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Angelina_Jolie [31]
2 years ago
14

Suppose that you began a one-year study of tuberculosis (TB) in a subsidized housing community in the Lower East Side of New Yor

k City on January 1st, 2010. You enrolled 500 residents in your study and checked on their TB status on a monthly basis. At the start of your study on January 1st, you screened all 500 residents. Upon screening, you found that 20 of the healthy residents were immigrants who were vaccinated for TB and so were not at risk. Another 30 residents already had existing cases of TB on January 1st. On February 1st, 5 residents developed TB. On April 1st, 5 more residents developed TB. On June 1st, 10 healthy residents moved away from New York City were lost to follow-up. On July 1st, 10 of the residents who had existing TB on January 1st died from their disease. The study ended on December 31, 2010. Assume that once a person gets TB, they have it for the duration of the study, and assume that all remaining residents stayed healthy and were not lost to follow-up.
Is the subsidized housing community in the Lower East Side of New York City a dynamic or fixed population? Briefly explain the rationale for your answer.

Dynamic population it can changeable people can move out and move in into the population of a subsidized housing

What was the prevalence of TB in the screened community on January 1st?
Mathematics
1 answer:
Rufina [12.5K]2 years ago
3 0

Answer:

The subsidized housing community in the lower East side of New York City is a dynamic population because people moved out of the population and others died while the survey was still being conducted.

The prevalence of TB on the screened community on January 1st is \frac{30}{500}  or = 0.06

Step-by-step explanation:

Prevalence is given by; the number of persons affected by TB/ total number of people in the the study or survey.

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Answer:

h=7

Step-by-step explanation:

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Simplify to get:

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We expand to get:

h+5+2h-10=16

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scale factor=0.4

</span><span><u>Explanation </u>
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That is the ratio of distance from the origin to the image to the distance from the origin to the object.
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3 \sqrt{256 \ \times x^{10} y^{7} }

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square root of 256 is 16 :
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= 48 \sqrt{x^{10} y^{7}}

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√10¹⁰ = 10⁵ :
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= 48x^5 \sqrt{ y^{7}}

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