5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10
And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.
<span>4x-6y=24
2x-3y=12
2x-12=3y
y=2/3*x-4</span>
There will be 50 gift bags.
The first set of 25 bags
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Because there are 25 thumb drives, the first 25 bags will each have 1 thumb drive.
Also, each bag will have 5 times as many key chains as a thumb drive. Therefore each of the first 25 bags will have
1 thumb drive, 5 key chains.
After the first 25 bags, we have used
25 thumb drives, 5*25 = 125 key chains.
The second set of 25 bags
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We have a total of 200 key chains, so we have 200-125 = 75 key chains left.
Distribute them equally among the remaining 25 bags, so each bag has
75/25 = 3 key chains.
Answer:
25 bags, each with 1 thumb drive and 5 key chains.
Another 25 bags, each with 3 key chains.
Usando un sistema de ecuaciones, se encuentra que
- Cada manzana cuesta $3.
- Cada pera cuesta $1.
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- Un sistema de ecuaciones soluciona esta pergunta.
- El custo de una manzana es x.
- El custo de una pera es y.
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- <u>Seis manzanas y 8 peras cuestan $26</u>, o sea,
- <u>Cada manzana cuesta el triple de cada pera</u>, o sea,
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Primero, encontramos el cuesto de una pera, substituyendo la segunda en la primera ecuación.
Cada pera cuesta $1.
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<u>Cada manzana cuesta el triple de cada pera</u>, o sea, 
.
Cada manzana cuesta $3.
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