<span>A Winter Storm Watch is issued when there is the potential for significant and hazardous winter weather within 48 hours.</span>
Answer:
1 out of 16 or 1/16
Explanation:
Let:
R = Round seed r = wrinkled seed
Y = Yellow seed y = green seed
Starting from the parent generation or the P generation, the genotype of the parents would be:
RRYY x rryy
This is because the cross is between a parents that are "purebred" for specific traits. This means that the parents are homozygous for the traits which in genotype form is represented by either two capital letters or two lower case letters.
Attached is the F1 generation results between a cross between the P generation.
As you can see the genotype of all the offsprings is RrYy.
Now also below is the cross between the F1 generation offsprings.
For recessive traits, it is only expressed when the dominant gene is not in the genotype, or in other words, the genotype is all in small letters. Since we are looking at two traits, you need to look for how many squares out of the total have all lowercase letters.
There is only 1 rryy out of the 16 predictions. So the answer would be 1/16.
That it dissolves either when the aicd hit or it takes time
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume