<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:
27,804
Step-by-step explanation:
28x993= 27,804
Answer:
a) slope intercept form: y = -5/14 x + 7/2
b) point slope form: y - 1 = 3/2(x - 4)
Step-by-step explanation:
a) Line A: (4,1) and (0, 7/2)
Slope = (1 - 7/2)/(4 - 0)
Slope = - 5/2 * 1/4
Slope = - 5/14
Equation in slope intercept form:
y = -5/14 x + 7/2
b)
(4,1) and (0, -5)
Slope = (1 + 5)/(4 - 0) = 6/4 = 3/2
Equation in point slope form:
y - 1 = 3/2(x - 4)
113.5
Step-by-step explanation:
454g / 32G = 14.1875 x 8 = 113.5
Answer:
Step-by-step explanation:
13.
x³-x²-x-2=0
x³-2x²+x^2-2x+x-2=0
x²(x-2)+x(x-2)+1(x-2)=0
(x-2)(x²+x+1)=0
x-2=0,x=2
x²+x+1=0

<em>12.</em>
<em>3x^4-11x³+15x²-9x+2=0</em>
<em>\[3x^4-3x^3-8x^3+8x²+7x²-7x-2x+2=0\]</em>
<em>3x³(x-1)-8x²(x-1)+7x(x-1)-2(x-1)=0</em>
<em>(x-1)[3x³-8x²+7x-2]=0</em>
<em>x-1=0,gives x=1</em>
<em>3x³-8x²+7x-2=0</em>
<em>3x³-3x²-5x²+5x+2x-2=0</em>
<em>3x²(x-1)-5x(x-1)+2(x-1)=0</em>
<em>(x-1)(3x²-5x+2)=0</em>
<em>x-1=0,gives x=1</em>
<em>3x²-5x+2=0</em>
<em>3x²-3x-2x+2=0</em>
<em>3x(x-1)-2(x-1)=0</em>
<em>(x-1)(3x-2)=0</em>
<em>x-1=0,gives x=1</em>
<em>3x-2=0, gives x=2/3</em>
<em>so roots are 2/3,1,1,1</em>
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