<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 a ratio problem. If dry ing : water = 5 : 2, what is water when dry ing = 15?
The new ratio is 15 : water. How do we get from the first ratio to the new one? We multiply by 5. Meaning, we also need to multiply the 2 quarts of water by 5, giving us 10 quarts of water.
Step-by-step explanation:
G= T+3/7
G-3/7=T
T= G-3/7
Answer:
Step-by-step explanation:
Let x be the portion of 60%
180 - x is the portion of 40%
0.60x + 0.40(180 - x) = 0.55(180)
0.60x + 72 - 0.40x = 99
0.20x = 27
x = 135
180 - 135 = 45
First brand = 45 gallons of 40%
Second Brand = 135 gallons of 60%
