Probability

5/13

If you want the parameters a, b and c then you must know the values of x and y. Even in this case you will not be able to calculate unique values for a,b and c but rather get a set of answers which describes the relationship necessary between a,b and c that will satisfy your values for x and y.

Here is how to do that:

Start with the usual parabolic equation: y = ax^2 + bx + c

There is a standard formula to solve this equation for x when you want y to be zero (also called the 'roots' of the equation). That formula is x = (-b +/- sqrt ( 4ac - b^2 ) ) / 2a).

The sqrt means take the square root of what in parentheses. and +/- means the formula must be calculated twice, once using a plus in place of +/- and once using a minus.

Instead of y being zero we want to be able to make it any value we like but our formula only works for y being zero. So we need to subtract the y value from the original parabolic equation to make one side zero, that give us 0 = ax^2+bx + (c-y).

Now we can use the standard formula again , but this time with c-y in place of just c.

x = (-b +/- sqrt( 4a(c-y) - b^2 ) / 2a

Using this new formula, if you know x and y you can calculate a relationship between a,b and c that will always give you that value of x and y in a parabolic equation.

69 lolz

The smallest percentage point of any distribution is 0% and the largest is 100%.

1 to 1.Yep have good day it's my birthday today!

not sure

i want to know also

There is not enough information to answer your question. To determine a Z-Score, the mean and standard deviation is also required.

Total number of possibilities on two dice = 36.

Number of ways to get 2, 3, or 12 = 4.

1, 1

1, 2

2, 1

6, 6

Probability = 4/36 = 1 chance in 9 = 0.111 = 11.1%

Chances=50% because on a 6 sided dice, as you have said, the sides are numbered 1,2,3,4,5 and 6. There are 3 even numbers and 3 odd numbers. Chances would be 3/6=50%

Quadratic functions are used to describe free fall.

An imprecation is a curse. To imprecate is to call a curse upon someone or something.

Probability density function (PDF) of a continuous random variable is a function

that describes the relative likelihood for this random variable to occur

at a point in the observation space.

The PDF is the derivative of the probability distribution (also known as

cummulative distriubution function (CDF)) which described the enitre range of values

(distrubition) a continuous random variable takes in a domain.

The CDF is used to determine the probability a continuous random variable occurs any (measurable) subset of that range.

This is performed by integrating the PDF over some range (i.e., taking the area under of CDF curve between two values).

NOTE: Over the entire domain the total area under the CDF curve is equal to 1.

NOTE: A continuous random variable can take on an infinite number of values. The probability that it will equal a specific value is always zero.

eg. Example of CDF of a normal distribution:

If test scores are normal distributed with mean 100 and standard deviation 10. The probability

a score is between 90 and 110 is:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )

= 0.84 - 0.16 = 0.68.

ie. AProximately 68%.

The GCF is 2.

I assume you mean how many possible outcomes when looking at all 13 results.

It would be 2^13 = 8192

24

None.

Two ordinary dice can only sum to 12 and so you cannot roll a 36.

The odds of hitting 2 straights in a row at 5 card stud is about 5%.

Since the deck is reshuffled between deals, the two hands are sequentially unrelated, so the odds of two deals with two straights is simply th odds of one times the odds of another.

In order to draw a straight, you need to draw an ace, 2, 3, 4, 5, or a 10, jack, queen, king, ace, assuming that ace can be high or low.

The odds of drawing one of these cards is 10 in 13, but you have 5 chances to do so, so the odds of drawing the first card is 50 in 13.

The odds of drawing the second card is 39 in 51, but you have 4 chances to do so, so the odds of drawing the second card is 156 in 51.

Similarly, the odds of drawing the third, fourth, and fifith card is 38 in 50, 37 in 49, and 36 in 48.

Multiply these odds and you get 19,740,240 in 77,968,800. Reducing that to lowest terms, you get 246,753 in 1,112,110, which is the odds of drawing a straight. The numeric value of the odds is 0.22187823146990855221156180593646, or about 22%

Square that and you get 60,887,043,009 in 1,236,788,652,100. That is aready in lowest terms, so that is the odds of drawing two sequential straights. The numeric value of the odds is 0.049229949600214317813758989938262, or about 5%.

war

The second and third marbles depend upon the first, so they are dependent. Even if they are all being drawn together, because there is no replacement, each removal affects the original group of marbles.

eight

they sit on top of eachover

and the 3rd guy gives the 1st guy a reach-around

and the 2nd guy gives the 3rd guy a blowjob

and the 4th guy videos it and puts it on youporn.

hope that helped.

Assuming you mean a sum of 4, there are three ways this can be done: (1, 3), (2, 2) and (3, 1).

well lets look at the probability on your first pick you will get a blue eyed girl

2 have blue out of 5, so yo uhave a 2/5 chance, or 40%

on your second pick, there are 4 girls left, and one of them will have blue eyes now

1 have blue out of 4, so 1/4 or 25%

so you multiply the two probabilities to get your final sequential probability

2/5 * 1/4

find a common denominattor, for this we will use 20

8/20 * 5/20 = 40/400

simplify it you get 1/10 0.10, or 10% chance

The frequency can be determined by recording the sum for each toss and producing a tally chart for the outcomes.