The cafeteria uses rectangular trays with a raised edge around them. The area of the flat surface of each tray is 228 Inches squ
2 answers:
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Answer: a) P = 0.5, b) P = 0.07
Step-by-step explanation:
Hi!
Lets call X₁ the time at which you arrive, and X₂ the time at which Bob arrives. Both are random variables with uniform density in the interval [0, 60] (in minutes). Their joint distribuition is uniform over the square in the image, with value P = 1/(60*60) = 1/3600.
a) For you to get more cake than Bob, you should arrive earlier. This event is A = { X₁ < X₂ }, the shaded triangle in the figure.The area of this event (set) is half the total area of the square, so P(A) = 0.5.
It makes sense, beacuse its equally probable for you or Bob to arrive earlier, as both have uniform density over the time interval.
b) In this case you arrive later than Bob, but less than 5 minutes later. So the event is B = { X₂ < X₁ < (X₂ + 5) } . This is the gray shaded area in b) part of the image. Its area is the difference two triangles (half square - blue triangle), then the probability is:
Question
During one waiters shift he delivered 13 appetizers, 17 entrées, and 10 desserts what percentage of the dishes he delivered were: A. Desserts B. Appetizers C. entrees
Answer:
A)Desserts = 25 %
B)Appetizers = 32.5 %
C)Entrées = 32.5 %
Step-by-step explanation:
<u>Given:</u>
Number of appetisers = 13
Number of entrees = 17
Number of desserts = 10
<em><u>A. percentage of the dishes he delivered were Desserts</u></em>
Percentage of Desserts =
Total number of dishes = 13 + 17 + 10 = 40
Percentage of Desserts =
Percentage of Desserts =
Percentage of Desserts = 25 %
<em><u>B. percentage of the dishes he delivered were Appetizers </u></em>
Percentage of Desserts =
Percentage of Desserts =
Percentage of Desserts =
Percentage of Desserts = 32.5 %
<em><u>C. percentage of the dishes he delivered were Entrées </u></em>
Percentage of Desserts =
Percentage of Desserts =
Percentage of Desserts =
Percentage of Desserts = 42.5 %