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lisabon 2012 [21]
3 years ago
6

A menu offers a choice of 22 ​salads, 77 main​ dishes, and 55 desserts. How many different meals consisting of one​ salad, one m

ain​ dish, and one dessert are​ possible?
Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

70 different meals

Step-by-step explanation:

A menu offers a choice of 2 ​salads, 7 main​ dishes, and 5 desserts

There are 3 types of choices are given

each one have different choices. we need to select one salad, one main dish and one dessert

To find the number of different meals we multiply the different choices

2 \cdot 7 \cdot 5=70

so 70 different meals

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Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel
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The unusual X values ​​for this model are: X = 0, 1, 2, 7, 8

Step-by-step explanation:

A binomial random variable X represents the number of successes obtained in a repetition of n Bernoulli-type trials with probability of success p. In this particular case, n = 8, and p = 0.53, therefore, the model is {8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}. So, you have:

P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024

P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215

P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848

P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912

P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695

P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431

P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371

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P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062

The unusual X values ​​for this model are: X = 0, 1, 7, 8

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