A travel agent is booking trips for tourists who travel from New York to Chicago. Tourists have three choices for how to travel from New York to Chicago. They can take an airplane for $350, a bus for $150, or a train for $225. Once they arrive in Chicago, they can travel by van to their hotel for $60 or take a cab for $40. If each option is equally likely to occur, what is the probability that a tourist will spend more than $275 on these 2 legs of the trip?

P = 1/2

Step-by-step explanation:

If the tourist spends more than 275$, they must not arrive in Chicago by bus.

( 150 + 60 < 275, 150 + 40 < 275)

The total options the tourist can make:

3 x 2 = 6

(1st leg: 3 possible options, 2nd leg: 2 possible options)

The number of options the tourist can make after excluding bus option:

2 x 2 = 4

(1st leg: 2 remaining possible options, 2nd leg: 2 possible options)

The number of options the tourist can make after excluding the bus option and spend more than 275$:

4 - 1 = 3

(excluding the case of selecting train and cab, because 225 + 40 < 275)

=> The probability that the tourist will spend more than 275$ on these 2 legs of the trip:

P = 3/6 = 1/2

choose (5, 1.5) and (-1.5,-3.5)

step-by-step explanation:

5 - 1.5 =3.5

-1.5 - (-3.5) = 2

3.5 + 2 =5.5

hope this !