Answer: option c
Step-by-step explanation: with a two-tailed hypothesis she can find the way to go deep on the study, she will be able to increase the statistical power. She will be able to separate in specifics groups.
Notice that
13 - 9 = 4
17 - 13 = 4
so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :
149 = 9 + 4k
140 = 4k
k = 140/4
k = 35
This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.
Put another way, we have
but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.
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:
I messed up and forgot the six at the end dont put that answer as your answer sorry
Answer:
4 feet of ribbon ÷ 1/3 foot riboon = 12 pockets
