The average baggage-related revenue per passenger is $16.30 per passenger.
<h3>Expected value</h3>
Expected value formula: x×p(x)
First step
No passenger=0×.54
No passenger=0
Second step
One checked luggage for first bag=.30×$25
One checked luggage for first bag=$7.50
Third step
Two piece for the first and second bag=.16×($25+$30)
Two piece for the first and second bag=.16×$55
Two piece for the first and second bag=$8.80
Last step
Expected value=$7.50+$8.80
Expected value=$16.30
Therefore the average baggage-related revenue per passenger is $16.30 per passenger.
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General term of a geometric sequence is
a(n)=a(1)×r^(n-1)
a1=5
a2=5×6=30
a3=5×6²=180
a4=5×6³=1080
$10 (Teacher will purchase supplies) ... (2) bottles Elmer's Washable. Glue. (2) Elmer's Glue Sticks. (1) Expo Dry Erase marker ... 1 pk Crayola markers ... ALL, ITEMS will become ... pencil box (approx. size 5x8) ... 2 boxes of disinfectant wipes. Kindergarten - Ms. Amanda Hamilton ... Grades 3, 4, 5 - Mrs. Carol Drummond ... sticks 4 pocket folders with tabs-1 of each color (no designs) ... 1 glue stick. Art Box Ruler Colored pencils & sharpener 24 pack of crayons ... All inquiries will be responded to in a timely manner, and inquiries will be replied to via e-mail unless ...
Answer:
D
Step-by-step explanation:
You would have to eliminate A because the students have not yet been to other school districts especially if it's every third. You would also have to eliminate B because well, there really isnt any other place to eat but the lunchroom because otherwise it might pose a problem, and with C it can be difficult but again, if we were to try to change something the time they take really has no impact. D would have to be the correct answer because every third student will have various favorite lunch items and will ultimately tell the school that the gross macaroni that is like...unedible is bad and needs to be taken out...immediatley.
Step-by-step explanation:
To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the 
variable when we add the two equations together.
Right now, there's a 
term in the first equation, and a 
term in the second equation, so if we add those together, we'll be able to eliminate the variable altogether and solve for 
.
However, when we also have a 
term in the first equation and 
term in the second equation, so adding these together will also eliminate the term, leaving a 
on the left-hand side of the equation.
If we add the two numbers on the right side of the equation, we get 
, which does not equal , meaning there are no solutions to this system of equations.
