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taurus [48]
3 years ago
10

Explain how scientific investigations that preceded Lamarck , contributed to the development of his theory of evolution

Biology
2 answers:
Andru [333]3 years ago
4 0

Lamarck noticed many similarities among a variety of animals and was interested in the fossil record.


This led Lamarck to argue that organisms changed as a result of changes in their environment.

mihalych1998 [28]3 years ago
3 0
Lamarck developed the theory of acquired characteristics that organisms changed during their lifetime to accommodate their environment and that these acquired characteristics became inherited in their offspring. This theory was preceded for example by Erasmus Darwin who believed that "improvements" to organisms could be passed on or inherited.
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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

8 0
3 years ago
In order to help you answer this true/false statement, which words should be underlined?:
juin [17]
A only in Northern America
3 0
3 years ago
Now it is time to do some word problems! In pea plants purple flower color is dominant over white flower color. A pea plant that
soldier1979 [14.2K]

Answer:

All the offsprings would be heterozygous purple colored flower

Explanation:

Given

Purple color flower is dominant over white color flower

Let the allele for purple color of flower be represented by "P"

and the allele for white color flower be represented by "p"

A pea plant that is homozygous purple-flowered is crossed with a pea plant that is white-flowered.

Genotype of homozygous purple-flower is "PP"

and Genotype of homozygous white-flower is "pp"

Cross between PP and pp produce the following offspring

PP * pp

Pp, Pp , Pp , Pp

All the offsprings would be heterozygous purple colored flower

5 0
3 years ago
MULTIPLE CHOICE
adoni [48]

Answer:

B

Explanation:

Its 23 in mitosis cell division

8 0
3 years ago
If the blood drops above were found in two locations of a crime scene, from which direction did the blood come in each case? Dro
lawyer [7]

Answer:

What is a. A; A

Explanation:

Drop 1 struck the surface from direction A, and drop 2 struck the surface from direction A.

6 0
3 years ago
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