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OLEGan [10]
2 years ago
13

Ano ang naging damdamin ni jacob nang nalaman niya ang kalagayan ni jose.

SAT
1 answer:
zheka24 [161]2 years ago
7 0

Answer:1.naging malungkot

you got this keep trying harder

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The correct answer is letter A.
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The Cancer Conference (Tumor Board) qualifies as a ............................. and therefore can share some PHI with other phy
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The reason why the Cancer Conference (Tumor Board) can share some PHI with other physicians is that it is a not covered entity.

<h3 /><h3>What is a not covered entity?</h3>

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1 year ago
WHO WANTS FREEEEE POINTSSSSSSSS
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2 years ago
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The average waist size for teenage males is 29 inches with a standard deviation of 1. 4 inch. If waist sizes are normally distri
sweet [91]

Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of \mu = 29.
  • The standard deviation is of \sigma = 1.4.

The proportion of teenagers who will have waist sizes greater than 31 inches is <u>1 subtracted by the p-value of Z when X = 31</u>, hence:

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

Z = \frac{31 - 29}{1.4}

Z = 1.43

Z = 1.43 has a p-value of 0.9236.

1 - 0.9236 = 0.0764.

0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.

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

8 0
2 years ago
Helppppppp
nadezda [96]

Answer:

B

Explanation:

Lifeguards have to work outside on the beach.

Hope this helped

8 0
3 years ago
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