Part B: Suppose Steven combined the table-tennis balls into one large bag instead of two individual bags. Would the probability change from Part A if Steven wanted to draw a “1” and a “B” without replacement? A.
No, the probability would not change because there is still one table-tennis ball marked with a “1” and one table-tennis ball marked with a “B”.

B.
No, the probability would not change because the outcome is the same.

C.
Yes, the probability would change because you are drawing two table-tennis balls.

D.
Yes, the probability would change because you are drawing from one bag and this changes the number of possible outcomes.

me! what did you put for this question?

step-by-step explanation:

their is no graph showing

step-by-step explanation:

you'll just earn points here

step-by-step explanation:

< 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)