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Lady_Fox [76]
3 years ago
6

Browning Labs is testing a new growth inhibitor for a certain type of bacteria. The bacteria naturally grows exponentially each

hour at a rate of 6.2%. The researchers know that the inhibitor will make the growth rate of the bacteria less than or equal to its natural growth rate. The sample currently contains 100 bacteria.The container holding the sample can hold only 300 bacteria, after which the sample will no longer grow. However, the researchers are increasing the size of the container at a constant rate allowing the container to hold 100 more bacteria each hour. They would like to determine the possible number of bacteria in the container over time.Create a system of inequalities to model the situation above, and use it to determine how many of the solutions are viable.

Mathematics
2 answers:
Hatshy [7]3 years ago
7 0

Answer:

Look at the attachment

Step-by-step explanation:

First we need to find out the equations that will represent each inequality:

For the bacteria:

This is an exponential growth equation, the formula is simple:

y≤n*(1+r)^{x}  where n is the starting point of the sample, r is the rate and x is the variable dependent on time so:

y≤100*(1+0.062)^{x}

y≤100*(1.062)^{x}

For the container:

This is a line equation, following the formula:

y<mx+b where m is the slope or growing rate (100 more per hour), and b is the starting point (300 bacteria)

y< 100x+300

The graph will be like is showed in the attachment, and the solution is the intersecting area to the right of both functions, since they are trying to find out if the inhibitor works, the rate of growth will be equal or smaller than 6.2% thus closing in to 100 bacterias as a constant in time if it works.

makvit [3.9K]3 years ago
3 0

Hey! I just answered this on plato. the answer is that it includes negative factors, which makes not all solutions viable.

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2 years ago
In a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of th
11111nata11111 [884]

Answer:

Required probability is 0.784 .

Step-by-step explanation:

We are given that in a large statistics course, 74% of the students passed the first exam, 72% of the students pass the second exam, and 58% of the students passed both exams.

Let Probability that the students passed the first exam = P(F) = 0.74

     Probability that the students passed the second exam = P(S) = 0.72

     Probability that the students passed both exams = P(F \bigcap S) = 0.58

Now, if the student passed the first exam, probability that he passed the second exam is given by the conditional probability of P(S/F) ;

As we know that conditional probability, P(A/B) = \frac{P(A\bigcap B)}{P(B) }

Similarly, P(S/F) = \frac{P(S\bigcap F)}{P(F) } = \frac{P(F\bigcap S)}{P(F) }  {As P(F \bigcap S) is same as P(S \bigcap F) }

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Therefore, probability that he passed the second exam is 0.784 .

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Elenna [48]

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53  is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

a)

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

so

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

b)

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) =  2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53  is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.

6 0
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