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What is the probability of getting 3 red balls from an urn in 3 picks without replacement if the urn contains 5 red 3 blue?
The probability the first ball will be red is 5/8. The probability that the first and second balls will be red is 5/8 x 4/7. The probability that the first, second, and third balls will be red is 5/8 x 4/7 x 3/6, or overall 60/336 (about 17.86%).
What is the probability of 4 cards being red if all the red faced cards and black faced cards numbered 2-9 are put into a bag and randomly drawn?
The probability of the first card being red is 16 in 32. The probability of the second card being red is 15 in 31. The third is 14 in 30. The fourth is 13 in 29. Multiply these probabilities together and you get 16 x 15 x 14 x 13 in 32 x 31 x 30 x 29, which is equal to 43680 in 863040 or about 0.0506.
Negative x to the second power positive five x plus twenty-four?
x to the second power +11x+24 (x+_)(x+_) Is there a question here?
How does one identify a Red X?
A red X usually means the object is disconnected or not turned on. It looks like a big red X over the device. Click on the device with the red X and read the menu that pops up letting you know the problem.
What is the solution of the given system of linear equations 4x plus 8y equals 8 x plus 3y equals 13?
4x + 8y = 8 or x + 2y = 2and x + 3y = 13Subtracting the first from the second gives y = 11.Substituting this value in the second,x + 33 = 13 so that x = -20.4x + 8y = 8 or x + 2y = 2and x + 3y = 13Subtracting the first from the second gives y = 11.Substituting this value in the second,x + 33 = 13 so that x = -20.4x + 8y = 8 or x + 2y = 2and x + 3y = 13Subtracting the first from the second gives y = 11.Substituting this value in the second,x + 33 = 13 so that x = -20.4x + 8y = 8 or x + 2y = 2and x + 3y = 13Subtracting the first from the second gives y = 11.Substituting this value in the second,x + 33 = 13 so that x = -20.
