Answer:
Make them wear a mask, along with yourself, also have a face sheild in case they start coughing blood. Keep them in a room by themselve, only come in to bring food and water. Have a seperate bathroom that they alone use. Not 100% sure this will stop much, but put a towel under the door frame (again not sure how effective it will be.
Explanation:
Answer:
Mr. Smith is not justified, Mrs. Smith can not be blamed necessarily for being unfaithful
Explanation:
Widow's peak is a dominant trait which means that it will be expressed both in homozygous and heterozygous condition. If "A" is the dominant allele and "a" is the recessive allele, the trait can be represented as AA or Aa.
Both Mr. Smith and Mrs. smith expressed the trait so they can either have AA or Aa genotype. If even one of them had AA genotype all the offspring would have the trait. But if both of them are heterozygous for it:
A a
A AA Aa
a Aa aa
There is 75 % probability of the child to have the trait (AA or Aa) but there is also 25% probability that the child does not express the trait (aa). Their second child belongs to this category and hence Mrs. Smith can not be blamed for being unfaithful.
Answer:
b. After two hours have passed, stop eating foods left out at room temperature.
Explanation:
According to USDA (United States Department of Agriculture), cooked food that is left at room temperature has a higher risk of being unsafed than the one that is refrigerated. This is because, at room temperature (40F- 140 F approximately) bacteria grows in the food and therefore it is unsafe to eat.
According to this Department, the bacteria double its number every 20 minutes approximately so the Department considers that, after two hours standing at room temperature, food is no longer safe to eat anymore.
Hydration!!!!!!!!!!!I think
