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miskamm [114]
2 years ago
10

PLEASE HELP ASAP!! will mark brainlest. please solve & show work for number 8.

Mathematics
1 answer:
Studentka2010 [4]2 years ago
4 0

Answer:

1 and 4

Step-by-step explanation:

Because it says it is one degree forward which is the angle for number one and if you look at the measurementjust add four two DBC

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1:2

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If its just apples to bananas then it would be 1:2

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The savings account offering which of these APRs and compounding periods
Ann [662]

Answer:

The answer is "Option C"

Step-by-step explanation:

The using formula= (1+\frac{r}{n})^n -1

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→ n = compounded value

In choice a:

When compounded is monthly 4.0784\%  

n = 12

\to (1+ \frac{0.040784}{12})^{12} = 1.0403-1 = 0.0403

In choice b:

When compounded is quarterly4.0792\%

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In choice c:

Whenn compounded is daily 4.0730 \%

n = 365\\\\\to (1+ \frac{0.040730}{12})^{365}  = 3.328-1 = 2.328

In choice d:

When compounded is semiannually4.0798\%

n = 2\\\\\to (1+ \frac{0.040798}{12})^{2} = 1.0066-1 = 0.0066

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What is the product?
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Answer:

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1 year ago
Evaluate 2b^2-4a+4a^2 for a= 3 and b= -8
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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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