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77julia77 [94]
1 year ago
9

A local health care company wants to estimate the mean weekly elder day-care cost. A sample of 10 facilities shows a mean of $25

0 per week, with a standard deviation of $25. What is the 90% confidence interval for the population mean?
Mathematics
1 answer:
laiz [17]1 year ago
8 0

Answer:

Step-by-step explanation:

Mean "X"=250

s = 25

n=10

Because Standard deviation is unknown and sample size is small, we must employ the T-distribution

df=n-1 = 10-1

df=9

For 90% confidence, t = 1.833

E=t*s/\sqrt{n\\ = 1.833*(25/\sqrt{10})

E=14.49

The 90% confidence interval is X±E=250±14.49 or 235.51, 264.49

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ANSWER

\frac{x - 4}{2x - 2}

EXPLANATION

The given expression is;

\frac{2 {x}^{2}  + 5x + 3}{ {x}^{2}  - 3x - 4}  \div  \frac{4 {x}^{2}  + 2x - 6}{ {x}^{2}  - 8x + 16}

We factor to obtain;

\frac{(x + 1)(2x + 3)}{(x - 4)(x + 1)}  \div  \frac{2(x - 1)(2x + 3)}{(x - 4)(x - 4)}

Multiply by the reciprocal of the second fraction

\frac{(x + 1)(2x + 3)}{(x - 4)(x + 1)}  \times \frac{(x - 4)(x - 4)}{2(x - 1)(2x + 3)}

Cancel out common factors to get,

\frac{1}{1}  \times \frac{(x - 4)}{2(x - 1)}

\frac{x - 4}{2x - 2}

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Step-by-step explanation:

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1 year ago
A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained
sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

<u>Sample 1</u>

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.</em>

<em><u /></em>

<u>Sample 2 </u>

Min= 50.32

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Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both the max and min values of the sample are within the working range without warning.</em>

<em />

<u>Sample 3</u>

Min= 50.25

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Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.</em>

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<u>Sample 4</u>

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Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values are within working levels without warnings.</em>

<em />

Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:

Shutdown: Sample 1, 3 and 4

Do not shutdown: Sample 2 and 5.

I hope it helps!

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