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Contact [7]
3 years ago
11

A. When comparing the "# of Bacteria" each hour, what is the number being multiplied by each time? *

Mathematics
2 answers:
LenaWriter [7]3 years ago
6 0

Answer:

Part a) The number is 8

Part b) The initial number of bacteria is 4

Part c) y=4(8)^x

Part d) y=100(8)^x

Step-by-step explanation:

we know that

The equation of a exponential function is equal to

y=a(b)^x

where

y is the number of bacteria

x is the time in hours

b is the base of the exponential function

a is the initial number of bacteria

Part a) When comparing the "# of Bacteria" each hour, what is the number being multiplied by each time?

we know that

For x=1 h ----> y=32 bacteria

For x=2 h ----> y=256 bacteria

For x=3 h ----> y=2,048 bacteria

For x=4 h ----> y=16,384 bacteria

For x=5 h ----> y=131,072 bacteria

For x=6 h ----> y=1,048,576 bacteria

so

256/32=8

2,048/256=8

16,384\2,048=8

131,072/16,384=8

1,048,576\131,072=8

so

the base of the exponential function  b is 8

Part b) What is the initial number of bacteria? (at time zero)

we know that

The number of bacteria at time x=1 hour , divided by the number of bacteria at time x=0 must be equal to 8 (see part a)

so

\frac{32}{a}=8

solve for a

a=32/8=4\ bacteria

Part c) Write a rule for this table

we have

y=a(b)^x

we have

a=4\\b=8

substitute

y=4(8)^x

Part d) Suppose you started with 100 bacteria, but they still grew by the same growth factor. Write the function rule for this situation

In this case

a=100

b=8 ---> the growth factor is the same

so

y=100(8)^x

Gala2k [10]3 years ago
6 0

Answer:

Part A: B

Part B: C

Part C: A

Part D: D

Step-by-step explanation:

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Step-by-step explanation:

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time
Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

\frac{dP}{dt} = Pr

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

\frac{dP}{P} = r dt

\int \frac{dP}{P} = \int r dt

\ln{P} = rt + K

Applying the exponential to both sides:

P(t) = Ke^{rt}

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

P(t) = 300e^{rt}

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

P(t) = 300e^{rt}

1000 = 300e^{24r}

e^{24r} = \frac{1000}{300}

e^{24r} = \frac{10}{3}

\ln{e^{24r}} = \ln{\frac{10}{3}}

24r = \ln{\frac{10}{3}}

r = \frac{\ln{\frac{10}{3}}}{24}

r = 0.05

So

P(t) = 300e^{0.05t}

At what time t is the population 500?

This is t for which P(t) = 500. So

P(t) = 300e^{0.05t}

500 = 300e^{0.05t}

e^{0.05t} = \frac{500}{300}

e^{0.05t} = \frac{5}{3}

\ln{e^{0.05t}} = \ln{\frac{5}{3}}

0.05t = \ln{\frac{5}{3}}

t = \frac{\ln{\frac{5}{3}}}{0.05}

t = 10.22

The population is of 500 after 10.22 hours.

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