these are some logic problems that i just cant figure out?
967393
the second part,\n\nLet robinson's age be x\nbrother's age = x+2\nsister's age = x+6\nmother's age = x+20\n\nAverage age is 39\nTotal age is 39*4=156\n\nx + (x+2) + (x+6) + (x+20) = 156\n4x + 28 = 156\n4x = 128\nx = 32\n\nHe is 32 years old
522868
Take it to someone who knows what they are doing. \n\nWe've answered this question about 8 times for you. \n\nBring it to me, and I'll fix it for you - only if you quit posting the same question.
885691
Part (a) of problem 1 is pretty easy.  There are 10 marbles total, and 6 of them are red.  On your first draw, you have a 6/10, or 0.6 (60%) probability of drawing a red sock.  In other words, there are 6 ways of drawing a red sock out of a total of 10 possibilities.\n\nFor part (b), you need to know that the probability of two independent events happening is given by the product of the probabilities of each of the individual events.  In this case, the two independent events are that the first marble drawn is red AND the that the second marble is also red.  We already calculated the probability for drawing a red marble on the first draw in part (a).  After the first draw, the box contains 9 total marbles, 5 of which are red, so the probability that the second marble is red is 5/9.  The probability that both are red is given by the product, 6/10 * 5/9 = 30/90 = 0.3 (30%).\n\nThe reasoning for part (b) is the same as for part (a).  The probability that the first marble drawn is white is 4/10, and the probability that the second marble drawn is also white is 3/9.  The probability that both are white is given by the product, 4/10 * 3/9 = 12/90 = 0.133... (13.333...%).\n\nFor part c, there are two ways one can get a red and a white marble: either the first marble is red and the second white, OR the first marble is white and the second red.  When there is more than one way to achieve the desired result, the probability of getting that result is given by the SUM of the probabilities of each of the different ways to achieve that result.  In this case, the probability of drawing red-white is 6/10 * 4/9 = 24/90, while the probability of drawing white-red is 4/10 * 6/90 = 24/90.  The total probability of getting a red and a white marble is then 24/90 + 24/90 = 48/90 = 0.53333....\n\nUsing the same reasoning, you should now be able to solve parts (a) and (b) of the second problem.  You should find for part (a) that the probability is equal to 0.42424242...., and that the probability for part (b) is 0.09090909....\n\nFor part (c), you would use a similar logic as used for part (c) of the first problem.  The probability of getting a matched pair (either both white or both blue) is given by the sum of the probabilities found in parts (a) and (b).  You should find that the probability is 17/33 = 0.51515151....\n\nFor the third problem, you need to know that the expected "payoff" of a game is given by the sum of the products of the prize value times the probability of winning that prize.  In this case, the probabilities of winning the various prizes are:\n\n$100 : 1/500\n$50 : 3/500\n$25 : 5/500\n\nSo the expected payoff (i.e., the "mean" or "average" winnings for a single ticket) is given by:\n\nP = $100*(1/500) + $50*(3/500) + $25*(5/500)\nP = $100/500 + $150/500 + $125/500\nP = $375/500\nP = $0.75
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
968857
just work backwards to find out the logic in it. That way you'll get why it works that way it did
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
112971
There's a problem with equation. Go back and check it.\nAs it stands, no x would satisfy this.\nInfity is an abstraction, not a number.
79803
you will find everything you want here:\n\nhttp://en.wikipedia.org/wiki/Logic_gates
741948
First problem.\n\nf(x) = 10 x^(-2) - 7 x^(-6)\nF(x) = [10/(-1)] x^(-1) - [7/(-5)] x^(-5) + C\nF(x) = -10 / x + (7/5) / x^5 + C\nF(1) = 0 = -10 + (7/5) + C\nC = 43/5\nF(x) = -10 / x + (7/5) / x^5 + 43/5\n\nSecond problem.\n\nf(t) = 7 sec^2(t) - 4 t^2\nF(t) = 7 tan t - (4/3) t^3 + C\nF(0) = 0 = C\nF(t) = 7 tan t - (4/3) t^3\n\nThird problem.\n\nf"(x) = 8 x + 3\nf'(x) = 4 x^2 + 3 x + C\nf'(-2) = 6 = 4(-2)^2 + 3(-2) + C = 16 - 6 + C\nC = -4\nf'(x) = 4 x^2 + 3x - 4\nf(x) = (4/3) x^3 + (3/2) x^2 - 4x + K\nf(-2) = -5 = (4/3)(-8) + (3/2)(4) + 8 + K\nK = -25/3\nf(x) = (4/3) x^3 + (3/2) x^2 - 4x - 25/3\nf(2) = (4)(8)/3 + (3)(4)/2 - 4(2) - 25/3\nf(2) = 1/3
1056055
1) You should try making a flow-chart to see the issues here...\n\n2) Your comparison:  if %1==%1 (and also for %2) will ALWAYS BE TRUE and will never get to parts of your code!
356197
OK, here is what is going on and then the answer:\n\nFirst, assuming that we have the array a (which we assume we do according the problem), then this algorithm is going to look at each item in an array and will assign B a value. The value B holds will be true only if B was true to start and the current item in the array is less than the next.\n\nTherefore if the array A was defined as follows:\n\na[0] = 1\na[1] = 2\na[2] = 3\na[3] = 5\na[4] = 6\na[5] = 8\n\nThe algorithm checks if a[0] <= a[1], and since it is and B is true as well, then B is set to true;\nthen we check a[1] <= a[2], and this is true. Since B is true and this is true, B is set to true. \nthen we check a[2] <= a[3], and this is true. Since B is true and this is true, B is set to true.\nthen we check a[3] <= a[4], and this is true. Since B is true and this is true, B is set to true.\nthen we check a[4] <= a[5], and this is true. Since B is true and this is true, B is set to true.\n\nTherefore it can be determined that as long as the items in the array keep increasing or staying the same (or non-decreasing) then the algorithm will end with B true. So, Answer D is the correct answer.\n\n\nGood luck!
1154228
In this table, column 1 has the values of "A," column 2 has the values of "B," column 3 has the values of "A>B" (the conditional), and column 4 has the values of "~(A>B)."\n\n1 2 3 4\nT T T F\nT F F T\nF T T F\nF F T F\n(I had to build the table like this because it wouldn't keep everything spaced out correctly if I did it otherwise.  Sorry.)\n\nThe first two columns simply list all of the possibilities.  The third column lists the truth values for (A conditional B)--or "If A then B."  In this column, B must be true whenever A is true, so the only time this is false is the second row.\n\nThe last column simply takes the negation--or the reverse--of the third column.  Since the conditional is only false in the second row (true everywhere else), the negation will only be true in the second row (false everywhere else).\n\nDoes that help?
917052
16/24....it's really not that hard...you should be able to solve it on your own.\nps. you might want to simplify it down to 2/3.
700951
4x + 6 is not quite right\nwatch the order of operations and distributive properties\n\nrepost your answer and I'll tell you if you got it\n\n(you didn't think I'd just give up the answer, did you ?)
509467
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
475920
Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"