Different starting lineups can be created - 120+60+120=300. A certain college team has on its roster 4 centers, 5 guards, 5 forwards, and one individual (x) who can play either guard or forward.
1. A lineup without x is an example. You must decide.
- two of the three forwards;
- from 4 centers, 1 center.
It can be made as -
2C5 × 2C3× 1C4 = 10× 3×4 = 120 ways
2. Consider starting lineups with guard x. You must decide.
- 2 guards chosen from 6 quards (at least one of them must be x);
- two of the three forwards;
- from 4 centers, 1 center.
1C5×2C3×1C4 = 5×3×4 = 60 ways
3. Think about lineups where x is the forward. You must decide.
- 5 quards, 2 guards;
- 2 forwards from 4 forwards (x must be one of them);
- from 4 centers, 1 center.
2C5×1C3×1C4 = 10×3×4 = 120 ways.
Therefore, total number different lineups is - 120+60+120=300.
To learn more about combinations from given link
brainly.com/question/8781187
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Answer:5) 58*
6) 87*
7) 72*
Step-by-step explanation:
5) right angle is 90* minus 32* so X= 58*
6) half is 180* minuses 61* then take another 32* from that which gives you 87*
7) another half so 180* minus 108* so p is 72*
Answer:
the first one is true:)
Step-by-step explanation:
We are given the angle WON. The letters W, O and N represents points included in the angle. One of these points is the vertex of the angle. It is the point which an angle is measured. For this case, the vertex is point O.
ABF= 28 degrees
BCD= 28 degrees since vertical angle are congruent
ABC= 180 -28 = 152 degrees
