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.
(-3 7following is equivalent
__.
4
<u>answers for the blank:</u> 23 and 25
<u>explanation:</u> plug in the n to the equation and solve! :)
hope this helps ❤ from peachimin
Answer:
No
Step-by-step explanation:
The sum of the smaller sides is not bigger than the largest side, 122. ( 31 + 89 = 120 < 122)
Answer:
Probability at least one car will get punctured: 0.39347
Step-by-step explanation:
B(10,000 , 0.00005)
P(X ≥ 1) = 1 - P(X = 0)
= 1 - (1 - 0.00005)^10,000
= 1 - (0.99995)^10,000
= 1 - 0.60652...
= 0.39347 (probability that at least one car will get punctured)
As you can tell P(X ≥ 1) as we have to solve for the probability that at least one car will get punctured. That is of course 1 - [ P(X = 0) ].
