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3 years ago

Solve the system algebraically.

1 answer:

7 0

1x - 1y + 2z = -3 → 1x - 1y + 2z = -3 → 2x - 2y + 4z = -6
2x + 2y - 1z = -3 → 2x + 2y - 1z = -3 → 2x + 2y - 1z = -3
3x + 2y + 0z = 8    4x + 3z = -9

1x - 1y + 2z = -3
2x + 2y - 1z = -3 → 2x + 2y - 1z = -3 → -2x - 2y + 1z = 3
3x + 2y + 0z = 8 → 3x + 2y + 0z = 8 → 3x + 2y + 0z = 8
  x + z = 11

4x + 3z = -9 → 4x + 3z = -9
1x + 1z = 11 → 4x + 4z = 44
  -z = -53
-1   -1
z = 53
x +   z = 11
x + 53 = 11
- 53 - 53
x = -42

  3x + 2y = 8
3(-42) + 2y = 8
  -126 + 2y = 8
+ 126 + 126
2y = 134
2 2
y = 67
(x, y, z) = (-42, 67, 53)

The answer is B.

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3 years ago

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Sample 1:

Mean x₁ = 88

Sample size n₁ = 173

Sample standard deviation s₁ = 12,22

Sample 2:

Mean x₂ = 53,5

Sample size n₂ = 127

Sample standard deviation s₂ = 13,59

CI 95 % α = 5% α = 0,05 α/2 = 0,025

We need to find:

( x₁ - x₂ ) - tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂) < μ₁ - μ₂ < ( x₁ - x₂ ) + tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂)

v = degree of fredom

v = [ ( s₁²/n₁ + s₂²/n₂)² / (s₁²/n₁)² /n₁-1 + (s₂²/n₂)²/n₂-1

v = [ (0,86 + 1,45 ) / 0,0043 + 0,017

v = 2,31 / 0,0213

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Then t 0,025, 108 from t table is: We will take v = 100

t = 1,984

Now

√ (s₁²/n₁ ) + (s₂²/n₂) = √ 0,86 + 1,43

√ 2,3 = 1,51

Then: CI:

(173 - 127 ) - 1,984*1,51 < μ₁ - μ₂ < ( 173 - 127 ) + 1,984*1,51

46 - 3 < μ₁ - μ₂ < 46 + 3

43 < μ₁ - μ₂ < 49

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