The half life of Carbon 14 is 5730 years. By substituting the given figures to the formula x=I(1/2)^n where x is the amount of the remaining material, I is the initial amount and n is the quotient between the time elapsed and carbon 14's halflife. You will arise with the formula x=20(1/2)^1.745. You will get 5.97 mg.
The urinary or the excretion system.
Complete question:
You will find the complete problem in the attached files, where all the proposed crosses are detailed.
Answer:
1. Which genotype(s) and phenotype for fur length are dominant?
- <em>Genotype: FF + Ff</em>
- <em>Phenotype: Short fur</em>
2. Which genotype(s) and phenotype for fur length are recessive?
- <em>Genotype: ff</em>
- <em>Phenotype: Long fur</em>
Explanation:
<u>Available data:</u>
- F is the dominant allele that codifies for short fur
- f is the recessive allele that codifies for long fur
CROSSES:
<u>1) Test One</u>: If I breed a short fur, FF female with a short fur, Ff male, then I will expect to see what type of offspring?
- all short fur;
- some short and some long fur;
- all long fur offspring.
Cross: FF x Ff
Gametes) F F F f
Punnet square) F F
F FF FF
f Ff Ff
F1) Genotype:
2/4 = 1/2 = 50% FF
2/4 = 1/2 = 50% Ff
Phenotype:
4/4 = 100% Short-fured individuals
<u>2) Test Two:</u> If I breed a short fur, Ff female with a short fur, Ff male, then I will expect to see what type of offspring? Pick 1)
- all short fur;
- some short and some long fur;
- all long fur offspring.
Cross: Ff x Ff
Gametes) F f F f
Punnet square) F f
F FF Ff
f Ff ff
F1) Genotype:
1/4 = 25% FF
2/4 = 1/2 = 50% Ff
1/4 = 25% ff
Phenotype:
3/4 = 75% Short-fured individuals
1/4 = 25% long-fured individuals
<u>3) Test Three</u>: If I breed a long fur, ff female with a long fur, ff male, then I will expect to see what type of offspring? Pick 1)
- all short fur;
- some short and some long fur;
- all long fur offspring.
Cross: ff x ff
Gametes) f f f f
Punnet square) f f
f ff ff
f ff ff
F1) Genotype:
4/4 = 100% ff
Phenotype:
4/4 = 100% long-fured individuals
c) P(5<X<15); (d) the value of the constant c such that P(|X −10|≥c) ≤ 0.04
Answer:
a) 0.4444
b) 0.5556
c) 0.8400
d) 10
Explanation:
Chebyshev’s theorem states that
P(|X - μ| ≥ kσ) = 1/k²
μ = Mean = 10
σ = standard deviation = √variance = √4 = 2
a) P(|X −10|≥3)
Comparing this with P(|X - μ| ≥ kσ)
It is evident that kσ = 3
2k = 3
k = 3/2
k² = 2.25
1/k² = 0.444
P(|X −10|≥3) = 0.444
b) P(|X −10| < 3) = 1 - P(|X −10|≥3)
And P(|X −10|≥3) was found in (a) to be 0.444
So,
P(|X −10| < 3) = 1 - P(|X −10|≥3) = 1 - 0.4444 = 0.5556
c) P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5)
P(|X-10|≥5) = 1/k²
Comparing with P(|X - μ| ≥ kσ) = 1/k²
where 2k = 5
k = 5/2
k² = 6.25
1/k² = 0.16
P(|X-10|≥5) = 0.16
P(5<X<15) = P(|X-10|<5) = 1 - P(|X-10|≥5) = 1 - 0.16 = 0.8400
d) P(|X −10|≥c) ≤ 0.04
1/k² = 0.04
k = 5
c = kσ = 5×2 = 10
