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Leni [432]
2 years ago
14

the amount of gasoline at a gas station changes from 58,432.4 gallon to 56,475.2 gallon from 2:00 pm to 4:00 pm. what rational n

umber represent the average​
Mathematics
1 answer:
jarptica [38.1K]2 years ago
4 0

Answer:

I take it you mean the average gallons lost an hour, in which case you would do 58,432.4 - 56,475.2 to get 1,597.2 to which you divide by 2 getting your final answer of 978.6

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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

Max= 52.56

Mean= 51.16

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

Max= 53.12

Mean= 51.83

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>

<em />

<u>Sample 4</u>

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values surpass the warning interval but do not reach action levels</em>.

<u>Sample 5</u>

Min= 50.35

Max= 52.71

Mean= 51.37

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!

8 0
2 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
2 years ago
Two pounds of sugar cost $1.40. How much sugar do you get per dollar? Round your answer to the nearest hundredth, if necessary.
Vsevolod [243]

Answer:

1.429 lbs per dollar of sugar

Step-by-step explanation:

multiply by 5 get 10 lbs of sugar for 7 dollars divide by 7 get 1.4285 round up

5 0
2 years ago
What does collinear mean
storchak [24]
Lying in the same straight line i think..

5 0
1 year ago
Read 2 more answers
Two standard dice are rolled and their face values multiplied. What is the probability that the product is prime or ends in 0?
Trava [24]
The answer is 10/36 or 27.77%
here’s the working out

8 0
2 years ago
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