The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in
Hardy-Weinberg equilibrium.
(a) Calculate the percentage of heterozygous individuals in the population.
According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq
term. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 × 0.19 × 0.81 = 0.31 or 31%
(b) Calculate the percentage of homozygous recessives in the population.

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy

i'm not really sure how to answer that

step-by-step explanation:

simplifying

7ln(x + 2.8) = 11.2

reorder the terms:

7ln(2.8 + x) = 11.2

(2.8 * 7ln + x * 7ln) = 11.2

(19.6ln + 7lnx) = 11.2

solving

19.6ln + 7lnx = 11.2

solving for variable 'l'.

move all terms containing l to the left, all other terms to the right.

reorder the terms:

-11.2 + 19.6ln + 7lnx = 11.2 + -11.2

combine like terms: 11.2 + -11.2 = 0.0

-11.2 + 19.6ln + 7lnx = 0.0

the solution to this equation could not be determined.

step-by-step explanation:

i searched it on google but it says that the solution to this equation could not be determined but i hope that it you a little.