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Pepsi [2]
3 years ago
12

Within the national incident management system characteristics the concept of common terminology covers all of the following exc

ept:
A. Resource Descriptions
B. Technical Specifications
C. Incident Facilities
D. Organizational Functions
SAT
2 answers:
Lelu [443]3 years ago
8 0

Answer:

B. <em>Technical Specifications</em>  

777dan777 [17]3 years ago
7 0

<u>ANSWER: </u>

Within the "National Incident Management System characteristics", the concept of the common terminology covers all except "Technical specifications".

<u>EXPLANATION: </u>

National Incident Management System is a widespread, national methodology to incident management, and it applies to all the jurisdiction functionalities.

Common terminology is one of the features in (NIMS), that promotes the usage of common terminology for efficient communication purposes. It includes Resource description, Incident Facilities, and Organizational Functions whereas Technical Specification is an uncommon terminology.

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In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 A
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Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:

\mu = np = 200(0.22) = 44

\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583

The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{30.5 - 44}{5.8583}

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

0.0107 = 1.07% probability that more than 30 are single.

More can be learned about the normal distribution at brainly.com/question/24663213

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