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Ilya [14]
1 year ago
13

Claire lives 2.2 miles away from the library. She decides to walk to the library with her friend. They have already walked 0.53

miles. How much further do they need to walk to reach the library? Show your work.
Mathematics
1 answer:
Dmitry [639]1 year ago
8 0

Answer:

2/3

Step-by-step explanation:

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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
Which is greater 9000cm or 2 km?
satela [25.4K]
Convert 2 km into cm

1 km = 100,000 cm

2 km = 200,000 cm

therefore 9,000 cm is greater
6 0
1 year ago
The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the
Harman [31]

THE QUESTION IS NOT PROPERLY WRITTEN

The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If t1 is the time (in seconds) it takes for the object to strike the water, then t1 obeys the equation s = 16t1², where s is the distance (in feet) Solving for t1, we have t1 = √s/4. Let t2 be the time it takes for the sound of the impact to reach your ears. Since sound waves travel at a speed of approximately 1100 feet per second, the time to travel the distance s is t2 = s/1100. Now t1 +t2 is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation; Total elapsed time √s/4 + s/1100.

Find the distance to the water’s surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Answer:

229.94 feet

Step-by-step explanation:

Given

t1 = √s/4

t2 = s/1100

From the question, we understand that

t1 + t2 = 4 seconds

Let y = √s

So,

t1 + t2 = 4 becomes

√s/4 + s/1100 = 4

y/4 + y²/1100 = 4 ------ Solve the fractions

(275y + y²)/1100 = 4

275y + y² = 4 * 1100

275y + y² = 4400 ------- Rearrange

y² + 275y - 4400 = 0 (Quadratic Equation)

The standard form of a quadratic equation is ax² + bx + c = 0

Where x =

frac{-b+-\sqrt{bx^{2} - 4ac } }{2a}

Here a = 1, b = 275 and c = -4400

So,

y = (-275+- √(275² - 4*1*-4400))/2

y = (-275+- √75625 + 17600))/2

y = (-275+- √(93225))/2

y = (-275 +- 305.328)/2

y = (-275 + 305.328)/2 or (-275-305.328)/2

y = 30.328/2 or -580.328/2

y = 15.164 or -290.164 -------Negative is not applicable

So, y = 15.164

But y = √s

So, s = y²

S = 15.164²

s = 229.94 ft

6 0
2 years ago
Sheila has $47 to buy graphing pads for her math class. Each pad costs $8,
bija089 [108]

Answer:

5 tablets can be bought

Step-by-step explanation:

47/8= 5

8 0
2 years ago
You work two jobs. You earn $11.90 per hour as a salesperson and $10.50 per hour stocking shelves. Your combined earnings this m
Delicious77 [7]
178.50 = 11.90y + 10.50x ⇒ equation in standard form.

4 0
2 years ago
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