<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:
-7 7/15
Step-by-step explanation:
-11 2/3 -(-4 1/5) = -11 2/3 + 4 1/5. Now, we need to find the LCM of 3 and 5. The LCM of 3 and 5 is 15 because 15 is the lowest number that can divide both 3 and 5 separately and both results will still be whole numbers.
Now, we have -11 10/15 + 4 3/15.
-11 10/15 + 4 3/15 = -7 7/15.
To give a brainliest after someone answers or multiple people answer there is a button near it that says pick as brainliest which gives that person a gold crown...sorry i dont know what each of this symbols.mean hope this help though
Answer:
150*2/3=100
take school size then multiply it by the ratio to get the amount of student who prefer such toothpaste.
Step-by-step explanation:
