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SpyIntel [72]
3 years ago
10

There are 70,000 bacteria present in a culture. An antibiotic is introduced to the culture and the number of bacteria is reduced

by half every 4 hours. Which of the following statements are true? Select all that apply

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

24 hours after the antibiotic is introduced, the number of bacteria in the culture is reduced to 1093; The amount of bacteria present in the culture during the 24 hours after the antibiotic is introduced is an example of exponential decay; The function B(x) = 70000(0.5)ˣ, where x represents the number of 4 hour periods, represents the amount of bacteria present after the antibiotic is introduced.

Step-by-step explanation:

In any situation where an amount grows or declines by a set percent every time period can be represented using an exponential function.  Since the amount of bacteria is reduced by half, or 50%, every 4 hours, this is exponential decay.

Exponential functions such as this are of the form

y = a(1+r)ˣ, where a represents the initial population, r is the percent of growth or decrease, and x is the number of time periods.  In this situation, the initial population, a, is 70000.  Since the number of bacteria decreases by half, r = -0.50.  This gives us

y = 70000(1+-0.50)ˣ, or y = 70000(0.50)ˣ.

In 24 hours, there are 6 4-hour periods.  This means x = 6; this gives us

y = 70000(0.50)⁶ = 1093.75.

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vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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