Answer:
B. m
Step-by-step explanation:
Answer:
A - 0%
B- 50%
C- 50%
D- 100%
Step-by-step explanation:
Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.
Performing the cross of Aa x AA, we can see that:
a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.
b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.
c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.
d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.
Answer:
i believe it would take 16. let me know if thats right?
Step-by-step explanation:
Answer:
Step-by-step explanation:
put it on a coorinate grid,
circle circumference = 2*pi*r = 100.53
so the arc 19.2/100.53 = about 0.15 times the circumference of the circle
0.15*2pi = angle AOB in radians = 0.97
length DE is like 7.44
angle DOE is approx 0.485 radians beacuse it's half of angle AOB
area of shaded region is 804.2477 probably. I'm not sure about this last one sorry I think my calculator messed up
Answer:
part a
The probability of getting a 2 is 1/6 and the probability of getting a head is 1/2. The probability of getting both is thus 1/6 x 1/2= 1/12
part b
The probability of getting an even number is 3/6 = 1/2. The probability of getting a tail is 1/2. The probability of getting both is thus 1/2 x 1/2= 1/4.
