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bazaltina [42]
3 years ago
13

How Many solutions are there for the following system. y = x² - 5x + 3 Y = x - 6

Mathematics
1 answer:
emmasim [6.3K]3 years ago
7 0
Only one solution, x=3
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1) -2/2i
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Consider the game of independently throwing three fair six-sided dice. There are six combi- nations in which the three resulting
Murrr4er [49]

Answer:

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

  1. 1+3+ 6,
  2. 1+ 4+ 5,
  3. 2+2+6,
  4. 2+3+5,
  5. 2 + 4 + 4,
  6. 3+3+4

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

  1. 1+3+ 6 : 6 permutations
  2. 1+ 4+ 5 : 6 permutations
  3. 2+2+6: 3 permutations
  4. 2+3+5: 6 permutations
  5. 2 + 4 + 4: 3 permutations
  6. 3+3+4: 3 permutations

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

  1. 1 + 2 + 6: 6 permutations
  2. 1+3+5: 6 permutations
  3. 1+4+4: 3 permutations
  4. 2+ 3+ 4: 6 permutations
  5. 2+2 +5: 3 permutations
  6. 3+3+3: 1 permutation.

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

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2x + y = 10
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