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2 years ago

(5 points)

Mathematics

1 answer:

Schach [20]2 years ago

3 0

Answer:

The original Slope is y=-1/3x-2

The perpendicular slope would be y=3x+b

Step-by-step explanation:

Original Slope:

You already have the y-interecpet, (0,-2), and you know two points (0,-2) and (-6,0), so you can find the slope with (y1-y2)/(x1-x2) so in this case, it would be (0-(-2))/(-6-0)= -1/3 and you can find y=1/3x-2.

Perpendicular Slope:

The slope of any perpendicular line would be the reciprocal of the original. So the reciprocal of -1/3 is 3. The b can be any real number because no matter the y-intercept the slope would still be perpendicular

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Answer:

Options (1), (3), and (5)

Step-by-step explanation:

In the figure attached,

An inequality has been graphed and the shaded region shows its solutions.

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3 years ago

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2 years ago

Management at Webster Chemical Company is concerned as to whether caulking tubes are being properly capped. If a significant pro

marta [7]

Answer:

$UCL = 0.078$

$LCL = 0$

Step-by-step explanation:

<em>Poorly formatted question (see attachment)</em>

From the attachment, we have:

$n = 135$ --- the sample size

$Samples = 20$ --- The number of samples

$\sum np = 87$ --- Number of leaking tubes

First, we calculate the number of observations

$\sum n = n * Samples$

$\sum n= 135 * 20$

$\sum n= 2700$

Using 3 sigma control limit, we have:

$z = 3$

Calculate $\bar p$

$\bar p = \frac{\sum np}{\sum n}$

So, we have:

$\bar p = \frac{87}{2700}$

$\bar p = 0.0322$

Next, calculate the standard deviation

$s_p = \sqrt{\frac{\bar p* (1-\bar p)}{n}}$

So, we have:

$s_p = \sqrt{\frac{0.0322* (1-0.0322)}{135}}$

$s_p = \sqrt{\frac{0.03116}{135}}$

$s_p = \sqrt{0.0002308}$

$s_p = 0.0152$

The control limits is then calculated as:

$UCL = \bar p + z * s_p$ --- upper control limit

$LCL = \bar p - z * s_p$ --- lower control limits

So, we have:

$UCL = \bar p + z * s_p$

$UCL = 0.0322 +3 * 0.0152$

$UCL = 0.078$

$LCL = \bar p - z * s_p$

$LCL = 0.0322 - 3 * 0.0152$

$LCL = -0.013$

Since the calculated LCL is less than 0, we simply set it to 0

$LCL = 0$

So, the p chart control limits are:

$UCL = 0.078$

$LCL = 0$

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2 years ago