Answer:
(-7, -12)
Step-by-step explanation:
4x-3y=8
5x-2y=-11
Is there any of the like terms can be added and the result will be 0? No, so we have to multiple one OR both of the equations to make that one number do that.
(I will try to remove the y like terms so i will multiple both of them by the opposite so both of the ys will be 6)
2(4x-3y=8)
-3(5x-2y=-11)
8x-6y=16
-15x+6y=33
(now the easy part… cancel the 6s and add the equations)
8x+(-15x)=-7x
16+33=49
-7x=49
(divide 49 by -7)
x=-7
Replace x in any of the equations and you’ll get the y value.
4x-3y=8
4(-7)-3y=8
-28-3y=8
-3y=36
y=12
Threfore, there is one solution which is….. (-7,-12)
4n² - 16n - 84 = 0
Multiply both sides by 1/4 :
n² - 4n - 21 = 0
Add 21 to both sides:
n² - 4n = 21
Complete the square by adding 4 to both sides:
n² - 4n + 4 = 25
(n - 2)² = 25
Solve for n :
n - 2 = ± √25
n - 2 = ± 5
n = 2 ± 5
Then n = 2 + 5 = 7 or n = 2 - 5 = -3.
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Answer:
The expected net winnings for the bet are -$1.0526
Step-by-step explanation:
P(x =+$20) = P(Black outcome) = 18/38
P(x =-$20) = P(red outcome) + P(green outcome)
= 18/38 + 2/38 = 20/38
Hence the probability distribution of x = $20 , P(x) = 18/38
x = -$20, P(x) = 20/38
Expected value of the random variable x is given by ;
miu = Summation [xP(x)] = 20(18/38) - 20( 20/38)
= -$1.0526
hence, the expected net winnings for the bet are -$1.0526
This implies that if a player bet on a very large number of games, the player would on the average lose $1.0526 per single bet
Answer:
the questions is not too correct
