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3 years ago

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1. Solve the system of equations algebraically.

2 answers:

goldfiish [28.3K]3 years ago

8 0

Hello,
Please, see the attached files.
Thanks.

Andrew [12]3 years ago

4 0

1. We can double the second equation to get: 2x - 6y = 6; and then add the equations together to get -y = 4, y = -4. We plug y in to get x - 3(-4) = 3, x = 15. The answer would be (15, -4)
2. We can subtract the two equations to eliminate y and get 2x = 12, x = 6. We can plug x in to the second equation to get 9 + 3y = 18, and 3y = 9, so y = 3. The answer would be (6, 3)
3. We add the first two equations to get 8x - 5y = -34, and then subtract the last two equations to get 3x - 2y = -13. We multiply the first by 2, and the second by 5, and then add them together. We get x = -3. We plug the x-value into the second equation to get -9 - 2y = -13, so y = 2. We plug both values into the third equation to get -9 - 2 - z = -6. -z = 5, z = -5.
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Hope this helps!
==jding713==

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0.003 = 0.3% probability that Sarah, Jamal, Kate, and Mai are chosen in any order.

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The students are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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N is the size of the population.

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k is the total number of desired outcomes.

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$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

11 students means that $N = 11$

4 are Sarah, Jamal, Kate, and Mai, so $k = 4$

4 are chosen, which means that $n = 4$

What is the probability that Sarah, Jamal, Kate, and Mai are chosen in any order?

This is P(X = 4). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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