I think it's the home health aide, because they can be there to see when and what medicane is given. The client could be too sick to write, and the physician isn't always there. And the nurse sees so many other patients to have enough time. I think it's the home health aide.
Answer:
1. The parents genotypes could have been BO and AO
2. wire-hair
Explanation:
There are four possible blood types which are type A, B, AB, O. blood group is the classification of blood based on the presence or the absence of inherited antigenic substances on the surface of the red blood cells. They have hereditary basis and also rely on a series of alternative genes sometimes used in solving dispute of parental heritage. With the four possible blood groups, there are six possible genotypes and these are:
Blood type possible genotypes
Type A AA, AO
Type B BB, BO
Type AB AB
Type O OO
Thus, for parents with blood type B and A to give birth to a child with blood type O, it means their genotype could have been both BO and AO for them to be able to produce a child with OO. a cross between these two could give rise to OO.
Question 2
Wire hair is dominant (S) to smooth (s), thus wire hair could be in the homozygous (SS) and heterozygous form (Ss) and the smooth hair can only be expressed in the homozygous recessive form (ss).
thus, in a cross between homozygous wire haired and smooth haired, we will have:
homozygous wire haired homozygous smooth haired
P gen SS x ss
F1 gen. Ss
phenotype: wire haired
Given:
The solution decreased by 
ounces every hour for 5 hours.
Remaining solution at the end of the experiment = 
ounces.
To find:
The initial amount of solution.
Solution:
Let the initial amount of the solution be x.
The solution decreased by ounces every hour for 5 hours.
Total decreased amount 
Initial amount = Remining amount + Decreased amount of solution.
Therefore, the initial amount of solution is 18 ounces.
Ovulation is the process in which eggs are released from the ovary.
I’m assuming it’s all of the above
