Based on the information given the percentage of the brown geese that are heterozygous is 48%.
Random sample = 250
Brown color = 210
Gray color = 250 - 210 = 40
First step is to calculate the proportion of a alleles in the population
Proportion (a alleles)=√40/250
Proportion (a alleles)= √0.16
Proportion (a alleles)= 0.4
Second step is to calculate the proportion of A alleles in the population
p(A) = 1 - 0.4
p(A)= 0.6
Third step is to calculate brown heterozygous percentage using this formula
Brown heterozygous percentage=2pq(Aa)
Let plug in the formula
Brown heterozygous percentage= 2 x 0.6 x 0.4
Brown heterozygous percentage= 0.48 ×100
Brown heterozygous percentage= 48%
Inconclusion the percentage of the brown geese that are heterozygous is 48%.
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For non simplest form 2 3/9 for simplest form 2 1/3
Answer:
24feet 1 inches
Step-by-step explanation:
Answer:
Diana's score is better.
Step-by-step explanation:
In Diana case :
Score = 92
Mean score , μ = 71
Standard deviation σ = 15
So,
P( x ≥ 92 ) = 1 - P( x ≤ 92 )
= 1 - P(z ≤ )
= 1 - P( z ≤ )
= 1 - P( z ≤ 1.4 )
= 1 - 0.9192
= 0.0808 = 8.08 %
⇒P( x ≥ 92 ) = 8.08%
∴ we get
Diana score is > 91.92% other students in the class.
In Micheal case :
Score = 688
Mean score , μ = 493
Standard deviation σ = 150
So,
P( x ≥ 688 ) = 1 - P( x ≤ 688 )
= 1 - P(z ≤ )
= 1 - P( z ≤ )
= 1 - P( z ≤ 1.3 )
= 1 - 0.9032
= 0.0968 = 9.68 %
⇒P( x ≥ 688 ) = 9.68%
∴ we get
Micheal score is > 90.32% other students in the class.
From both scores , we can conclude that
Diana's score is better.