Parking on campus is often expensive, and it is tempting to park illegally and take your chances on getting a ticket. From past experience, the probability of getting a ticket is about 0.30 on any given day, and because you always change your parking location, each day's outcome is independent of the previous day's outcome. The probability that you will NOT get a ticket on two days is

P(not getting a ticket for 2 days) = 0.49

Step-by-step explanation:

We are told that the probability of getting a ticket on any given day is 0.3.

Thus;

P(getting a ticket a day) = 0.3

This means that;

P(not getting a ticket a day) = 1 - 0.3 = 0.7

Now, 0.7 is probability of not getting a ticket for 1 day. Since each day's outcome is independent of the previous day's outcome, it means that for n days, probability of not getting a ticket for n days will be;

P(not getting a ticket for n days) = 0.7ⁿ

For 2 days, its = 0.7² = 0.49

Thus;

P(not getting a ticket for 2 days) = 0.49

which expression represents the approximate length of bc?

step-by-step explanation:

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72