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AnnyKZ [126]
3 years ago
12

This partial circle has a radius of 11 inches.

Mathematics
2 answers:
shutvik [7]3 years ago
7 0

im pretty sure the answer is 284.96 in2

hope this helps

Alecsey [184]3 years ago
4 0
First, find the area of the circle
 pie*radius to the power of 2(radius*raduis)
(11*11=121) 121*3.14=379.94
take your answer and divide it by 4 we would divide it by 2 if we had a semicircle but since we have a 3/4 circle we divide by 4 and then times the Quotient by 3
379.94/4=94.985
94.985*3=284.955
Round ur answer to the nearest tenth so it equals 284.96
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Answer:

with 0.10 level of significance the P-VALUE that would be used in the hypothesis claim is 0.05%

Step-by-step explanation:

In hypothesis testing in statistics, we can say that the p-value is a probability of obtaining test results when we assume that the null hypothesis is correct.

The p-value is the probability that the null hypothesis is true.

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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

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Answer:

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Step-by-step explanation:


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Step-by-step explanation:

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