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saveliy_v [14]
3 years ago
12

In 2004,there were 320 endangered mammals in the world about 75% of these mammals reside outside of the United States.about how

many endangered mammals are not in the United States?
Mathematics
1 answer:
Troyanec [42]3 years ago
4 0
240 endangered mammals do not live in the united stated
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C) 30 minutes

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X + 5 ≤ 13 what's the answer for this question?​
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Step-by-step explanation:

x + 5 ≤ 13

Subtract 5 on both sides...

13 - 5 = 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
3 years ago
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