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svetoff [14.1K]
3 years ago
5

Pls answer this fast will give brainliest

Mathematics
2 answers:
maria [59]3 years ago
5 0

Answer:

12

Step-by-step explanation:

n_{C} r = \frac{n!}{(n-r)!(r!)} \\\\

n is the <u>number </u>of choices

r is the <u>number</u> of Selections we can make

For Sandwiches:

3 Choices w/ 1 Selections

n = 3\\r = 1\\

n_{C} r = \frac{3!}{(3-1)!(1!)} = 3

For Veg.:

2 Choices w/ 1 Selections

n = 2\\r = 1\\

n_{C} r = \frac{2!}{(2-1)!(1!)} = 2

For Dessert.:

2 Choices w/ 1 Selections

n = 2\\r = 1\\

n_{C} r = \frac{2!}{(2-1)!(1!)} = 2

Multiply All Combinations ∴

3 * 2 * 2 = 12

stira [4]3 years ago
3 0

Answer:

12

Step-by-step explanation:

4 options per sandwich, 4 times 3 =12

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Step-by-step explanation:

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Hitman42 [59]

The number of cookies packs needed if the treasurer buys 1 package of brownies bites is 5 packs

<h3>Equation</h3>

  • Total attendance = 75 people
  • Cookies in each pack = 12
  • Brownies in each pack = 15
  • Number of cookies pack = c
  • Number of brownies pack = b

15b + 12c = 75

If 1 package of brownies is bought

15b + 12c = 75

15(1) + 12c = 75

15 + 12c = 75

12c = 75 - 15

12c = 60

c = 60/12

c = 5 packs

  • It is a reasonable value in this context

If 9 packages of brownies are bought

15b + 12c = 75

15(9) + 12c = 75

135 + 12c = 75

12c = 75 - 135

12c = -60

c = -60/12

c = -5 packs

  • The value is not reasonable in this context

Learn more about equation:

brainly.com/question/4344214

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