Answer:
A - 0%
B- 50%
C- 50%
D- 100%
Step-by-step explanation:
Cystic fibrosis is inherited in an autosomal recessive form, meaning that a person has to inherit two abnormal genes for the disease to manifest. In the case of this question, one parent is a gene carrier, so his genotype is Aa, while the other does not have the cystic fibrosis gene, so AA.
Performing the cross of Aa x AA, we can see that:
a.) The probability of a child would have cystic fibrosis is 0%, since the disease is recessive and to be affected it should receive a recessive gene from each parent.
b.) The probability of a child would be a carrier is 50%, as 50% of the crossing phenotypes are Aa.
c.) The probability of a child would not have cystic fibrosis and is not a carrier is 50%, as 50% of the child's genotype is AA.
d.) The probability of a child would be healthy is 100%, as of all possible phenotypes none is affected.
The answer is D since is the only one that has the correct y intercept
Need to take 25 mg times 18% then times it by 4 then times it again by .01
Given that the total number of students that sent messages = 150 students
a) To obtain the equation to represent the number of students who send text messages, we will sum up the variables in the Venn diagram and equate it to 150.
Hence, the equation is
b) Solving for x
Therefore, x = 15.
c) The total number of student that uses cell phone = 75 + x = 75 + 15= 90students
The total number of students that sent messages = 150students
The formula for probability is,
Hence,
Therefore, the probability that a randomly chosen student uses their cell phone to send text messages is 3/5.