Answer: 6/12 are white, 3/12 are colored and 3/12 are albino.
Step-by-step explanation: If the horses are white and their parents are ccww (albino) and CCWw (white horse), according to Mendel's premises, they both must be CcWw, since the crossing provides one C from one parent and other c from the other parent, one W and the other w. Using Mendel's chess and the principle of independent segregation, the crossing between CcWw results in the following fenotypical ratio:
1/16 CCWW (lethal)
2/16 CCWw (white)
2/16 CcWW (lethal)
4/16 CcWw (white)
1/16 CCww (normal)
2/16 Ccww (normal)
2/16 ccWw (albino)
1/16 ccWW (lethal)
1/16 ccww (albino)
Excluding the 4 individuals that have the lethal locus, we have 6/12 that are white (2/12 + 4/12) and 3/12 (1/12 + 2/12) that are colored. Also, there are 3/12 of albino individuals as well.
Your answer should be the 3rd choice. There is no solution.
I think that the answer you are looking for is 0.3
Step-by-step explanation:
answer: 135 degrees for the missing side
Answer + Step-by-step explanation:
1) The probability of getting 2 white balls is equal to:
2) the probability of getting 2 white balls is equal to:
3) The probability of getting at least 72 white balls is:
![=\sum^{150}_{k=72} [C^{k}_{150}\times \left( \frac{8}{15} \right)^{k} \times \left( \frac{7}{15} \right)^{150-k}]](https://tex.z-dn.net/?f=%3D%5Csum%5E%7B150%7D_%7Bk%3D72%7D%20%5BC%5E%7Bk%7D_%7B150%7D%5Ctimes%20%20%5Cleft%28%20%5Cfrac%7B8%7D%7B15%7D%20%5Cright%29%5E%7Bk%7D%20%20%5Ctimes%20%5Cleft%28%20%5Cfrac%7B7%7D%7B15%7D%20%5Cright%29%5E%7B150-k%7D%5D)
