Answer:
a) 99.93%
b) 99.98%
Explanation:
Given that:
Total population size was only around 18 individuals ( for black-footed ferret)
If the annual survival rate in the wild was 0.4%.
a)
What is the probability that all 18 would have died in a single year?
To find that; we multiply the annual survival rate in the wild with the total population; which is:
(0.4%×18) = 0.072%
Then; we subtract it from a total of 100% in order to determine the probability that all 18 would have died in a single year.
= (100.00% - 0.072%)
= 99.93%
b)
What is the probability that all 18 would have died in a single year if canine distemper was present?
Given that;
Because of the presence of canine distemper disease, the annual mortality rate might have been as high as 0.89.
To determine the probability; we have:
1 - Annual mortality rate = annual survival rate
1 - 0.89 = 0.11 %
Therefore, 0.11% of 18 individuals = (0.0011 × 18)
= 0.0198%
Probability that all 18 would have died in a single year if canine distemper was present = 100 - probability of annual survival rate
= (100 - 0.0198)%
= 99.98%
Answer:
SS+SS=SS
Ff+ff=ff
Explanation:
Dominant alleles are capital
Answer:
c lunar and solar eclipse
Explanation:
The answer would be true
Iron is an important substance that needed to create hemoglobin for red blood cells. A pregnant woman will produce more blood as prevention for the blood loss when laboring. The baby also need iron to makes their red blood cells. That is why the iron requirement increases in pregnancy.
About half pregnant woman doesn't meet the requirement iron from diet alone.
Answer:
Plants use carbon dioxide for the process of photosynthesis.
