<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
This is not a divison problem but if it was the answer to n would be the quotient.
72% of 25=18
25 there are
18 are good at math
7 aren't good at math
5 pounds for $12.95 is the better choice since it is cheaper per pound. ($2.59) versus 3 pounds which is ($2.77)
Answer:
See below for answers
Step-by-step explanation:
1st angle = 2x
2nd angle = x
3rd angle = x - 8
Sum of the angles in a triangle = 180°
1st + 2nd + 3rd = 180°
2x + x + x - 8 = 180°
4x - 8 = 180°
4x = 188
x = 47°
Now substitute
1st angle = 2x = 2(47) = 94°
2nd angle = x = 47°
3rd angle = x - 8 = 47 - 8 = 39
Check: 94 + 47 + 39 = 180°
