Answer: 16.7%
There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.
Second Answer : 6/36
Out of the 6^2 = 36 outcomes, there is only 6 that are doubles (double 1,double 2,… double 6). Therefore the probability is 6/36 i 1/6.
Answer:
BC=10
Step-by-step explanation:
BD-BC=14-4=10
Answer:
x2-5x-8=0
Step-by-step explanation:
Two solutions were found :
x =(5-√57)/2=-1.275
x =(5+√57)/2= 6.275
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x2" was replaced by "x^2".
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-5x-8
The first term is, x2 its coefficient is 1 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -8
Step-1 : Multiply the coefficient of the first term by the constant 1 • -8 = -8
Step-2 : Find two factors of -8 whose sum equals the coefficient of the middle term, which is -5 .
-8 + 1 = -7
-4+ 2 = -2
-2+ 4 = 2
-1 + 8 = 7
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
