Answer:
600$
Step-by-step explanation:
I think this is right
Answer:
12000
Step-by-step explanation:
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
Answer:
c = 3√5 ≈ 6.708
Step-by-step explanation:
Use the Pythagorean Theorem because this is a right triangle.
a^2 + b^2 = c^2
3^2 + 6^2 = c^2
9 + 36 = c^2
45 = c^2
√45 = c
<u>c = 3√5 ≈ 6.708</u>
