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
The answer to your question is 13<2/3
Answer:
The answer is 7.
Step-by-step explanation:
4 divided by 7 is 0.571428 these just keep repeating over and over again so the closest dividable number of 6 (6 is how many digits there are before it repeats) is 1998 so the 2nd digit in .571428 is 7 hope this helps.
Based on your problem, you only have one given. So, you can't make an equation for this because there are not limits to the equation. The only thing that you know is that the three numbers are consecutive even integers. My way of solution for this is trial-and-error. However, it's really quite easy.
For example: 42 + 46 = 88. I have to increase the numbers more to reach 136. Suppose: 82 + 86 = 168. That exceeded 136. So, it must be between 46 and 82. Suppose again: 66 + 70 = 136. Therefore, the sequence of the consecutive even integers are 66, 68, and 70.
Its 60 all you really had to do is 6 times 10