Consider a typical die, each side with one more dot than the previous side, beginning with one dot. There are six sides to the die (a ube) . You throw a three on your first throw . If you picked up the die and threw it again , what is the chance of getting a three again ? If you enter your answer as a decimal, round to the thousandths place.

0.028

Step-by-step explanation:

Probability of throwing the first three = 1/6

The probability of throwing the second three = 1/6

1/6 * 1/6 = 0.028

ryan had a head start of 10 meters.

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.