Answer:
C. y₂ = (1 + (t/n))²
Step-by-step explanation:
yₙ₊₁ = yₙ + Δt F(tₙ, yₙ)
yₙ₊₁ = yₙ + Δt yₙ
yₙ₊₁ = yₙ + (t/n) yₙ
When n=0:
y₁ = y₀ + (t/n) y₀
y₁ = 1 + (t/n)
When n=1:
y₂ = y₁ + (t/n) y₁
y₂ = 1 + (t/n) + (t/n) (1 + (t/n))
y₂ = 1 + (t/n) + (t/n) + (t/n)²
y₂ = 1 + 2(t/n) + (t/n)²
y₂ = (1 + (t/n))²
30 + x0.15=y is the first equation
15+x0.20=y is the second equation
Where in both x equals the amount of days it takes and y is your out come
You can solve this problem two ways one by finding the common denominator or just multiply x until you get the same awnser between the both of them
Answer:
For bbff we have only 6.3% probability
Step-by-step explanation:
If the parents are heterozygous for both traits, them they are represented by:
BbFf × BbFf
Parent 1: BbFf
Parent 2: BbFf
We have to find the percentage of occurence of bb × ff, which is a child that has blue eyes and no freckles, with no dominant factor.
By distributing the possibilities in a Punnett square, <em>vide</em> picture. We have the following possibilities:
Genotype Count Percent
bBfF 4 25
BBfF 2 12.5
bBFF 2 12.5
bBff 2 12.5
bbfF 2 12.5
BBFF 1 6.3
BBff 1 6.3
bbFF 1 6.3
bbff 1 6.3
For bbff we have only 6.3% probability
It isn't easy really but you can always divide the number out. For example 50% of 200 is 100 by dividing 200 by 2 or 25% of 100 is 25 by dividing 100 by 4 since 25% is equivalent to 1/4