How do I solve this math question?
First let's try to develop a general theory for these kinds of questions. So lets consider a general equation of Conic section which is [math]ax^2+2hxy+by^2+2fx+2gy+c = 0[/math]Rearrange this equation as [math]ax^2+2x(hy+f)+(by^2+2gy+c) = 0 [/math]Solving for the [math] x[/math] we have [math]x= \dfrac{-2(hy+f)\pm2\sqrt{(hy+f)^2-a(by^2+2gy+c)}}{2a} [/math]Now if we want that given equation should have linear factors then this also means that above expression should have linear relationship between [math]x[/math] and [math]y[/math]. Now for this to happen the term under square-root should be a perfect square.[math]\implies (hy+f)^2-a(by^2+2gy+c)[/math] should be perfect square Simplifying this equation and we will get [math](h^2-ab)y^2+2y(gh-af)+(g^2-ac) = 0 [/math]Now this equation will be a perfect square if it has only one root i.e it's discriminant is zero.[math]\implies 4(gh-af)^2 - 4(g^2-ac)(h^2-ab) = 0[/math] Simplify this and we will get [math]\boxed{af^2+ch^2+bg^2-2fgh-abc = 0}[/math] Hence this condition should be true if you want to factorize any such equation into two linear factor. More neat and aesthetic form to express this condition is [math] \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 [/math]. This is also called discriminant of Conic section . (It decides whether the given equation represent an ellipse or parabola or circle or point or hyperbola or pair of straight line. Here it represents the pair of straight line).Now come to this particular equation. (I am changing [math]a[/math] to [math]P[/math] and [math]b[/math] to [math]Q[/math] to avoid confusion).So given equation is [math]3x^2 + 2Qxy + 2y^2 + 2Px - 4y + 1[/math]. If you compare this to standard form you will find that ,[math]a= 3 , h = Q , g = P , b = 2 , f = -2[/math] and [math]c = 1[/math] So discriminant of this equation is [math] \begin{vmatrix} 3 & P & Q \\ Q & 2 & -2 \\ P & -2 & 1 \end{vmatrix} = 0[/math]Solving and rearranging this we have ,[math]Q^2+4PQ + 2P^2 + 6 = 0 [/math]Clearly [math]q[/math] is a root of quadratic equation given by [math]x^2+4Px+2P^2+6[/math]
Statistic question urgent help!?
a is True. Think of it this way, picking any number is an independent event. Picking the number is also an independent event, and the probability of choosing any number is 1/10. You get this probability from the fact you are picking 1 number, and there are 10 to choose from (0-9). So what do you do when you want to find the combined probability of independent events? You multiply them. So probability of any pair of numbers, be it 33, 44, 02 or whatever, the probability, p=1/10*1/10 =(1/10)^2 =1/10^2 =1/100 b is false. Since the probability of any number being 4 is 1/10, that does not mean that for every 10 digits, that there is definitely a 4. In fact, there may never be a 4 in an entire random digit table. But that is unlikely. The probability of there not being a 4 for 40 digits is (9/10)^40=.0148. c is false. In fact we can calculate the probability happening, which is (1/10)^5, which is small, 10^-5. Essentially nothing is impossible, just varying levels of likely. The law of averages shows you how that if you took trials of 50 numbers and checked how many times 4 shows up, if you took a million groups of 50 numbers, the average of the number of times 4 showed up will be about 5. But that's different. The distinction is important.
Which of the following statements are true of a table of random digits, and which are false?
1. False - if digits are truely random there can be anything from 0 - 40 zeros. The most likely scenario is 4 but any number of zeros can occur. In any set of 40 digits some digits are likely to appear more frequently than others. 2. True. The probability of any digit being a zero = 1/10. As they are random numbers the result of the first number being selected does not affect the probability of the second number also being a zero. therefore 1/10 x 1/10 = 1/100 3. False - With random numbers any combination is possible. In theory you could get 40 zeros in a row. Very very very unlikely, but nevertheless theoretically possible.
A number consists of two digits. The sum of the digits is 9. If 63 is subtracted from the number, its digits are interchanged. What is the number?
In order to solve this let’s follow your description carefully:(1) “A number consists of two digits” : Let that number be X such that X is informally represented as xy which formally can be represented as 10*x + 1*y. From this, it follows that X = 10x + y.(2) “The sum of the digits is 9” : This statement translates to x + y = 9(3) “If 63 is subtracted from the number, its digits are interchanged.” : This statement translates to X - 63 = yx (informally). But we know from (1) that X = 10x + y and that yx (informal) translates to 10y + x. Hence, this statement becomes 10x + y - 63 = 10y + x which can be simplified further to 9x - 9y = 63.Using the above equations we can build a simple 2 x 2 system {x + y = 9 , 9x - 9y = 63}. Solving this using Cramer’s method, we get that x = 8 and y = 1.So to answer your question, since X = xy informally, then it must be that X = 81.
How many combination of 6 numbers can you make from 49 numbers if they are 1 to 49?
I assume you’re planning to win the lottery. I’ll answer your question in exchage for 1% out of the jackpot.:))The 6 winning numbers can come in any order, therefore…If you pick 1 number, and the Lottery draws only 1 winning number, the odds of you guessing that number are… well, 1:49. But the Lottery draws 6 numbers, which increases your odds 6 times. Now your odds are 6:49.If you pick 2 numbers, the odds of guessing both are 6:49 (as explained above) times 5:48. “Why 5:48?” — because you guessed the 1st number (6:49 odds), and your 2nd must match any of the other 5 winning numbers which are drawn from the remaining 48. Thus it’s 5:48.Following the same logic, if you pick 3 numbers, the odds for the 3rd are 4:47.4th number = 3:465th number = 2:456th number = 1:44So, the odds of guessing all 6 numbers are:6:49 X 5:48 X 4:47 X 3:46 X 2:45 X 1:44= 1 in 13,983,816Write me when you have the money.:))
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100 people standing in a circle in an order 1 to 100. No. 1 has a sword. He kills the next person (i.e. No. 2) and gives the sword to the next (i.e. No. 3). All people do the same until only 1 survives. Which number survives at the last? I want C++ program for this.
The straight forward way of solving this could be manually writing down 1 to 100 and then striking out the numbers as per the problem.In such a scenario it would go like this;Round 1: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99Round 2: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97Round 3: 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97Round 4: 9, 25, 41, 57, 73, 89Round 5: 9, 41, 73Round 6: 9, 73Round 7: 73so answer is 73what if their were 1000 soldiers??would it be possible to solve it by this method??Yes but it would consume a lot of time.Here’s the correct way of solving this riddle.This is actually a classic Josephus Problem.The original Josephus Problem goes like this;“The problem is named after Flavius Josephus, a Jewish historian living in the 1st century. According to Josephus’ account of the siege of Yodfat, he and his 40 soldiers were trapped in a cave, the exit of which was blocked by Romans. They chose suicide over capture and decided that they would form a circle and start killing themselves using a step of three. Josephus states that by luck or possibly by the hand of God, he and another man remained the last and gave up to the Romans.”Solution to the Riddle using Josephus Problem approach:The solution requires getting the nearest smaller number that is the power of 2, in this case 64 and subtract it with the given number.100-64=36.Now we apply the formula;2n+1 = 2*36 + 1 = 72 + 1 = 73.Hence answer = 73
How do you calculate 1+2+3+4+5+6+7+8+9+10 quickly?
The series “1+2+3+4+5+6+7+8+9+10” can be considered as the sum of first 10 Natural numbers.The generalized formula to find the sum of first N natural numbers is(N * (N + 1) ) /2For the series given, the value of N is 10So substituting value of N in the generalized formula( 10 * (10 + 1) ) / 2=> (10 * 11 ) / 2=> 110/2=> 55Hence the answer is 55Note:The generalized formula is only to find the sum of first N consecutive numbers. It can be also be solved using summation formula of AP(Arithmetic Progression).For any series starting from a value and having the following numbers repeating after a common difference, the summation formula of AP (Arithmetic Progression) can be applied.The summation formula of AP if the last term of the series is knownS= (n/2)* (a1 + an)where n=number of terms in the seriesa1=first terman=last termIn this case the all the 3 variables are known and they a common difference(say d) of 1 so formula of AP can be applied,n=10a1=1an=10Substituting it in the AP summation formulaS= (10/2) * (1+10)S= 5 * 11S=55Both the formula will give you the required result.If the last term is not known but the number of terms and the common difference (d) is known then the alternate formula to find the sum of an AP series can be used, which isS=(n/2)*[(2*a1) + (n - 1)* d]where n=number of terms in the seriesa1=first terman=last termd=common difference
How many numbers between 100 and 200 (including both 100 and 200), are not divisible by any of these numbers: 2, 3 and 5?
There are 101 terms from 100 to 200 including them.So if we need the numbers not divisible by 2,3,5 in them we get a series…..which is like…For 2,100,102,104,106…….200For 3,102,105,108……198For 5,100,105,110,115……200We can write the terms for 5 as105,115,125 ……195 cause the terms ending in zeroes are already contained in set of 2.Now there is 51 terms for set of 2 and 10 for 5 so if reduce them from total we get 101–51–10=40So now we need to remove even terms and terms ending in 5 from set of 3. Clearly first three terms are to be excluded so if we check from 111…every alternate number will be even 111,114,117,120…..From 111…….198 there are n terms(for meanwhile)So last term which is 198 is198= 111 +(n-1)3 - simple A.P. formula87/3=n-129+1=nn=30 and if every alternate term had to be excluded which half …..then we will be left with 15 terms….which are….111,117,123,129,135……195But terms ending in 5 are still there so exclude them , we do something like juggad135 is first such no. Clearly to get next no. we subtract it by 135 and check if remainder is divisible by 3 if it is then that no. Is in this set and we got remove it.To get no. Ending in 5 we have add 10 or 20 or 30 but 10 and 20 are not divisible by 3 so we consider 30 only.135+30=165165+30=195So there are 3 no. To be excluded15–3=12Finally we earlier had 40 so40–12=28There are 28 numbers which are not divisible by any of 2,3,5 from 100 to 200!!!!