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3 years ago

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A production manager randomly sampled production lines at a factory that produces automobiles. She wanted to find out how many p

roduction lines caused defects in newly produced automobiles. The proportion of production lines that caused defects was 0.08, with a margin of error of 0.01. Construct a confidence interval for the proportion of production lines that caused defects

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

Step-by-step explanation:

A confidence interval for a population proportion is a function of the sample proportion and the margin of error.

The interval has two bounds, a lower bound and an upper bound.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this problem, we have that:

Sample proportion 0.08

Margin of error 0.01

0.08 - 0.01 = 0.07

0.08 + 0.01 = 0.09

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

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Parallel / Perpendicular Practice

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The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are <u>neither</u> parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are <u>parallel</u>

║

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are <u>Neither</u> parallel nor perpendicular

<u>Neither</u>

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are <u>perpendicular</u>

⊥

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are <u>parallel</u>

║

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are <u>neither</u> parallel nor perpendicular

<u>Neither</u>

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

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Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

<u>Neither</u>

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

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$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is <u>Neither</u>

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Answer:

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Step-by-step explanation:

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The side lengths of a right triangle must satisfy the <u>Pythagorean Theorem. </u>

Pythagorean Theorem: $Hypotenuse^2=Opposite^2+Adjacent^2$

<u>Option A:</u> $3, 6, \sqrt{45}$

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True

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