If the frequency of a recessive allele is 0.4, then what will be the frequency of individuals with dominant phenotype in the Hardy-Weinberg population?

q = 0.4, q² = 0.16

p = 0.6, p² = 0.36

2pq = 2 $\times$0.4 $\times$ 0.6 = 0.48

Frequency of individuals with dominant phenotype = p² $+$2pq = 0.36 $+$ 0.48 = 0.84

Alternate method: 

The frequency of recessive phenotype (q²) = 0.16

Since, p² $+$ 2pq $+$ q² = 1

p² $+$ 2pq = 1 $-$ q² = 1 $-$ 0.16 = 0.84