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Naya [18.7K]
3 years ago
6

A gallon of paint covers 400 square feet how many square feet will 2 3 8th gallons of paint cover

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
6 0
2 \frac{3}{8} /1=2 \frac{3}{8} 

400*2\frac{3}{8}=950 square feet

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0.21

Step-by-step explanation:

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and if that is made to a decimal it would be 0.21 .

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Which of the following expressions represents a function?
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What one of the shape has a flat surface
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Could someone help and explain their answer? (NO LINKS!)<br> Thanks!
Nonamiya [84]

Answer:

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Step-by-step explanation:

To find the prime factorization of a number, start by dividing the number by each prime number that you can divide by as many times as possible, starting with the smallest prime number and moving up. Then move on to the next prime number. The last division must give a quotient of 1. Then the prime factorization is the product of all the prime factors you divided the number by.

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2, 3, 5, 7, 11, 13, 17, 19, 23, 29

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Suppose the following number of defects has been found in successive samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6
Brut [27]

Answer:

Given the data in the question;

Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.

a)

For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;

UCL = p" + 3√[ (p"(1-P")) / n ]

CL = p"

LCL = p" - 3√[ (p"(1-P")) / n ]

here, p" is the average fraction defective

Now, with the 30 samples of size 100

p" =  [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]

p" = 234 / 3000

p" = 0.078

so the trial control limits for the fraction-defective control chart are;

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]

UCL = 0.078 + ( 3 × 0.026817 )

UCL = 0.078 + 0.080451

UCL = 0.1585

LCL = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]

LCL = 0.078 - ( 3 × 0.026817 )

LCL = 0.078 - 0.080451

LCL =  0 ( SET TO ZERO )

Diagram of the Chart uploaded below

b)

from the p chart for a) below, sample 28 violated the first western electric rule,

summary report from Minitab;

TEST 1. One point more than 3.00 standard deviations from the center line.

Test failed at points: 28

Hence, we conclude that the process is out of statistical control

Lets Assume that assignable causes can be found to eliminate out of control points.

Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.

so

p" = 0.0745

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]

UCL = 0.0745 + ( 3 × 0.026258 )

UCL = 0.0745 + 0.078774

UCL = 0.1532

LCL  = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]

LCL = 0.0745 - ( 3 × 0.026258 )

LCL = 0.0745 - 0.078774

UCL = 0  ( SET TO ZERO )

The second p chart diagram is upload below;

NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits

7 0
2 years ago
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