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
What you can do in this case is the following rule of three to find the result:
1 light year ---> 5.88 * 10 ^ 12 miles
3.2 * 10 ^ 2 light year ---> x
Clearing x we have:
x = ((3.2 * 10 ^ 2) / (1)) * (5.88 * 10 ^ 12)
x = 1.88 * 10 ^ 15 miles
answer:
In scientific notation, approximately it is 1.88 * 10 ^ 15 miles
X^4=256
x^4-256=0
(x^2)^2-(16)^2=0
(x^2+16)(x^2-16)=0
either x^2+16=0
it gives complex roots.
or x^2-16=0
or x^2-4^2=0
(x+4)(x-4)=0
either x+4=0,x=-4
or x-4=0
x=4
<span>perpendicular slope :
m1 * m2 = -1
1/2 * m2 = -1
m2 = -2
y = -2x + b
2 = -4 +b
b = 6
y = -2x + 6
</span>
Answer:
BC= 29
Step-by-step explanation:
If the total, AC, is 54 and AB is 25 subtract those two to find the last part, BC.
