Answer:
its B. <em><u>thoracic</u></em><em><u> </u></em><em><u>cavity</u></em><em><u>.</u></em>
Explanation:
it is the ventral body chamber that contains the pericardial cavity (the heart) and the pleural cavity (the lungs).
Answer:A scientific law is something that will happen 100 percent of the time
but a thoery is something u think will happen like a hypothsis
Explanation:
<span>polysaccharide bc the rest are lipids</span>
Answer:
Used the C1V1 = C2V2 method for calculations
Explanation:
Component Calculation Volume of Stock to be added
FBS: 100*V1 = 10*25 2.5 mL
Glutamine: 200*V1 = 2*25 0.25 mL
Penicillin: 10000V1 = 100*25 0.25 mL
Streptomycin: 10000V1 = 100*25 0.25 mL
NB: C1 = Stock Concentration; V1 = Volume of stock required; C2 = Final concentration required; V2 = Final volume required
Answer:
When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0
Explanation:
According to Levin's model (1969):
<em>dP/dt = c - e</em>
where P represents the proportion of occupied patches.
<em>c</em><em> </em>and <em>e </em>are the local immigration and extinction probabilities per patch.
Thus, the rate of change of P, written as dP/dt, tells you whether P will increase, decrease or stay the same:
- if dP/dt >0, then P is increasing with time
- if dP/dt <0, then P is decreasing with time
- if dP/dt = 0, then P is remaining the same with time.
The rate dP/dt is calculated by the difference between colonization or occupancy rate (<em>c</em>) and extinction rate (<em>e</em>).
c is then calculated as the number of successful colonizations of unoccupied patches as a proportion of all available patches, while e is the proportion of patches becoming empty. Notice that P can range between 0 and 1.
As a result, if the patch occupancy rate (c) equals the patch extinction rate (e), then patch occupancy P equals to 0.
