MHB How Do You Calculate Average Revenue Per Group?

AI Thread Summary
To calculate average revenue per group, the probabilities and prices of each item ordered must be multiplied and summed. For Group A, the average revenue is calculated as (5*0.5) + (6*0.4) + (7*0.1), resulting in 5.6. For Group B, the calculation is (5*0.3) + (6*0.3) + (7*0.1), which needs to be completed for the final average revenue. The discussion highlights the importance of accurately applying probability to revenue calculations. Understanding these calculations is crucial for determining average revenue effectively.
humm0s
Messages
1
Reaction score
0
Hi, I'm currently stuck on a homework question and I was hoping if I could get some help.

Group A has 50% chance of ordering french fries (price: 5),40%chanceoforderingmilkshake(price:6), and 10% chance of ordering a burger (price: 7).GroupBhas30%chanceoforderingfrenchfries(price:5), 30% of chance of ordering a milkshake (price: 6),and10%chanceoforderingaburger(price:7) .

What is the average revenue per group?


Not sure if my work is correct: Avg revenue for Group A: (5∗50%)+(6 * 40%) + (7∗10%)=5.6.
Avg revenue for Group B: (5∗30%)+(6 * 30%) + ($7 *10%)
 
Mathematics news on Phys.org
Re: Avg Revenue Per Group

humm0s said:
Hi, I'm currently stuck on a homework question and I was hoping if I could get some help.

Group A has 50% chance of ordering french fries (price: 5),40%chanceoforderingmilkshake(price:6), and 10% chance of ordering a burger (price: 7).GroupBhas30%chanceoforderingfrenchfries(price:5), 30% of chance of ordering a milkshake (price: 6),and10%chanceoforderingaburger(price:7) .

What is the average revenue per group?


Not sure if my work is correct: Avg revenue for Group A: (5∗50%)+(6 * 40%) + (7∗10%)=5.6.
Avg revenue for Group B: (5∗30%)+(6 * 30%) + ($7 *10%)

Did you have a question about the response already supplied?

https://www.freemathhelp.com/forum/threads/110278-Average-Revenue-Per-Group
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top