Answer:
<em>The large sample n = 117.07</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the estimate error (M.E) = 0.08
The proportion (p) = 0.75
q =1-p = 1- 0.75 =0.25
Level of significance = 0.05
Z₀.₀₅ = 1.96≅ 2
<u><em>Step(ii):-</em></u>
The Marginal error is determined by
M.E = 
Cross multiplication , we get
√n = 
squaring on both sides , we get
n = 117.07
<u><em>Final answer:-</em></u>
<em>The large sample n = 117.07</em>
Answer:
C) 4
Step-by-step explanation:
Given equation:
The above equation represents proportional relationship.
To find the constant of proportionality.
Solution:
<em>The equation representing proportional relationship is given by:</em>
<em>
</em>
<em>where 
represents constant of proportionality.</em>
So, in order to find the value of for the given proportionality relationship, we will solve for 
We have:
Solving for
Dividing both sides by 2.
∴ 
Thus, the constant of proportionality = 4.
Answer:
16 cans per minute
Step-by-step explanation:
Its simple 80 divided by 5 is 16 so ever minute 16 cans are packed.
Answer:
8 were yoused because 8 times 3 =24 so if you subtract you get 8
Step-by-step explanation:
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
