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:
the answer of that question is Letter C
Answer:
I believe that it is graph B
Step-by-step explanation:
Suppose the three sides are a, b, and c, with c being the longest, a being the shortest
first equation: c=(a+b)-66
second equation: 2a=c-55=> c=2a+55
plug c=2a+55 into the first equation: 2a+55= a+b-66 => b=a+121
a+b+c=a+(a+121)+(2a+55)=7272 =>4a+176=7272 =>a=1774
a=1774, b=1895, c=3603
8 × 16 = 128
128 ÷ 2 = 64
64cm squared
