three distinct vertices of a cube are to be randomly chosen what is the probability that they will be the vertices of an equilateral triangle

The triangle formed by the vertices would be equilateral if the sides of the triangle coincide with the diagonals of the square face that defines the cube. See the attached sketch if that description is unclear.

Count the number of equilateral triangles that are possible. Such a triangle will always use up 3 vertices of the cube, and separate 1 vertex from the remaining 4. This means there are as many equilateral triangles as there are corners in the cube: 8.

Count the total number of triangles that can be made. From the 8 available vertices, we only need 3. If we fix 3 vertices, it doesn't matter in which order we connect them to make a triangle, so we count the number of combinations of 8 vertices taking 3 at a time, which is

Then the probability of forming an equilateral triangle is 8/56 = 1/7.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !