The evaluation of their DNA and short tandem repeats will let them know if the birds are related or not. The genetic linkage will provide information about their common ancestry as well as if they belong to the same species, or are closely related species.
51.5- AB
Since in the ABO blood system AB is a blood type as well as B. The rest of the population will have either of these types so exclusively for B type then it will ne the remaining - AB
Answer:
(B) Energy transfer between trophic levels is almost always less than 20% efficient.
Explanation:
The ultimate source of energy on the Earth is the Sun. The energy coming from the Sun is captured by green plants by the photosynthesis. During photosynthesis sun energy is fixed into chemical energy (carbohydrate). So, in an ecosystem energy flow is unidirectional (from sun to the green plants). The fixed chemical energy from green plants is transferred to the herbivores then to carnivores through food. When one organism eats another organisms, only 10 % of the energy present in the organism is transferred as a food for the next organism and a large amount of energy is lost as heat into the environment. Thus, energy keeps on decreasing when stored energy moves from producers to top consumers. Thus, less than 20% energy transfer limits the trophic levels in most of the ecosystem.
<span>Let the equation be : at^2 + bt + c = P,
where t = time (hrs),
P = population (1000's).
When t = 1, P = 3.03.
When t = 2, P = 1.72.
When t = 3, P = 1.17.
Substitute these into the equation to obtain these 3 simultaneous equations :
a + b + c = 3.03
4a + 2b + c = 1.72
9a + 3b = c = 1.17
Solving gives :
a = 0.38,
b = -2.45,
c = 5.1.
The equation is therefore,
P = 0.38t^2 - 2.45t + 5.1
Testing with t = 0 to 6 gives the population values as provided,
so it seems to be a valid model.
At t = 9 hrs,
P = 0.38*9^2 - 2.45*9 + 5.1
= 13.83.</span>
Answer:
0.05 mg/mL ( B )
Explanation:
Given data:
20 mg/ml starch
2% solution = 2g of solute is in 100g of solvent
<u>Determine the new concentration in mg/ml </u>
Dilution equation = C1V1 = C2V2
new concentration ; applying the dilution factor
dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every 100 mL
∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL
