NBA announces first all-star team

Nikola Jokic (Google Images)

According to the NBA on Tuesday, the Association announced its first All-Star team. The team consists of Milwaukee Bucks forward Giannis Antetokounmpo, Dallas Mavericks guard Luka Doncic, Denver Nuggets center and reigning MVP Nikola Jokic, Phoenix Suns guard Devin Booker, and Boston Celtics forward Jayson Tatum.

Antetokounmpo of Athens, Greece, averaged 29.8 points per game, 11.6 rebounds per game, 5.8 assists per game, and had a field goal percentage of .553, three-point percentage of .293, and a free throw percentage of .722. He had a season-high 50 points in a 128-119 Bucks win over the Indiana Pacers on February 15.

Doncic of Ljubljana, Slovenia, averaged 28.4 points per game, 9.1 rebounds per game, 8.7 assists per game, and had a field goal percentage of .457, three-point percentage of .353, and a free throw percentage of .744. He had a season-high 51 points in a 112-105 win overt the Los Angeles Clippers on February 10.

Jokic, a native of Sombor, Serbia, averaged 27.1 points per game, 13.8 rebounds per game, 7.9 assists per game, and had a .583 field goal percentage, .337 three-point percentage, and .810 free throw percentage. Booker, a native of Grand Rapids, Michigan, averaged 26.8 points per game, five rebounds per game, 4.8 assists per game, a field goal percentage of .466, a three-point percentage of .383, and a free throw percentage of .868.

Tatum, a native of St. Louis, Missouri, averaged 26.9 points per game, eight rebounds per game, and 4.4 assists per game. He had a field goal percentage of .453, three-point percentage of .353, and a free-throw percentage of .853. Tatum had a 54-point game this season. It came in a 126-120 Celtics win over the Brooklyn Nets on March 6. 

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