Of the cartons produced by a company, 66% have a puncture, 44% have a smashed corner, and 0.50.5% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

0.095 or 9.5%

Step-by-step explanation:

Puncture → P(P) = 0.06

Smashed corner → P(S) = 0.04

Puncture and smashed corner → P(P and S) = 0.005

The probability that a randomly selected carton has a puncture or a smashed corner is given by the probability of a puncture, added to the probability of a smashed corner, subtracted by the probability of both:

The probability is 0.095 or 9.5%.

john earns $14.00 for two hours of work.

< 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)