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ki77a [65]
3 years ago
11

18. Brown eyes (B) are dominant to blue eyes (b). The presence of freckles (F) is dominant to the absence of freckles (f). What

is the probability that two people who are heterozygous for both traits will have a child that has blue eyes and no freckles?

Mathematics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

For bbff we have only 6.3% probability

Step-by-step explanation:

If the parents are heterozygous for both traits, them they are represented by:

BbFf × BbFf

Parent 1: BbFf

Parent 2: BbFf

We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.

By distributing the possibilities in a Punnett square, <em>vide</em> picture. We have the following possibilities:

Genotype   Count     Percent  

bBfF  4     25

BBfF  2     12.5

bBFF  2     12.5

bBff          2     12.5

bbfF  2     12.5

BBFF  1     6.3

BBff      1            6.3

bbFF  1     6.3

bbff      1     6.3

For bbff we have only 6.3% probability

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Given one zero of the polynomial function, find the other zeros.
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Answer:

Step-by-step explanation:

You need to use synthetic division to do all of these.  The thing to remember with these is that when you start off with a certain degree polyomial, what you get on the bottom line after the division is called the depressed polynomial (NOT because it has to math all summer!) because it is a degree lesser than what you started.

a.  3I   1   3   -34   48

I'm going to do this one in its entirety so you get the idea of how to do it, then you'll be able to do it on your own.

First step is to bring down the first number after the bold line, 1.

3I   1    3    -34    48

   _____________

      1

then multiply it by the 3 and put it up under the 3.  Add those together:

3I    1    3    -34    48

           3

----------------------------

      1     6

Now I'm going to multiply the 6 by the 3 after the bold line and add:

3I    1     3     -34     48                                                                                                    

             3      18

_________________

      1      6     -16

Same process, I'm going to multiply the -16 by the 3 after the bold line and add:

3I      1      3      -34      48

                3       18     -48

___________________

        1       6      -16       0

That last zero tells me that x-3 is a factor of that polynomial, AND that the depressed polynomial is one degree lesser and those numbers there under that line represent the leading coefficients of the depressed polynomial:

x^2+6x-16=0

Factoring that depressed polynomial will give you the remaining zeros.  Because this was originally a third degree polynomial, there are 3 zeros as solutions.  Factoring that depressed polynomial gives you the remaining zeros of x = -8 and x = 2

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Do the remaining problems like that one; all of them come out to a 0 as the last "number" under the line.

You got this!        

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