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Pachacha [2.7K]
2 years ago
8

The members of a book club spend the same amount of money for refreshments at each meeting. At one meeting they bought 6 bagels

and spent $9.00 on beverages. At the next meeting they bought 8 bagels and spent $8.20 on beverages. What was the cost of each bagel?
Mathematics
1 answer:
Tom [10]2 years ago
8 0

The members of a book club spend $0.4 for each Bagel as refreshment.

Let x represent the cost of each bagel.

At a meeting, they bought 6 bagels and spent $9.00 on beverages, hence:

Cost = 6x + 9

At another meeting, they bought 8 bagels and spent $8.20 on beverages, hence:

Cost = 8x + 8.20

Since the same money is spent on each meeting, hence:

6x + 9 = 8x + 8.2

2x = 0.8

x = $0.4

Therefore the cost of each bagel is $0.4

Find out more at: brainly.com/question/13911928

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The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

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The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

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Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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