Abusiness man is flying from san francisco to new york with a connection in chicago. he is concerned about having his luggage arriving safely in new york, but this will depend on whether or not his flight from san francisco leaves on time. using past flight information, he estimates the probability that the flight leaves on time is 0.15. if the flight leaves on time, the probability his luggage arrives safely is 0.95, but if the flight does not leave on time the probability his luggage arrived safely is only 0.65.
a. what is the probability that his luggage arrives safely in new york? b. when the businessman arrived in new york, his luggage was not there. what is the probability his flight from san frcisco left on time?

a. The probability is 0.695

b. The probability is 0.0246

Step-by-step explanation:

Let's call T the event that the flight leave on time, NT the event that the flight doesn't leave on time, S the luggage arrived safely and NS the luggage doesn't arrived safely.

So, the probability P(S) that his luggage arrives safely in New York is:

P(S) = P(T∩S) + P(NT∩S)

P(S) = (0.15*0.95) + (0.85*0.65)

P(S) = 0.695

Where P(T∩S) is the multiplication of the probability that the flight leave on time and the probability that his luggage arrive safely given that the flight leave on time. At the same way, P(NT∩S) is the multiplication of the probability that the flight doesn't leave on time and the probability that his luggage arrive safely given that the flight doesn't leave on time.

The probability P(T/NS) that his flight from San Francisco left on time given that his luggage was not there is:

P(T/NS) =  P(T∩NS)/P(NS)

Where P(NS) = P(T∩NS) + P(NT∩NS)

So, Calculating  P(T∩NS), P(NT∩NS) and P(NS) at the same way that we calculate P(S), we get:

P(T∩NS) = 0.15*(1-0.95) = 0.0075

P(NT∩NS) = 0.85*(1-0.65) = 0.2975

P(NS) = 0.0075 + 0.2975 = 0.305

Then, if the businessman arrived in New York and his luggage was not there, the probability P(T/NS) that his flight from San Francisco left on time is:

P(T/NS) = 0.0075/0.305

P(T/NS) = 0.0246

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

astronomers is your final answer! : )