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iren2701 [21]
4 years ago
13

What is the source of the pressure that has caused coyotes to adjust to a more nocturnal behavior

Biology
1 answer:
butalik [34]4 years ago
4 0
<h2>Nocturnal Behavior</h2>

Explanation:

  • The wellspring of the weight that has caused coyotes, which were once basically diurnal, to change in accordance with a progressively nighttime conduct is the nearness of increasingly forceful diurnal predators, <em>all the more ordinarily, lions. Lions just chase in the day which constrains the movement of the coyotes.  </em>
  • When living in nearness to people, coyotes will in general be nighttime yet may likewise be dynamic in the early morning and at dusk.
  • In zones with next to zero human movement, coyotes will chase during the day, and when a litter of puppies should be taken care of, they may need to chase nonstop.  
  • <em>Coyotes normal 25 pounds, yet can arrive at 50 pounds, and measure 40 to 50 creeps from nose to tail tip</em>.
  • They can be dynamic day or night, yet are regularly generally dynamic at day break and nightfall.
  • That is the point at which you can hear them s<em>peaking with barks, howls and cries.</em>
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Select ALL the correct answers.
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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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