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earnstyle [38]
3 years ago
13

Two ad agencies want to understand issues that are important to students. Agency A will interview 50 students whose names are ra

ndomly chosen from school enrollment data. Agency B will interview 50 randomly selected people from those who respond to a social media posting. Which method will produce a fair sample of the student population and why?
A. Agency A, because students are convenient and easy to interview
B. Agency A, because names randomly selected from school data will be student names
C. Agency B, because people who respond to social media postings have strong opinions
D. Agency B, because people who respond to social media postings are usually students
Mathematics
1 answer:
cluponka [151]3 years ago
7 0
These ad agencies must focus on their target audience, which are the students. Hence, they should gather data on the pool that will surely comprise of students. For agency B, social media posting is not a good source pool. It's true that students are very participative and opinionated in social media. However, they can't be sure that these are students. Some parents are in social media, as well. Some are working individuals, and some are out of school youth. Unlike agency A, agency B has to sort out profiles first and identify which ones are students. Hence, agency A will produce a fair sample of the student population because it is unarguably true that everyone in the school enrollment data are students.

The answer is B.
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A bacteria culture starts with 660 bacteria and grows at a rate proportional to its size. After 6 hours there will be 3960 bacte
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Answer:

a) P(t) = 660 e^{\frac{ln(6)}{6} t}

b) P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617

Step-by-step explanation:

For this case we have the following info:

P(0) = 660 represent the initial amount of bacteria

P(6) = 3960 represent the population of bacteria after 6 hours

Part a

For this case we use the proportional model given by:

\frac{dP}{dt}= kP

Where P represent the population at time t and k is a constant of proportionality.

We can rewrite the model like this:

\frac{dP}{P} = k dt

And if we integrate both sides we got:

ln |P|= kt + C

Now if we apply exponential we got:

P = e^{kt +c}= e^{kt} e^C

P(t)= P_o e^{kt}

For this case using the initial condition:

660 = P_o e^{0k}, P_o = 660

We have the following model:

P(t) = 660 e^{kt}

Using the second condition P(6) = 3960 we have this:

3960 = 660e^{6k}

We can divide both sides by 660 and we got:

6= e^{6k}

Now we can apply natural log on both sides and we got:

ln (6)= 6k

k = \frac{ln(6)}{6}=0.2986265782

So then the model would be given by:

P(t) = 660 e^{\frac{ln(6)}{6} t}

Part b

For this case we just need to replace t=3 into the model and we got:

P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617

4 0
3 years ago
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