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11111nata11111 [884]
3 years ago
6

Mammals have approximately the same density as fresh water. (a) Find the volume in liters of a 75 kg woman and (b) the volume in

cubic meters of a 121,165 kg blue whale. (Note: 1 liter = 10−3 m3 = 0.001 m3.) Round all your answers to the nearest integer.
Mathematics
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

(a) The volume of the woman in liters is 75 L

(b) The volume of the blue whale in cubic meters is 121 m³

Step-by-step explanation:

From the question,

Mammals have approximately the same density as fresh water.

The density of fresh water is 1000 kg/m³,

∴ The density of mammals is approximately 1000 kg/m³.

Now, we will convert this to kg/L

1L = 0.001 m³

∴ 1 m³ = 1/0.001 L = 1000L

Hence, 1000 kg/m³ = 1000kg/1000L = 1 kg/L

The density of mammals is 1 kg/L

(a) To find the volume in liters of a 75 kg woman,

From the formula

Density = Mass / Volume

Then, Volume = Mass / Density

Mass of the woman = 75 kg

Density = 1 kg/L

∴ Volume = 75kg / 1kg/L = 75 L

Hence, the volume of the woman in liters is 75 L

(b) To find the volume in cubic meters of a 121,165 kg blue whale.

Using the same formula

Volume = Mass / Density

Mass = 121165 kg

Density = 1000 kg/m³

∴ Volume = 121165kg / 1000 kg/m³ = 121.165 m³ ≅ 121 m³

Hence, the volume of the blue whale in cubic meters is 121 m³.

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Hello!

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Now, we can multiply in accordance with the FOIL method.

First: 7x · 3x = 21x²

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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

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Step-by-step explanation:

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

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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

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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

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