What is the probability that two people selected at random have the same birthday? ignore leap years. 1/365 2/365 1/133,225

 The correct option is (A)

Step-by-step explanation:  We are given to find the probability that two people selected at random have the same birthday.

Since there are 365 days in a non-leap year, so we will be considering the number of days as 365.

Now, whatever be the birth day of the first person, we need to select another person that has the same birth day as the first one.

Let, 'E' be the event that the second person has the same birthday as the first one.

So, n(E) = 1, because the birthday can be only one day out of 365 days.

Let, 'S' be the sample space for the experiment.

So, n(S) = 365.

Therefore, the probability  the second person has the same birthday as the first one will be

thus, the required probability is

Option (A) is correct.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)