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Math question
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最佳解答:
1a) (x-5)^2+(8-x-5)^2 = 4 (x-5)^2+(3-x)^2 = 4 x^2-10x+25+ (9-6x+x^2) = 4 2x^2 -16x + 30 = 0 x^2 - 8x + 15 = 0 (x-5)(x-3) = 0 x = 5 or 3 2a) there are total 90 no. 2-digit number (from 10 - 99) and there are 9 from them can be divided by 11 (ie. 11,22,33,44,55,66,77,88,99) P = 9/90 =1/10 b)(i) Let x be the no. of red bands, 3x be the no. of green bands x + 3x + 100 = 156 4x = 56 x = 14 red 14 no., green = 14*3 = 42 no. (ii) P(red) = 14/156 = 7/78 (iii) P(red) = 10/81 ie. 10/81 = 14+y/ 156+y 1560+10y = 1134+81y 426 = 71y y = 6 c) P(open) + P(lock & open by 2 keys) = 1/3 + [2/3*(1/5+ 4/5*1/5)] = 1/3 + (2/3 * 9/25) = 1/3 + 6/25 = 25+18/75 = 43/75 2007-06-21 23:48:49 補充： c) should beP(open) P(lock & open by 2 keys)= 1/3 [2/3*(1/5 4/5*1/4)]= 1/3 (2/3 * 2/5)= 1/3 4/15= (5 4)/15= 3/5
其他解答:
1a) 唔知你有無打錯野,我跟你條式做 (x-5)^2+(8-x-5)^2 = 4 (x-5)^2+(-x+3)^2 = 4 x^2 - 10x + 25 + x^2 - 6x + 9 = 4 2x^2 - 16x + 30 = 0 x^2 - 8x + 15 = 0 (x - 5)(x - 3) = 0 x = 3 or x = 5 2a) 慢慢數都計到 2-digit number 有 10 至 99 其中 11, 22, 33 ... 99 divisible by 11 p(2-digit number chosen at random is divisible by 11) = 9/90 = 1/10 b) i) (156 - 100) (3/4) = 42 ii) number of red bands = 156 - 42 - 100 = 14 p(red) = 14/156 = 7/78 iii) let y be number of red bands added (14+y)/(156+y) = 10/81 81 (14+y) = 10 (156+y) y = 6 c) p(door unlocked) + p(門鎖左,第一次選錯 key,第二次選岩) + p(門鎖左,第一次就選岩 key) = 1/3 + (2/3)(4/5)(1/4) + (2/3)(1/5) = 9/15D1B39E804036C6BD
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1.a)(x-5)^2+(8-x-5)^2 = 4 我唔知點樣分解 2.關於Probability a)What is the probability that a 2-digit number chosen at random is divisible by 11? b)A box contains 156 rubber bands.There are 100 yellow rubber bands and the rest are green or red.The green ones are 3 times as many as the red one. i)Find the number of green... 顯示更多 1.a)(x-5)^2+(8-x-5)^2 = 4 我唔知點樣分解 2.關於Probability a)What is the probability that a 2-digit number chosen at random is divisible by 11? b)A box contains 156 rubber bands.There are 100 yellow rubber bands and the rest are green or red.The green ones are 3 times as many as the red one. i)Find the number of green bands ii)Find the probability of getting a red band is randomly chosen from from the box iii)y red bands are added tothe box.The probability of part b)becomes 10/81.Find y. 更新: c)the probability that a door is locked is 2/3.In a bunch of 5 keys,1 key can unlock the door.What is the probability that a man can open the door by choosing 2 keys randomly?最佳解答:
1a) (x-5)^2+(8-x-5)^2 = 4 (x-5)^2+(3-x)^2 = 4 x^2-10x+25+ (9-6x+x^2) = 4 2x^2 -16x + 30 = 0 x^2 - 8x + 15 = 0 (x-5)(x-3) = 0 x = 5 or 3 2a) there are total 90 no. 2-digit number (from 10 - 99) and there are 9 from them can be divided by 11 (ie. 11,22,33,44,55,66,77,88,99) P = 9/90 =1/10 b)(i) Let x be the no. of red bands, 3x be the no. of green bands x + 3x + 100 = 156 4x = 56 x = 14 red 14 no., green = 14*3 = 42 no. (ii) P(red) = 14/156 = 7/78 (iii) P(red) = 10/81 ie. 10/81 = 14+y/ 156+y 1560+10y = 1134+81y 426 = 71y y = 6 c) P(open) + P(lock & open by 2 keys) = 1/3 + [2/3*(1/5+ 4/5*1/5)] = 1/3 + (2/3 * 9/25) = 1/3 + 6/25 = 25+18/75 = 43/75 2007-06-21 23:48:49 補充： c) should beP(open) P(lock & open by 2 keys)= 1/3 [2/3*(1/5 4/5*1/4)]= 1/3 (2/3 * 2/5)= 1/3 4/15= (5 4)/15= 3/5
其他解答:
1a) 唔知你有無打錯野,我跟你條式做 (x-5)^2+(8-x-5)^2 = 4 (x-5)^2+(-x+3)^2 = 4 x^2 - 10x + 25 + x^2 - 6x + 9 = 4 2x^2 - 16x + 30 = 0 x^2 - 8x + 15 = 0 (x - 5)(x - 3) = 0 x = 3 or x = 5 2a) 慢慢數都計到 2-digit number 有 10 至 99 其中 11, 22, 33 ... 99 divisible by 11 p(2-digit number chosen at random is divisible by 11) = 9/90 = 1/10 b) i) (156 - 100) (3/4) = 42 ii) number of red bands = 156 - 42 - 100 = 14 p(red) = 14/156 = 7/78 iii) let y be number of red bands added (14+y)/(156+y) = 10/81 81 (14+y) = 10 (156+y) y = 6 c) p(door unlocked) + p(門鎖左,第一次選錯 key,第二次選岩) + p(門鎖左,第一次就選岩 key) = 1/3 + (2/3)(4/5)(1/4) + (2/3)(1/5) = 9/15D1B39E804036C6BD
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