5. At the local burger joint, 45% of the customers purchase cheeseburgers and 80% purchase fries. If a customer is selected at randomlet the event that they bought a burger be A and the event that they purchased fries be B. If A and B are independent what is P(A or B)?

P (A or B) =  0.53

Step-by-step explanation:

P(A or B) is the probability where one of the events happened but not both of them.

Probability of event A = 0.45

Probability of not event A = 1 - 0.45 = 0.55

Probabiity of event B = 0.80

Probability of not event B = 1 - 0.80 = 0.20

Then, the probability that one customer buys a burger but does not buy fries is:

P1 = 0.45*0.20 = 0.09

Probability of purchasing fries but not a burger:

P2 = 0.80*0.55 = 0.44

P (A or B) = P1 + P2 = 0.09 + 0.44 = 0.53

answer:

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

explanation:

service only with 20% tip  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

in the opening of the question the word 'service' is associated only with the pretax value.

the closing part of the question asks: how much did she pay for the  

service

−−−−−−  

so technically the answer is  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

however, the intention of the question provider may have been that your answer is to be  

$

87.12

(

1

+

8

100

)

=

$

94.09

to 2 decimal places.

'

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

tony b

mar 21, 2017

answer:

trying to explain it another way

explanation:

service charge  

.

→

$

72.60

20% tip  

20

100

×

$

72.60

=

$

14.52

−−−−−−  

←

add

$

87.12

tax  

8

100

×

$

87.12

=

$

6.97

−−−−−  

←

add

$

94.09

related questions

lo.lll

step-by-step explanation: