At an autosomal gene locus in humans, the allele for brown eyes is dominant over the allele for blue eyes. At another gene locus, located on the X chromosome, a recessive allele produces colorblindness while the dominant allele produces normal color vision. A heterozygous brown-eyed woman who is a carrier of colorblindness has a child with a blue-eyed man who is not colorblind. An ultrasound test shows that the child is a girl. What is the probability that she will be colorblind?

so chance that their daughter is colorblind= 0%

Step-by-step explanation:

colorblindness is a sex-linked recessive trait,

let the alleles be XC- normal, Xc- colour-blind

the woman is heterozygous for this gene, so her genotype is XCXc

man is normal so his genotype is XCY

( man is XY and woman is XX)

XCY * XCXc

see table at the attachment corner below

probability of getting a colorblind progeny given that progeny is a girl= number of colorblind girl/total number of girls=0/2=0

so chance that their daughter is colorblind= 0%

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.