The carbon 14 left from this animal is 3 grams of 24 grams which is: 3g/24g= 1/8 times of its original mass. Then, the age of the animal fossil would be:
final weight= initial weight * (1/2) ^ time/half-live
1/8 initial weight = initial weight * (1/2) ^ time/half-live
log2 1/8= log2 (1/2) ^ time/half-live
-3= -time/half-live
time= -3 * -(half-live)
time= 3 half-live
If the half-live is 5730 years, the age would be: 3* 5730= 17.190 years.
The answer would be 'b'-Active transport and osmosis
The correct answer is Karl will get sick and Jose will not this time, but neither will get this influenza again since they both will have acquired active immunity to it.
In the case of Jose artificially active immunity has been induced due to vaccination. A vaccine instigates a primary response against the antigen without resulting in any signs of the disease. On the other hand, Karl will acquire active immunity naturally in future, as when an individual get exposed to a live pathogen, he or she develops the disease, and turns immune as an outcome of the primary immune response.
The density of a population of living organisms is usually measured in individuals on one square km. In here we have 50 earthworms on an area of 5 square meters, thus we have 10 earthworms on every square meter. In order to get to the result we need to see first how many square meters there are in one square km. One square km has one thousand meters of length and one thousand meters of width so:
1,000 x 1,000 = 1,000,000 km²
Since we established that we have 10 earthworms on every one square meter, we just need to multiply the number of square meters with the amount of earthworms on every square meter:
1,000,000 x 10 = 10,000,000
So we have a density of 10 million earthworms per square km.
