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erica [24]
3 years ago
15

Solve this system of equations: y = x2 – 3x + 12 y = –2x + 14 4. Substitute the values of x, –1 and 2, into either original equa

tion to solve for the values of y. What are the solutions of the system of equations? One solution is (–1, ). The second solution (2, ).
Mathematics
2 answers:
Dvinal [7]3 years ago
5 0

Answer:

(-1, 16) and (2, 10)

Step-by-step explanation:

Substituting -1 in place of x in the first equation, we have

y = (-1)²-3(-1)+12 = 1--3+12 = 1+3+12 = 4+12 = 16.

Substituting -1 in place of x in the second equation gives us

y=-2(-1)+14 = 2+14 = 16

This makes the first ordered pair (-1, 16).

Substituting 2 in place of x in the first equation gives us

y=2²-3(2)+12 = 4-6+12 = -2+12 = 10

Substituting 2 in place of x in the second equation gives us

y=-2(2)+14 = -4+14 = 10

This means the ordered pair (2, 10) is the second solution.

SOVA2 [1]3 years ago
4 0

One solution is (–1, <span> ⇒ 16</span>).

The second solution (2, <span> ⇒ 10</span>). your welcome 

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Answer:

d. The interval contains only negative numbers. We cannot say at the required confidence level that one region is more interesting than the other.

Step-by-step explanation:

Hello!

You have the data of the chemical measurements in two independent regions. The chemical concentration in both regions has a Gaussian distribution.

Be X₁: Chemical measurement in region 1 (ppm)

Sample 1

n= 12

981 726 686 496 657 627 815 504 950 605 570 520

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Sample mean X[bar]₁= 678.08

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Sample 2

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1024 830 526 502 539 373 888 685 868 1093 1132 792 1081 722 1092 844

μ₂= 812

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Sample mean X[bar]₂= 811.94

Using the information of both samples you have to determina a 90% CI for μ₁ - μ₂.

Since both populations are normal and the population variances are known, you can use a pooled standard normal to estimate the difference between the two population means.

[(X[bar]₁-X[bar]₂)±Z_{1-\alpha /2}* \sqrt{\frac{Sigma^2_1}{n1}+\frac{Sigma^2_2}{n_2}  }]

Z_{1-\alpha /2}= Z_{0.95}= 1.648

[(678.08-811.94)±1.648*\sqrt{\frac{164^2}{12}+\frac{239^2}{16}  }]

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You cannot be sure without doing a hypothesis test but it may seem that the chemical measurements in region 1 are lower than the chemical measurements in region 2.

I hope it helps!

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