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amm1812
2 years ago
10

Hemophilia is an X-linked recessive trait. Using the genotypes above, complete a

Biology
1 answer:
horsena [70]2 years ago
3 0

Answer:

The answer is 50%..........

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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
What causes waves to bend?
N76 [4]
Gravity is one reason but also it depends in the speed, if the wave is fast, it will go higher and then bend.   hope i helped :D
6 0
3 years ago
7.
Vlada [557]

Answer:

The template DNA strand, from which the mRNA is synthesized, is 5' ... so the template strand and the mRNA will be complementary to each other) ... ATG Met (M) ... The following codons can be mutated by one base to produce an amber codon: ... which amino acid should be incorporated in the growing polypeptide chain.

Explanation:

3 0
2 years ago
Microbiological contaminants are best described as
andreyandreev [35.5K]

Answer:

Microbiological contamination refers to the non-intended or accidental introduction of infectious material like bacteria, yeast, mould, fungi, virus, prions, protozoa or their toxins and by-products. 1,2

Explanation:

if this answer helpful can you make me as brainlist

thank you :)

8 0
2 years ago
Each child born to a particular set of parents has probability 0.25 of having blood type o. if these parents have 5 children, wh
Taya2010 [7]
Use binomial distribution with p=0.25, n=5, x=2
P(X=x)=C(n,x)p^x (1-p)^(n-x)

P(X=2)=C(5,2) 0.25^2 0.75^(5-2)
=10*0.0625*0.421875
=0.26367

So the probability of 2 of 5 children having type-O blood from these parents is 0.26367.

7 0
3 years ago
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