Answer:
0.3745
Step-by-step explanation:
Data
mean (
): 80
standard deviation (
): 4
Normal distribution table is in figure attached
.
Z = (85 - 80)/4 = 1.25
Area below curve = 0.3944
Z = (82 - 80)/4 = 0.5
Area below curve = 0.0199
Proportion of exam scores between 82 and 85?
0.3944 - 0.0199 = 0.3745
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.
Answer:
slope = 1
Step-by-step explanation:
Calculate the slope m using the slope formula
m = 
with (x₁, y₁ ) = (1, 1) and (x₂, y₂ ) = (2, 2) ← 2 ordered pairs from the table
m = 
= 
= 1
Step-by-step explanation:
it is negative 20 I'm pre sure
Hannah surveyed 1/9th of the population.
The sample wouldn't really be representative of the population, because of the
small sample size.
Based on the table, you could expect around 60% of the students to own a cat or a dog. This is because there are 30 students with a pet, and 20 without, which makes a percentage of 60% students that own a pet.
