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Aloiza [94]
2 years ago
11

Please help will give brainliest

Mathematics
2 answers:
natulia [17]2 years ago
7 0

Answer:

A) it wouldn't fit

Step-by-step explanation:

sveta [45]2 years ago
5 0

Answer:

its b i did that befroe and it was b

Step-by-step explanation:

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wel

Answer:Count all of the windows that look like rectangles and put your answer below and explain the method you used multiplication, addition.

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Solve the quadratic equation:<br> 3x^2+x-5=0
adell [148]

3x²+x-5=0

a = 3, b = 1, c= -5

-> ∆ ( delta ) = b²-4ac = 61 > 0

-> x1 =( -b+√∆ )÷ 2a =...

x2 = (-b-√∆)÷2a =...

p/s: do your teachers teach you how to use ∆ ( delta ) in maths calculation ? i live in europe and our teachers teach us that way. however, it is a rịght and fast way. you should learn it.

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2 years ago
Write the coordinate rule to describe The translasion​
sergejj [24]

Answer:

The coordinates are:

A ( 4,7 )

B ( 8,10 )

c ( 7,7 )

Step-by-step explanation:

You could take the original coordinates and add the x value with 3 and y value with 6. Like for example...

The original coordinates of C is ( 4,1 ). Hence you could add the x value with 3 and y value with 6. 4 + 3 and 1 + 6 giving you the coordinates after translation = ( 7,7 )

6 0
2 years ago
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

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