Under ideal conditions a certain bacteria population is known to double every 2 hours. suppose there are initially 500 bacteria. 1. what is the size of the population after 6 hours?2. what is the size of the population after t hours?3. estimate the size of the population after 19 hours.

A)4000bacteria b) 500*2^(t/2) c) 362039 bacteria

Explanation:

The bacteria population doubles every 2hrs.

Initial population = 500 bacteria

After 6hours, it will undergo 3 generations since

6hours /2hrs = 3.

That means it has doubled its population three times

Initial is 500

After 2hrs = 1000bacteria, 4hrs = 2000 bacteria,

6hrs = 4000 bacteria

B) after 2hrs, the population is 500*2^1,

After 4hrs, the population is 500* 2^2,

After 6hrs, the population is 500*2^3

Which is the same as 500*2^(6/2)

Then after t hours, the population is 500*2^(t/2)

C) after 19 hours, the populations = 500*2^(19/2)

= 362038.67 approx 362039 bacteria

helicase, i think thats it

it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.

explanation:

the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.

countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.