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3 years ago

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5} + \sqrt{4} " alt="1 \frac{3}{5} + \sqrt{4} " align="absmiddle" class="latex-formula">

1 answer:

topjm [15]3 years ago

3 0

Step-by-step explanation:

Check out the pic, if it doesn't help you can ask me whichever part you don’t understand

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4 years ago

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3 years ago

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3 years ago

For 113 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for servin

RideAnS [48]

Answer:

upper control limit $UCL=7.1632$

Concept :

$UCL=\bar{\bar{x}}+A_2\bar{R}$ where $A_2=0.34$

$\bar{\bar{x}}=\frac{\bar{x}}{n}$

$S=\frac{\bar{R}}{d}\Rightarrow \bar{R}=S\times d$ where $S=0.5, d=2.97$

Given :

She took a sample of per day $(n)$ is $9$ bottles.

The standard deviation $S$ of all bottles is $0.5$ degree.

Average across all $1017$ bottles $\bar{x}$ are $60$ degree Fahrenheit.

To find :

The value of the upper control limit (UCL)

Explanation :

Here, firstly we find the value of $\bar{\bar{x}}$ .

$\because\bar{\bar{x}}=\frac{\bar{x}}{n}$

$\therefore \bar{\bar{x}}=\frac{60}{9}$

$\Rightarrow \bar{\bar{x}}=6.66$ and $\bar{R}=S\times d$

$\Rightarrow \bar{R}=0.5\times2.97=1.48$

$\Rightarrow \bar{R}=1.48$

Now, $UCL=\bar{\bar{x}}+A_2\bar{R}$

$\Rightarrow UCL=6.66+0.34\times1.48$

$\Rightarrow UCL=6.66+0.5032$

$\Rightarrow UCL=7.1632$ degrees.

4 0

3 years ago