Answer:
it depends its in between 4 and 8. what are you writing about?
<span>You are given a roster with 3 guards, 5 forwards, 3 centers, and 2 "swing players" (x and y) who can play either guard or forward. You are given a condition that 5 of the 13 players are randomly selected. You are asked to find the probability that they constitute a legitimate starting lineup.
</span>For these cases, there are a lot so,
legitimate ways:
two swings and one gurad = 2C2*5C2*3C1 = 30
two swings used as forwards = 3C2*2C2*3C1 = 9
two swings used one guard = 2C1*3C1*1C1*5C1*3C1 = 90
one swing used as forward = 3C2*2C1*5C1*3C1 = 90
zero swing used = 3C2*5C2*3C1 = 90
total of legitimate ways = 489
Total ways = 13C5 = 1287
The probability that they constitute a legitimate lineup is = 489/128 = 0.38
Answer:
ntersecting lines DA and CE.
To find:
Each pair of adjacent angles and vertical angles.
Solution:
Adjacent angles are in the same straight line.
Pair of adjacent angles:
(1) ∠EBD and ∠DBC
(2) ∠DBC and ∠CBA
(3) ∠CBA and ∠ABE
(4) ∠ABE and ∠EBD
Vertical angles are opposite angles in the same vertex.
Pair of vertical angles:
(1) ∠EBD and ∠CBA
(2) ∠DBC and ∠EBA
You should expect to win at least 3 times.
Hope this helps!
I believe the correct answer is 45 degrees. The smallest positive angle <span>θ in QI for which tan θ = 1 would be 45 degrees. We obtain it as follows:
</span>tan θ = 1
<span>θ = tan^-1 (1)
</span>θ = 45 degrees
Hope this answers the question. Have a nice day.
