4 yellow cards. An experiment consists of drawing/picking a card at random from Pack 1, then mixing it up with the cards in Pack 2, shuffling the cards and then picking a card at random from Pack 2. Find the probability of picking (1.) a red card (2.) a yellow card at the end of the experiment. Options: 1 (A) 1/2 (B) 5/8 (C) 3/7 (D) 31/56 (E) NOTA 2 (A) 1/2 (B) 3/8 (C) 25/56 (D) 4/7 (E) NOTA tags: Mrshape naturalwaves dejt4u Richiez mathefaro martinez39s dejt4u |
Re: Nairaland Mathematics Clinic by Mechanics96(m): 1:23pm On May 18, 2020 |
phemolisti: I"ll leave you with the no. 25, you should get it now.
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Re: Nairaland Mathematics Clinic by naturalwaves: 3:56pm On May 18, 2020 |
jackpot:I've been so busy teaching online. Will attempt this in my free time. |
Re: Nairaland Mathematics Clinic by naturalwaves: 4:43pm On May 18, 2020 |
jackpot:My solution Data pack 1...4 RED and 3 YELLOW cards pack 2.....3 RED and 4 yellow cards 1) Probability of picking a red card at the end of the experiment implies R,R or Y,R = [4/7 x 4/8] + [3/7 x 3/8] = 2/7 + 9/56 = 25/56 2) Probability of picking a yellow card at the end of the experiment implies Y,Y or R,Y = [3/7 x 5/8] + [4/7 x 4/8] =15/56 + 2/7 =31/56 2 Likes |
Re: Nairaland Mathematics Clinic by Martinez39s(m): 5:15pm On May 18, 2020 |
jackpot:SOLUTION A pair of choices must be made to produce an outcome in this experiment. (A) If we pick the 1st red ball in the 1st bag, there are 4 possible outcomes of a red ball, and 4 possible outcomes of a yellow ball. This is also true if we pick the 2nd, 3rd, or 4th red ball in the first bag. Hence, (B) If we pick the 1st yellow ball in the 1st bag, there are 3 possible outcomes of a red ball, and there are 5 possible outcomes of a yellow ball. This is also true if we start with the 2nd or 3rd yellow ball in the first bag. Hence, (C) From (A) & (B), --- it is clear that the number of the pairs of choices leading to an outcome a red ball is ORR + OYR = 16 + 9 = 25. --- It is also clear that the number of the pairs of choices leading to an outcome of a yellow ball is ORY + OYY = 16 + 15 = 31. --- it then follows that the number of the pairs of choices resulting into an outcome of the experiment is ORR + OYR + ORR + OYR = 56 The probability of picking a red ball is 25/56. The probability of picking a yellow ball is 31/56. 1 Like |
Re: Nairaland Mathematics Clinic by mathefaro(m): 9:13pm On May 18, 2020 |
The ancestors have spoken and they have spoken well. I've been overwhelmed by online classes these days. Never thought online classes could be as demanding as this. Well-done guys. May the mathematical force continue to be with us all |
Re: Nairaland Mathematics Clinic by naturalwaves: 9:56pm On May 18, 2020 |
mathefaro: 1 Like |
Re: Nairaland Mathematics Clinic by Aybalance: 11:22am On May 19, 2020 |
Pls I need help with this Cost price of 12 Oranges is equal to the selling price of 9 Oranges and the discount on 10 oranges is equal to the profit on 5 oranges.What is the percentage point difference between the profit percentage and discount percentage? A. 20 B. 22.22 C.16.66 D. 15 |
Re: Nairaland Mathematics Clinic by naturalwaves: 4:33pm On May 19, 2020 |
Aybalance: SOLUTION Let the cost price of 1 orange be = x Let the selling price of 1 orange = y From 1st statement, 12x = 9y x/y = 9/12 x:y = 9:12 x:y = 3:4 This implies that the ratio of cost price to selling price is 3:4 Thus, ratio of selling price to cost price (y:x) is 4:3 This will require a bit of thinking hereon. For every 3 things you buy, you sell at a value of 4. Profit is (4-3= 1).This means that your profit is actually 1/3 of your c.p Thus, % profit = 1/3 x 100 = 33.333% PART 2- Discount on 10 oranges Let the marked price be = z marked price of 10 oranges = 10z s.p of 10 oranges = 10y (Recall that s.p =y) Thus, Discount = 10z-10y PART 3- Profit on 5 oranges Profit on 5 oranges will be s.p of 5 oranges - c.p of 5 oranges. Thus, profit on 5 oranges = 5y-5x FROM QUESTION, Discount on 10 oranges = profit on 5 oranges Thus, 10z-10y = 5y- 5x Recall that, x= (9/12)y from equation 1 10z-10y=5y-5(9/12)y 10z= 5y-(45y/12) + 10y 120z= 135y 8z=9y y:z = 9:8 y-z = 9-8 = 1 This implies that discount value is 1/9 of the marked price. Thus, discount % = (1/9 x 100) = 11.111% Finally, % point difference between the profit % and discount % = 33.333% - 11.111% = 22.222%. Thus, the correct answer is option B 1 Like |
Re: Nairaland Mathematics Clinic by Aybalance: 8:37am On May 20, 2020 |
the correct answer is option B [/b] [/quote]Thanks natural waves.How come it is an objective question.Am asking because of how long it will take to do all these |
Re: Nairaland Mathematics Clinic by naturalwaves: 9:27am On May 20, 2020 |
Aybalance:You are welcome. Where did you get the question from? It depends I know what they do in some places is to memorise the process and then start solving any question of this form in a very short time. Do you know that because you know the method now, you can solve any question on this under 1 minute if it follows the same pettern? Just look for your sp:cp ratio and get your % do same for discount:sp and get your % and you are good to go. If one is just seeing such for the first time, it will take a while o. Do you know that I spent at least 15 minutes on this question trying to look for the best approach that will work? |
Re: Nairaland Mathematics Clinic by Aybalance: 9:53am On May 20, 2020 |
naturalwaves:It is a GMAT test past question.15 minutes....Lol that's almost the time for the whole test....seems the only green flag is to have crossed the question before. I also noticed in your solution that y:z=9:8 is z:y=9:8 and z-y=1 as there is a discount |
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