Elizabeth lives in San Francisco and works in Mountain View. In the morning, she has 3 transportation options (take a bus, a cab, or a train) to work, and in the evening she has the same 3 choices for her trip home. If Elizabeth randomly chooses her ride in the morning and in the evening, what is the probability that she'll use a cab exactly one time?

4/9

Step-by-step explanation:

The possibilities of transportation: (The first will be for morning, second will be for afternoon)

B, B

B, C

B, T

C, B

C, C

C, T

T, B

T, C

T, T

It is clearly seen that there are 9 transportation options.

(Using cab 1 time we have BC, CB, CT, TC. So four of the transportation methods use cab one time.)

Therefore, the probability that she will use a cab only once is 4/9.  

-4, -2, 1, 4

step-by-step explanation:

domain = x values

range = y values

127

whether it's diameter, or radius, i believe if i remember correctly from school, every amount has to always add up to 180. so if you take 53, and starting adding, long story short, taking 127 and adding it to 53, gives you 180.