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Probability
The probability of a certain event is a number expressing the likelihood that a specific event will occur, expressed as the ratio of the number of actual occurrences to the number of possible occurrences. In mathematics, it is a measure of how often an event will happen and is the basis of statistics.
14,643 Questions
14/42. You have to add the amount of marbles and then put the probabilityof answers you're looking for.
Ex:
14 white marbles
28 red marbles
14+28=42
?/42
your looking for the probability of white marbles, so you put in the amount of white marbles on the fraction.
=14/42
If you traveled 25 miles in 2 minutes, your average speed was 25 miles per 2 minutes or 750 miles per hour. If your rate of deceleration was constant, your initial speed was two times 750 miles per hour or 1500 mph.
I do not have enough information to determine your initial velocity because I don't know what direction you were going, and velocity is speed with direction.
Suggest a good math project based on triangles?
Mathematical designs and patterns using arithmetic progression.
What happens if you estamate 23975?
Tenths: 23980
Hundredths: 24000
Thousandths: 24000
Ten Thousandths: 20000
How do the width and height of a normal distribution curve?
By standard practice, the normal distribution curve should be normalized so that the area under the curve is 1. This results in a height, at the mean, of about 0.4, i.e. the probability of a sample value being equal to the mean is 40 percent.
The width of the normal distribution curve is infinite, as the tails are asymptotic to the X axis. It is easier to understand that the +/- one sigma area is 68.2 percent, the +/- two sigma area is 95.4 percent, and the +/- three sigma area is 99.6 percent.
How do you focus efforts on certain activities to make the best use of resources?
You apply econometric analysis to the range of possible activities. Economics is the study of allocation of [limited] resources so as to maximise benefit.
It is approaching unity that two in a random group of 30 people would have the same birthday. (Month and date)
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P(2 people share bd in 30) ≈ 38.0%
Using the following expression* that doesn't consider February 29 of a leap year but
gives good estimate;
P(2 people and only 2 people share a birthday, month and date, in a group of n people) = nC2 (1/365) Π1n-1(1 - (i-1)/365)
nCr = n!/(r!(n-r)!)
for n = 30, r = 2,
30C2 = 435
P(2 people share bd in 30) = 435(1/365)(1)(1-1/365)(1-2/365)∙∙∙
∙∙∙(1-27/375)(1-28/365) = 0.380215577...
P(2 people share bd in 30) ≈ 38.0%
*[For an expression that considers leap year see question "What is the probability that
in a room of 8 people 2 have the same birthday ?"]
P(rolls a ten)=3/36=1/12
P(rolls a three)=2/36=1/18
[I am going to assume that if he rolls a ten or three, he also gets his five dollars back, in addition to the winnings.]
Expected value=(1/12)*10+(1/18)*20-(31/36)*5=-85/36, or approximately -2.36
If he doesn't get his five dollars back when he wins,
Expected value=(1/12)*10+(1/18)*20-5=-110/36, or approximately -3.055
[I actually just had this on my statistics final.
I got this one wrong still don't know the answer, but the answer you have isn't one of the possible answers I had.... Maybe -2.78?]
Can you divide mean with standard error by mean with standard error If so how would you do it?
Your question is asking for a number to be divided by itself. Can you clarify your problem.
Two disjoint events in which one or the other must occur are called?
Complements or complementary events
What is the standard deviation of these set of numbers 6 7 7 12 12 11 9 9 8 4 6 13 14?
For 6 7 7 12 12 11 9 9 8 4 6 13 14: σ=3.0947
How many different ways can you arrange the letters in a six letter word?
If all the letters are unique in the set, there are 6 choices for the first letter, 5 for the second letter, 4 for the third letter, etc. This results in 6 X 5 X 4 X 3 X 2 = 720 arrangements. If some of the six letters are duplicated, there will be fewer distinct arrangements.
In a line: 13!/(6!*2!*2!*2!*1!) = 1081080
How do you derive the mean deviation about the mean of the pareto type 1 distributions?
Easy.
The mean deviation about the mean, for any distribution, MUST be 0.
