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Nataly [62]
3 years ago
11

A restaurant offers a special pizza with any 4 toppings. If the restaurant has 15 topping from which to choose, how many differe

nt special pizzas
are possible.​
Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer: 1,365 possible special pizzas

Step-by-step explanation:

For the first topping, there are 15 possibilities, for the second topping, there are 14 possibilities, for the third topping, there are 13 possibilities, and for the fourth topping, there are 12 possibilities. This is how you find the number of possible ways.

15 * 14 * 13 * 12 = 32,760

Now, you need to divide that by the number of toppings you are allowed to add each time you add a topping.

4 * 3 * 2 * 1 = 24

32,760 / 24 = 1,365

There are 1,365 possible special pizzas

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The Wilson family had 5 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wi
Dafna11 [192]

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0.813

0.500

Step-by-step explanation:

Use binomial probability.

P = nCr p^r q^(n−r)

where n is the number of trials,

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p is the probability of success,

and q is the probability of failure (1−p).

In this problem, n = 5, p = 0.5, and q = 0.5.

"At least 2 girls" means r = 2, 3, 4, or 5.

Or, we can use the complement.

P(at least 2 girls) = 1 − P(at most 1 girl)

P(at least 2 girls) = 1 − P(r=0 or r=1)

P(at least 2 girls) = 1 − ₅C₁ (0.5)¹ (0.5)⁵⁻¹ − ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰

P(at least 2 girls) = 1 − 5 (0.5) (0.5)⁴ − 1 (1) (0.5)⁵

P(at least 2 girls) = 1 − 6 (0.5)⁵

P(at least 2 girls) ≈ 0.813

"At most 2 girls" means r = 0, 1, or 2.

P(at most 2 girls) = P(r=0, r=1, or r=2)

P(at most 2 girls) = ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰ + ₅C₁ (0.5)¹ (0.5)⁵⁻¹ + ₅C₂ (0.5)² (0.5)⁵⁻²

P(at most 2 girls) = 1 (1) (0.5)⁵ + 5 (0.5) (0.5)⁴ + 10 (0.5)² (0.5)³

P(at most 2 girls) = 16 (0.5)⁵

P(at most 2 girls) = 0.500

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3 years ago
Thanks so much guys! I love your answers! A new question:<br><br> What is 55% of 93?
wlad13 [49]

Answer:

51.15

Step-by-step explanation:

93 * 0.55 = 51.15

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