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WITCHER [35]
3 years ago
13

2)Whitney is shopping for party supplies. She finds a package of 10 plates, 16 napkins, and a package of 8 cups. What is the lea

st number of packages of plates, napkins, and cups so that she has the same number of each? Write answer in sentence form.
Mathematics
1 answer:
Dominik [7]3 years ago
4 0

10=2\cdot5\\16=2^4\\8=2^3\\\\\text{lcm}(10,16,8)=2^4\cdot5=80

80/10=8 packages of plates

80/16=5 packages of napkins

80/8=10 packages of cups

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multiply 0.272 by 100

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A right triangle has a hypotenuse of 25 feet and a leg of 7 feet. Which of these is the length of the other leg?
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A^2 + b^2 = c^2...where a and b are ur legs and c is ur hypotenuse

7^2 + b^2 = 25^2
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Roulette is a very popular game in many American casinos. In roulette, a ball spins on a circular wheel that is divided into 38
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Answer:

a. P(AUB) = 0.74

b. P(A∩C) = 0.24

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d. P(Bc) = 0.55

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Step-by-step explanation:

n(U) = Total number of elements in the set = number of elements in the universal set = 38

A: (Outcome is an odd number (00 and 0 are considered neither odd nor even)]

An odd number referred to any integer, that is not a fraction, which is not possible to be divided exactly by 2. Therefore, we have:

A: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35

B: 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33, 35

C: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

a. Calculate the probability of AUB

AUB picks not more than one each of the elements in both A and B without repetition. Therefore, we have:

AUB = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 33, 35

n(AUB) = number of elements in AUB = 28

P(AUB) = probability of AUB = n(AUB) / n(U) = 28 / 38 = 0.736842105263158 = 0.74

b. Calculate the probability of A ∩ C  

A∩C picks only the elements that are common to both A and C. Therefore, we have:

A∩C: 1, 3, 5, 7, 9, 11, 13, 15, 17

n(A∩C) = number of elements in A∩C = 9

P(A∩C) = probability of A∩C = n(A∩C) / n(U) = 9 / 38 = 0.236842105263158 = 0.24

c. Calculate the probability of BUC

BUC picks not more than one each of the elements in both B and C without repetition. Therefore, we have:

BUC: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35

n(BUC) = number of elements in BUC = 27

P(BUC) = probability of A∩C = n(BUC) / n(U) = 27 / 38 = 0.710526315789474 = 0.71

d. Calculate the probability of Bc

Bc indicates B component it represents all the elements in the universal set excluding the elements in B. Therefore, we have:

Bc: 00, 0, 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36

n(Bc) = number of elements in Bc = 20

P(Bc) = probability of Bc = n(Bc) / n(U) = 20 / 38 = 0.526315789473684 = 0.55

e. Calculate the probability of A∩B∩C

A∩B∩C picks only the elements that are common to A, B and C. Therefore, we have:

A∩B∩C: 11, 13, 15, 17

n(Bc) = number of elements in A∩B∩C = 4

P(A∩B∩C) = probability of A∩B∩C = n(A∩B∩C) / n(U) = 4 / 38 = 0.105263157894737 = 0.11

Note: All the answers are rounded to 2 decimal places.

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