A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The experiment consists of observing the movement of a single vehicle through the intersection.
A) List the sample space for this experiment.
B) Assuming that all sample points are equally likely, find the probability that the vehicle turns.

a) Sample space (S) = {TR, TL, Cs}

b) Pr(Vehicle Turns) = 2/3.

Step-by-step explanation:

a) By sample space (S) we mean the list all possible outcome of an event. From the illustration in the question, only three (3) outcome is possible:

i) Vehicle Turn Right (TR)

ii) Vehicle Turn Left (TL)

iii) Continue Straight ahead (Cs)

And since the experiment consists of observing the movement of a single vehicle through the intersection, then, the sample space (S) is:

==> {TR, TL, Cs}

b) Since we are assuming that all sample points are equally likely, it imply that they have equal probability of occurring. And by probability:

Pr(TR) = 1/3

Pr(TL) = 1/3

Pr(Cs) = 1/3

Meanwhile, the question wants us to find the probability that the vehicle turns. This means, the vehicle turning either right or left. Thus;

Pr(vehicle turns) = Pr(TR) or Pr(TL) = Pr(TR) + Pr(TL)

Pr(vehicle turns) = (1/3) + (1/3) = 2/3

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

step-by-step explanation:

left be aligning the decimal points is required to not required see the number of digits to the right