Would really appreciate if you with any of these youu

A) Take the number with high IQ that have the gene present and divide it by the total number with the gene present. Thats 33/72, or 11/24.

B) Take the number without the gene and have high IQ, and divide it by the total number with high IQ. Thats 19/52.

C) Take the number with the gene and a normal IQ and divide it by the total children. Thats 39/102, or 13/34.

D) Take the number without the gene and have high IQ and divide it by the total children. Thats 19/102.

E) Take the number with normal IQ and have the gene present, and divide it by the total number with the gene present. Thats 39/72, or 13/24.

F) Take the number with the gene present and have high IQ and divide it by the total number with high IQ. Thats 33/52.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

table 3 represents a direct variation function. : )