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what three fractions are equivalent to 4/5

16/20 8/10 32/40
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Hello, Can I get this Answers?

8+3x=-7 3x=-15 x=-5
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How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.
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statistic probability

i. All the possible answers in Quiz 1 by chance can be represented by the terms in the expansion of (0.25+0.75)^8, where 0.25 is the probability of a correct choice of answer for each individual question and 0.75 is the probability of an incorrect choice of answer. The binomial expression has a value of 1, which is a probability of 1, i.e., certainty, made up of the sum of all the terms. Let p=0.25 and 1-p is 0.75. The expansion is p^8+... + 8Cr*p^r(1-p)^(8-r), where r indicates the rth term between r=0 and 8 and 8Cr is the combination function nCr for n=8. The value of 8Cr is given by (8*7*...*(8-r+1))/(1*2*...*r). The coefficients are, in order: 1, 8, 28, 56, 70, 56, 28, 8, 1, the 8th row of Pascal's triangle. The binomial terms start with the probability of 8 correct answers, then 7 correct 1 incorrect, 6 correct 2 incorrect, 5 correct 3 incorrect, and so on. The probability of answering incorrectly 3 question is (3/4)^3=27/64, so the probability of answering at least one question in 3 correctly by chance is 1-27/64=37/64, in other words, this is the probability that at most 3 questions are attempted before obtaining a correct answer. This is better than an evens chance (50%), so is more likely than unlikely. ii. In Quiz 2 p=1/5 or 0.20 and n=15. The probability of less than 4 correct answers is the sum of the last four terms of the binomial series for 15, which give all incorrect, 1, 2, 3 correct. The coefficients are respectively 1, 15, 105, 455. The p terms alone are 0.03518, 008796, 0.002199 and 0.0005498. When we multiply by the coefficients and add the terms together we get 0.6482 or 64.82% probability. iii. The probability of 6 or more correct answers in Quiz 1 is 1-(probability of up to 2 incorrect answers)=1-(0.75^8+8*0.75^7*0.25+28*0.75^6*0.25^2)=1-0.6785=0.3215 or 32.15%. This is less than 1/3, so the odds are 2:1 against Siti getting 6 or more correct answers by chance alone, which makes it reasonably unlikely. iv and v. The problem says "before the 8th question", so there need to be 5 correct answers within the first 7 questions. Using p=0.2, the probability of getting 5 correct answers out of 5 is 0.2^5=0.00032; the probability of getting 5 correct out of 6 is 6*0.2^5*0.8=0.001536; and the probability of 5 out of 7 is 21*0.2^5*0.8^2=0.0043008. Therefore the probability of getting 5 correct before the 8th question is the sum of these=0.0061568 or 0.62%, a most unlikely occurrence, less than 1 in 162.
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3(x+2) = 2x

hello..! 3(x+2)=2x 3x + 6 = 2x  ( combine the similar terms ) 3x - 2x = -6 x = -6
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times student will get 3 out of 10 questions right randomly guessing 4 choice questions

This is a binomial distribution, where p=1/4=0.25, the chance of getting the correct answer, and 1-p=0.75, the chance of getting the wrong answer. The distribution's last 4 terms are: (1-p)^10, 10(1-p)^9p, 45(1-p)^8p^2, 120(1-p)^7p ^3. These represent the chances of 0, 1, 2, 3 correct answers in 10. The coefficients are the number of ways of 0, 1, 2, 3 correct answers can be mixed with wrong answers. We need the last in this list which evaluates to 120*0.75^7*0.25^3=0.25 approx. and is the probability of getting exactly 3 right out of 10.
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What is 5/3 divide by 1/3?

If the question had been: what's 5 apples divided by one apple you would have said 5, wouldn't you? The question has thirds instead of apples, but the answer is still 5. Don't forget that to divide by a fraction you invert the fraction and multiply: 5/3 times 3 is 5, because 3 cancels out the denominator in 5/3. I'm sorry you didn't get your answer in 10 minutes as you wished, but I've only just seen it! There are thousands of questions coming in and questions cannot always be answered just as they come in, and you have to remember that the website is available all over the world, which is split into many time zones. Add to that the fact that the questions are answered by users, like me, who have a life outside answering and asking questions! There are meals to eat, shopping to be done, going to the office or factory, going to school, you know the sort of thing. And, of course, while your question is coming in, a user may be answering another question, perhaps a harder one. I'm sure everyone does their best to help people who have questions to answer. But, be assured that there are many users who will be only too pleased to help you. It's just a question of when. Here's a clue that may help you. Below your question you will usually find a list of similar questions, many of which will probably have been answered (shown in green with the number of answers), so if you look at them, you may well find out how to do your own question. Now that will save everyone's time!  
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In a mathematics emathematics, each correct answergets 2 marks and a wrong answer attract -1, ola got a total of 19 in aquiz with 20 questions. How many correct answers did he get?

If we call the number of right answers R and the number of wrong answers W, 2R-W=19. Also we know W+R=20. Add the two equations: 3R=39 so R=13, 13 correct answers. 13 correct answers gains 26 points; 7 incorrect answers loses 7 points and 26-7=19.
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The first die is 3 or the sum is 8

The answer posted above is correct. However the same answer can be arrived at in another way also. Tho get a total of 8 for 2 dies none of the dice must have a '1'. The probability that the first dice will not have a '1', or that it will have 2, 3, 4, 5, or 6, are: Probability of first dice not getting 1 = 5/6 This is because the probability of getting any number from 1 to 6 are equal. For each occurrence of the number 2 to 6 for dice the second dice must get a number given by: Required number if dice 2 = 8 - n Where: n = number of dice 1 As the probability of getting any one number is equal to 1/6, the probability that the second dice will have exactly the required number is: Probability of second dice getting the number (8 - n) = 1/6 The probability of getting sum of 8: Probability that sum is 8 = (Probability of first dice not getting 1)*(Probability of second dice getting the number (8 - n)) = (5/6)*(1/6) = 5/36 Answer: Probability of sum of two dice as 8 = 5/36
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Divide 2x4 - 5x3 - 8x2 - 17x - 4 by x + 4

I'm assuming the question reads: divide 2x^4-5x^3-8x^2-17x-4 by x+4, where the caret (^) means "to the power of". Long division in algebra is similar to long division in arithmetic. We lay out the dividend, which is the long expression, as we would lay out a number we're going to divide into. To the left of it we write the divisor, which is x+4. How many x's go into 2x^4? The answer is 2x^3 because 2x^3*x=2x^4. OK, the first term in the quotient (the answer) is 2x^3. We multiply x+4 by 2x^3, which gives us 2x^4+8x^3, and we write this under the dividend so that the powers of x line up. We should have 2x^4 under 2x^4 in the long expression and 8x^3 under 5x^3 in the long expression. We now have to work out 2x^4-5x^3-(2x^4+8x^3). Always beware of the + and - signs, it's so easy to make a mistake. The subtraction gives us -13x^3. We now look at this result and the next term in the long expression which is -8x^2. How many times does x go into -13x^3? The answer is -13x^2 times. That's the next term in the quotient following 2x^3. Multiply x+4 by -13x^2 and we get -13x^3-52x^2, subtract this from -13x^3-8x^2 and we get 44x^2. We continue the process bringing in successive terms from the dividend until we get to the end, when we'll perhaps be left with a remainder. The answer is the quotient=2x^3-13x^2+44x-193 remainder 768. Is that the answer you get?
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