Answer:
Cation-exchange capacity is a measure of how many cations can be retained on soil particle surfaces. Negative charges on the surfaces of soil particles bind positively-charged atoms or molecules, but allow these to exchange with other positively charged particles in the surrounding soil water
Answer:
The correct answer is b. prokaryotes store their genetic information in a nucleus
Explanation:
Prokaryotes are organisms that do not have a true nucleus. A true nucleus is an organelle that is membrane-bound and contains the cell's genetic material inside it.
In prokaryotes nucleus is not present do its genetic material is spread in the cytoplasm of the cell. Bacteria and archaea are considered to come under the prokaryotes group because they do not have a membrane-bound nucleus and other organelle.
Eukaryotes are the organisms that have membrane-bound nucleus and they store their genetic information in the nucleus. Therefore the correct answer is b. prokaryotes store their genetic information in a nucleus.
Because of the antibodies produced by the mother, inherent immunity..
Answer:
Atherosclerosis affects arteries, these blood vessels are elastic and flexible in order to allow good blood circulation
Explanation:
The blood vessels are classified into three types: arteries, veins, and capillaries. An artery is a muscular blood vessel that is able to carry blood away from the heart to different parts of the body. Healthy arteries need to be elastic and flexible in order to adjust their diameter and thus control blood flow by maintaining blood pressure. Arteriosclerosis is a chronic disease caused by the accumulation of fatty substances (especially cholesterol) in the artery walls, thereby difficulting blood circulation by thickening and hardening these walls.
Answer:
(a) 0.16
(b) 1
Explanation:
Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16
Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10
Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P(
) = 0.10
(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by
P(L 
H);
As we know that P(A 
B) = P(A) + P(B) - P(A B)
So, in our question;
P(L H) = P(L) + P(H) - P(L H)
0.10 = 0.16 + 0.10 - P(L H)
P(L H) = 0.16 + 0.10 - 0.10 = 0.16
Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .
(b) <em>Conditional Probability P(A/B) is given by</em> = 
So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)
P(H/L) = 
= 
= 1 .
