A red marble is selected at random from a bag containing 2 blues and 9 red marbles and is not replaced. the probability that a second marble selected will also be red is

 So the probability of drawing a second red marble is 8 out 10 or 0.8.

Step-by-step explanation:

It seems to me that if one red marble is already gone, there are 8 red marbles left out of a total of 10 marbles.

So the probability of drawing a second red marble is 8 out 10 or 0.8.

8/10 or 4/5 or 0.8

Step-by-step explanation:

since one red marble is already taken, there are only eight remaining now and the total number of marbles are ten now. So you have to divide.

Therefore, the probability is 8/10 or 4/5 or 0.8.

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.