There are eight marbles in a bag. Four marbles are blue (B), two marbles are red (R) and two marbles are green(G)
Steve takes a marble at random from the bag.
(i) What is the probability that Steve will take a blue marble.
(ii) What is the probability that Steve will take a green marble.
(iii) What is the probability that Steve will take a yellow marble.

i. 4/8 or 1/2, ii. 2/8 or 1/4, iii. 0/8

Step-by-step explanation:

In the problem it says that there are 8 marbles in total, so for number 1 it'll be 4/8 or 1/2 since 4 of the marbles are blue. Number 2 will be 2/8 because there are only 2 green marbles and you can simplify 2/8 to 1/4. The last one would be 0 since the problem didn't mention that there are yellow marbles. Hope this helps :)

answer: y = 3/2 x - 3

y = 1/8 x - 3/4 that would be your correct option

step-by-step explanation: w

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)