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lora16 [44]
3 years ago
15

Workland has a population of 10,000, of whom 7,000 work 8 hours a day to produce a total of 224,000 final goods. Laborland has a

population of 5,000, of whom 3,000 work 7 hours a day to produce a total of 105,000 final goods.
Mathematics
1 answer:
kenny6666 [7]3 years ago
7 0

Answer:

Workland has lower productivity but higher real GDP per person than Laborland.

Step-by-step explanation:

Consider the provided information.

Workland has a population of 10,000, of whom 7,000 work 8 hours a day to produce a total of 224,000 final goods.

Productivity = Output / Input

For Workland productivity is:

\frac{224,000}{7,000 \times 8}=\frac{224,000}{56,000}= 4

Laborland has a population of 5,000, of whom 3,000 work 7 hours a day to produce a total of 105,000 final goods.

For Laborland, productivity is:

\frac{105,000 }{3,000\times 7}=\frac{105,000 }{21,000 }= 5

Thus, Laborland has higher productivity .

To figure real GDP per person, divide output by population:

For Workland,

\frac{224,000}{10,000}= 22.4

for Laborland,

\frac{105,000 }{5,000 }= 21

Thus, Workland has a higher real GDP per person.

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A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t
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Answer:

  • <u>Question 1:</u>      dm/dt=0.2m<u />

<u />

  • <u>Question 2:</u>     m=Ae^{(0.2t)}<u />

<u />

  • <u>Question 3:</u>      m=10e^{(0.2t)}<u />

<u />

  • <u>Question 4:</u>      m=10g<u />

Explanation:

<u>Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m</u>

<u></u>

a) By definition:  m'=dm/dt

b)  Given:  rate=0.2m

c) By substitution:  dm/dt=0.2m

<u>Question 2: Find the general solution of this equation. Use A as a constant of integration.</u>

a) <u>Separate variables</u>

     dm/m=0.2dt

b)<u> Integrate</u>

           \int dm/m=\int 0.2dt

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c) <u>Antilogarithm</u>

       m=e^{0.2t+C}

       m=e^{0.2t}\cdot e^C

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<u>Question 3. Which particular solution matches the additional information?</u>

<u></u>

Use the measured rate of 4 grams per hour after 3 hours

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First, find the mass at t = 3 hours

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Now substitute in the general solution of the differential equation, to find A:

          m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976

Round A to 1 significant figure:

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<u>Particular solution:</u>

           

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<u>Question 4. What was the mass of the bacteria at time =0?</u>

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