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**Do Now: #16 on p.348 It will take about 10.319 years to reduce the**

Assume that the relative rate of spread of a certain disease is constant. Suppose also that in the course of any given year the number of cases of the disease is reduced by 20%. (a) If there are 10,000 cases today, how many years will it take to reduce the number to 1000? The variables: C = cases, t = years The model: It will take about years to reduce the number of cases to 1000.

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**It will take about 44.382 years to completely eradicate the disease.**

Do Now: #16 on p.348 Assume that the relative rate of spread of a certain disease is constant. Suppose also that in the course of any given year the number of cases of the disease is reduced by 20%. (b) How long will it take to eradicate the disease, that is, to reduce the number of cases to less than 1? It will take about years to completely eradicate the disease.

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**Applications with population Growth**

Section 6.5b

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Guided Practice A certain wild animal preserve can support no more than 250 lowland gorillas. Twenty-eight gorillas were known to be in the preserve in Assume that the rate of growth of the population is where t is time in years since 1970. (a) Find a formula for the gorilla population in terms of t.

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Guided Practice A certain wild animal preserve can support no more than 250 lowland gorillas. Twenty-eight gorillas were known to be in the preserve in Assume that the rate of growth of the population is where t is time in years since 1970. (a) Find a formula for the gorilla population in terms of t. Solve for A: The model:

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Guided Practice A certain wild animal preserve can support no more than 250 lowland gorillas. Twenty-eight gorillas were known to be in the preserve in Assume that the rate of growth of the population is where t is time in years since 1970. (b) How long will it take for the gorilla population to reach the carrying capacity of the preserve? years after 1970

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**Guided Practice Time to grow from 5000 to 10,000**

A population of honeybees grows at an annual rate equal to 1/4 of the number present when there are no more than 10,000 bees. If there are more than 10,000 bees but fewer than 50,000 bees, the growth rate is equal to 1/12 of the number present. If there are 5000 bees now, when will there be 25,000 bees? Time to grow from 5000 to 10,000

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**Guided Practice Time to grow from 10,000 to 25,000**

A population of honeybees grows at an annual rate equal to 1/4 of the number present when there are no more than 10,000 bees. If there are more than 10,000 bees but fewer than 50,000 bees, the growth rate is equal to 1/12 of the number present. If there are 5000 bees now, when will there be 25,000 bees? Time to grow from 10,000 to 25,000

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**Guided Practice years Total Time Required**

A population of honeybees grows at an annual rate equal to 1/4 of the number present when there are no more than 10,000 bees. If there are more than 10,000 bees but fewer than 50,000 bees, the growth rate is equal to 1/12 of the number present. If there are 5000 bees now, when will there be 25,000 bees? Total Time Required years

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**Guided Practice Solve the given initial value problem.**

Initial Condition: Solution:

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**Guided Practice Solve the given initial value problem.**

Initial Condition: Solution:

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1-1 Patterns and Inductive Reasoning

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