Ahockey all-star game has the eastern division all-stars play against the western division all-stars. on the eastern division team there are 8 united states-born players, 14 canadian-born players, and 2 european-born players. on the western division team there are 12 united states-born players, 8 canadian-born players, and 4 european-born players. if one player is selected at random from the eastern division team and one player is selected at random from the western division team, what is the probability that neither player will be a canadian-born player?

The probability of neither Canadian-born player would be selected from any Division All-Star Players roster is 5/18 or 27.8% (Rounding to the next tenth)

Explanation:

1. Let's review the information provided to us to answer the question correctly:

Eastern Division All-Star Players = 24

Western Division All-Star Players = 24

2. If one player is selected at random from the Eastern Division team and one player is selected at random from the Western Division team, what is the probability that neither player will be a Canadian-born player?

For answering the question, we should know the probability of neither Canadian-born player would be selected from each team independently, this way:

Probability of neither Canadian-born player would be selected from the Eastern Division All-Star Players roster = Non Canadian-born players/Eastern Division All-Star Players

Replacing with the real values, we have:

Probability of neither Canadian-born player would be selected from the Eastern Division All-Star Players roster = 10/24 = 5/12

Probability of neither Canadian-born player would be selected from the Western Division All-Star Players roster = Non Canadian-born players/Western Division All-Star Players

Replacing with the real values, we have:

Probability of neither Canadian-born player would be selected from the Western Division All-Star Players roster = 16/24 = 2/3

Now, we can find out the probability of neither Canadian-born player would be selected from any Division All-Star Players roster, this way:

These are independent events, therefore:

Probability of neither Canadian-born player would be selected from any Division All-Star Players roster = Probability of neither Canadian-born player would be selected from the Eastern Division All-Star Players roster * Probability of neither Canadian-born player would be selected from the Western Division All-Star Players roster

Replacing with the real values, we have:

Probability of neither Canadian-born player would be selected from any Division All-Star Players roster = 5/12 * 2/3 = 10/36 = 5/18

The probability of neither Canadian-born player would be selected from any Division All-Star Players roster is 5/18 or 27.8% (Rounding to the next tenth)

The probability that neither player will be a Canadian-born player is 27.77%.

Since a hockey all-star game has the Eastern Division all-stars play against the Western Division all-stars, and on the Eastern Division team there are 8 United States-born players, 14 Canadian-born players, and 2 European-born players, while on the Western Division team there are 12 United States-born players, 8 Canadian-born players, and 4 European-born players, to determine, if one player is selected at random from the Eastern Division team and one player is selected at random from the Western Division team, what is the probability that neither player will be a Canadian-born player, the following calculation should be performed:

Therefore, the probability that neither player will be a Canadian-born player is 27.77%.

Learn more in link?

if triangle abc is congruent to triangle def, their interior angles are congruent, because congruent parts of congruent triangles are congruent (cpctc).

that means ∠d is congruent to ∠a; ∠d is equal to 8x - 2.

interior angles of all triangles add to equal 180 degrees, so you can set the sum of these interior angle values equal to 180 and solve for x.

180 = (8x - 2) + 22 + 72

180 = 8x - 2 + 22 + 72

180 = 8x + 92

88 = 8x

11 = x

x = 11