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
Answer:
46
Step-by-step explanation:
230 divided by 5
Answer:
A. $141.75
B. 15% loss (803.25/945)
Step-by-step explanation:
We know that for every 5 red bricks there were 2 gray bricks.
The total amount of red bricks and grey bricks in this sample is 7.
5 red bricks + 2 grey bricks = 7 bricks
Now, we divide 175 "total number of bricks used" by 7 "5 red bricks + 2 grey bricks = 7 bricks" and we will get a quotient of 25.
Now we know that 25 bricks is 
of the wall. The gray bricks are 
so we can multiply 25 by 2 and we will get a product of 50. If 1/7 = 25 grey bricks so 2/7 would be the grey bricks.
175 - 50 = number of red bricks.
Therefore, there were 125 red bricks.
