7 ㅋ was
There were 53 players on a
team and the mean height
approx.
73.6. Draw the
normal distribution curve,
with percentages and 3 std,
deviations if the sta deviation
is 1.4.
I
SSS

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.

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1

umumumumum

im not sure