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oksian1 [2.3K]
3 years ago
6

to find the amount of space cube shaped bird cage occupies,find the cube of measure of one edge of the bird cage. express the am

ount of space occupied by the bird cage shown as a power. then find the amount in cubic units
Mathematics
1 answer:
Anton [14]3 years ago
4 0
Use this formula to help you find the answer V= LxWxH or V=BxH
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24% of what number is 72
nignag [31]

Answer:

300

Step-by-step explanation:

24% × x = 72

or,   24 /100

× x = 72

we have x = 72 ×  100 /24

x = 300

Hope it helps

4 0
3 years ago
Help me please <br> i need help lol
Dmitriy789 [7]

Answer:

<3 and <4 are supplementary

Step-by-step explanation:

<3 and <4 form a linear pair, so they are supplementary (=180 degrees)

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Makovka662 [10]
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3 years ago
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prisoha [69]

Answer:

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3 0
3 years ago
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

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
3 years ago
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