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Ymorist [56]
1 year ago
15

ABC High School is debating whether or not to write a policywhere all students musthave uniforms and wear them during school hou

rs. In a survey,45% of the studentswanted uniforms, and 55% did not.Calculate the probability that a person selected at random fromABC High School will want the school to have uniforms or willnot want the school to have uniforms.
Mathematics
1 answer:
larisa [96]1 year ago
6 0

If 45% don't want uniforms, it means that 45 out of 100 students don't want them, so the probability is 45/100 = 9/20

And 55% want unifrms, so the probability = 55/100 = 11/20

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Answer:

B

Step-by-step explanation:

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8 0
3 years ago
Here are yesterday's high temperatures (in Fahrenheit) in 12 U.S. cities. 48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80 Notice
irinina [24]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

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Upper quartile = (74+74)/2

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Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

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Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

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Learn more on Measures of a Data here: brainly.com/question/15097997

#SPJ1

5 0
2 years ago
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6 0
3 years ago
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omeli [17]

Answer:

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Step-by-step explanation:

6 0
2 years ago
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54%

Step-by-step explanation:

total  animals=27+23=50

dogs=\frac{27}{50} \times100=54

6 0
3 years ago
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