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viktelen [127]
3 years ago
9

How to find the area of a heptagon?

Mathematics
1 answer:
Leona [35]3 years ago
5 0
Regular heptagonSolve for <span>area

</span>A= 7/4 x a^2 x cot(180 degrees/7)

https://www.google.com/search?sourceid=chrome-psyapi2&rlz=1C1ZQQI_enUS697US697&ion=1&espv=2&ie=UTF-8&q=area%20of%20a%20heptagon%20formula&oq=area%20of%20a%20heptagon&aqs=chrome.1.69i57j0l5.4065j0j7

https://www.google.com/search?sourceid=chrome-psyapi2&rlz=1C1ZQQI_enUS697US697&ion=1&espv=2&ie=UTF-8<span>&q=area%20of%20a%20heptagon%20formula&oq=area%20of%20a%20heptagon&aqs=chrome.1.69i57j0l5.4065j0j7</span>
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Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
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Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
4 years ago
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