Eight percent of a population shows a recessive trait. what would be the frequency of the heterozygotes for that trait if the population is in hardy=w

Population that show recessive trait should be only homozygous recessive. If the population in Hardy Weinberg equilibrium, then the recessive trait gene frequency would be:

homozygous recessive = aa= 8%=0.08

a= √0.08=0.283

a+A=1

A= 1-a = 1- 0.283= 0.717

The percentage of heterozygote in population would be:

heterozygote= 2 Aa= (2) (0.717) ( 0.283)= 0.4058= 40.58%

the answer would be 3