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3 years ago

This hanger is in balance. There are two labeled weights of 8 grams and 20 grams.The four circles each have the same weight.

Mathematics

1 answer:

Svet_ta [14]3 years ago

5 0

Answer:

each circle weight 3 grams

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one. Please have a look at the attached photo,

<em>This hanger is in balance. There are two labeled weights of 8 grams and 20 grams.The four circles each have the same weight.</em>

<em>what is the weight of each circle.</em>

My answer:

Let x is the the weight of the circle

Because this hanger is in balance, so we can form the equation:

20 = 8 + 4x

<=> 4x = 20-8

<=> 4x = 12

<=> x = 3

So each circle weight 3 grams

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Translate the sentence into an equation.

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2 years ago

A number written in the form of p by Q is called

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Answer:

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3 years ago

Read 2 more answers

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:

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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4 years ago

List all the 4-digit even numbers that can be created by rearranging these number tiles.

Stels [109]

2000

4000

6000

8000

2460

2440

2420

4280

4260

4220

6666

Step-by-step explanation:

2000

2420

2440

2460

4000

4220

4260

4280

6000

6666

8000

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3 years ago