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Snowcat [4.5K]
1 year ago
9

Estimate the maximum time a person could listen to a sound of 87 decibels.

Biology
1 answer:
Kazeer [188]1 year ago
4 0

The maximum time a person could listen to a sound of 87 decibels is: 8 hours

<h3>Meaning of Decibels</h3>

Decibels can be defined as the unit used by scientists to measure sound level.The values of Decibel ranges from a very low value to a very high value and the higher we get the more dangerous it is to our human ear.

In conclusion, The maximum time a person could listen to a sound of 87 decibels is 8 hours

Learn more about Decibels and human hearing : brainly.com/question/1291668

#SPJ1

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Select ALL the correct answers.
tekilochka [14]

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}

10th\ run = 9

So:

B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}

Required

Select all true statements

(a) & (d) Median Comparisons

A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}                         B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}

n = 9                                                         n = 10

Arrange the data:

A = \{1, 1, 1, 1, 2, 2,  2,3, 4\}               B = \{1,1,1,1,2,2,2,3,4,9\}

                               Median = \frac{n + 1}{2}th

Median = \frac{9 + 1}{2}th                            Median = \frac{10 + 1}{2}th

Median = \frac{10}{2}th                              Median = \frac{11}{2}th

Median = 5th                                 Median = 5.5}th --- average of 5th and 6th

Median = 2                                    Median = \frac{2+2}{2} = 2

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

A = \{1, 1, 1, 1, 2, 2,  2,3, 4\}               B = \{1,1,1,1,2,2,2,3,4,9\}

n = 9                                                         n = 10

First, calculate the lower quartile (Q1)

Q_1 = \frac{n + 1}{4}th[Odd n]             Q_1 = \frac{n}{4}th [Even n]

Q_1 = \frac{9 + 1}{4}th                            Q_1 = \frac{10}{4}th

Q_1 = \frac{10}{4}th                              Q_1 = 2.5

Q_1 = 2.5th                              

This means that:

Q_1 = 2nd + 0.5(3rd - 2nd)              Q_1 = 2nd + 0.5(3rd - 2nd)

Q_1 = 1 + 0.5(1- 1)                   Q_1 = 1+ 0.5(1 - 1)                      

Q_1 = 1                                       Q_1 = 1

Next, calculate the upper quartile (Q3)

Q_3 = \frac{3}{4}(n + 1)th [Odd n]             Q_3 = \frac{3}{4}(n)th [Even n]

Q_3 = \frac{3}{4}(9 + 1)th                            Q_3 = \frac{30}{4}th

Q_3 = \frac{30}{4}th                                     Q_3 = 7.5th  

Q_3 = 7.5th                                    

This means that:

Q_3 = 7th + 0.5(8th- 7th)           Q_3 = 7th + 0.5(8th- 7th)

Q_3 = 2 + 0.5(3- 2)                       Q_3 = 2+ 0.5(4 - 2)                      

Q_3 = 2.5                                       Q_3 = 3

The interquartile range is  IQR = Q_3 - Q_1

So, we have:

IQR = 2.5 - 1                  IQR = 3 - 1

IQR = 1.5                       IQR  =2

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

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Hope this helps!
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Volcanic eruptions are caused by? A gravity. B Pressure C Explosions D earths rotation please help
IRISSAK [1]

Answer:

I'm pretty sure it's B, pressure.

Explanation:

I hope this helped :)

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