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Lilit [14]
3 years ago
9

Assume the exponential growth model ​A(t)equals=Upper A 0 e Superscript ktA0ekt and a world population of 5.95.9 billion in 2006

2006. If the population must stay below 2424 billion during the next 100​ years, what is the maximum acceptable annual rate of​ growth?
Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0

Answer:

1.4% is the maximum acceptable annual rate of​ growth such that the population must stay below 24 billion during the next 100​ years.

Step-by-step explanation:

We are given the following in the question:

The exponential growth model ​is given by:

A(t) = A_0e^{kt}

where k is the growth rate, t is time in years and A_0 is constant.

The world population is 5.9 billion in 2006.

Thus, t = 0 for 2006

A_0 = 5.9\text{ billions}

We have to find the maximum acceptable annual rate of​ growth such that the population must stay below 24 billion during the next 100​ years.

Putting these values in the growth model, we have,

24 = 5.9e^{100k}\\\\k = \dfrac{1}{100}\ln \bigg(\dfrac{24}{5.9}\bigg)\\\\k = 0.01403\\k = 0.01403\times 100\% = 1.4\%

1.4% is the maximum acceptable annual rate of​ growth such that the population must stay below 24 billion during the next 100​ years.

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An express train travels 80km/h from Ironton to Wildwood. A local train, traveling at 48km/h, takes 2 hours longer for the same
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Answer:

They are 240 kilometers apart.

Step-by-step explanation:

This question can be solved using the following relation:

v = \frac{d}{t}

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An express train travels 80km/h from Ironton to Wildwood.

Here we have v = 80, t = x. So

v = \frac{d}{t}

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A local train, traveling at 48km/h, takes 2 hours longer for the same trip.

This means that when v = 48, t = x + 2[tex]. So[tex]v = \frac{d}{t}

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