What is the probability of rolling a number greater than or equal to 8 with the sum of two dice, given that at least one of the dice must show 6?

a dice goes from 1-6

6+1=7

6+2=8

6+3=9

6+4=10

6+5=11

6+6=12

Of those possible choices, 5 are greater than or equal to 8. So the answer is 5/6.

mark brainiest please

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.

ok

step-by-step explanation: