Answer:
a) 33.33%)
b) 135 minutes
c) 8.66 min
d) 50%
Step-by-step explanation:
a) the probability for a uniform distribution is
P(b<X<a) = (a-b)/(c-d) , where c and d are the maximum and minimum values
therefore the probability that the flight is more than 140 minutes ( and less than 150 since it is the maximum value)
P(140<X<150) = (a-b)/(c-d) = (150-140)/(150-120) = 10/30 = 1/3 (33.33%)
b) the mean (expected value) for a uniform probability distribution is
E(X) = (c+d)/2 = (120+150)/2 = 135 minutes
c) the standard deviation for a uniform probability distribution is
σ²(X)= (c-d)²/12 = (150-120)²/12 = 75 min²
σ = √75 min² = 8.66 min
b) following the same procedure as in a)
P(120<X<135) = (a-b)/(c-d) = (135-120)/(150-120) = 15/30 = 1/2 (50%)
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
Answer:
Step-by-step explanation:both are 1/6 of a chance
Answer:
25&
Step-by-step explanation:
Answer:
<em>1690</em>
Step-by-step explanation:
V = LWH
V = 13 * 10 * 13
V = 1690