Arrivals of customers at a checkout counter follow a poisson distribution. it is known that, during a given 30-minute period, one customer arrived at the counter. then the actual time of arrival follows a uniform distribution over the interval of (0, 30). find the conditional probability that a customer arrives during the last 5 minutes of the 30-minute period if it is known that no one arrives during the first 10 minutes of the period.

1/4

Step-by-step explanation:

Let X be the time of arrival measured from the beginning of the 30-minute period.

Since the number of arrivals is Poisson, the time of the arrival is equally likely in any subinterval of time of a given size in the 30 minutes, and thus X has a uniform distribution in (0,30) and P(X > 25) =

where a = 30 and b = 25

Solving, we obtain P(X > 25) = 1/6

similarly, P(X>10) = 2/3 by substituting b for 10

Hence, for conditional probability we have

P(X>25 | X>10) = P(X>25) / P(X>10) = (1/6) / (2/3) = 1/4

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points

i need more info. are you looked to see there mph? how much they deliver? how far they go?

step-by-step explanation: