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Julli [10]
3 years ago
9

F( y ) = y + y 2 - 3

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
7 0
F(y) = y + y^2 - 3

f(-2) = -2 + (-2^2) - 3
f(-2) = -2 + 4 - 3
f(-2) = -1

f(-4) = -4 + (-4^2) - 3
f(-4) = -4 + 16 - 3
f(-4) = 9

f(0) = 0 + 0^2 - 3
f(0) = -3

f(2) = 2 + 2^2 - 3
f(2) = 2 + 4 - 3
f(2) = 3

f(4) = 4 + 4^2 - 3
f(4) = 4 + 16 - 3
f(4) = 17



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Which of the following terminating decimals is equivalent to Negative 1 and three-fourths?
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Solve the following equation. Remember to check for extraneous solutions 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).
xxMikexx [17]

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-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, \frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }

⇒\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }

⇒\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}

⇒\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}

⇒(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0

⇒ (x-2)(x-6) =0 or, [\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0

If, (x-2)(x-6) =0, then x=2 or x=6

If, [\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0, then x^{2} -5x-2=4

and (x-6)(x+1) =0

Therefore, x=6 or -1

So the solutions for x are -1, 2 6. (Answer)

4 0
3 years ago
Because gambling is a big​ business, calculating the odds of a gambler winning or losing in every game is crucial to the financi
IRISSAK [1]

Answer:

a) 0.1165

b) 0.0983

c) 0.000455

d) 0.787

e) 0.767

Step-by-step explanation:

5 ​bars, 4 ​lemons, 3 ​cherries, and a bell

Total = 5+4+3+1 = 13

The probability of getting a bar on a slot, P(Ba) = 5/13 = 0.385

A lemon, P(L) = 4/13 = 0.308

A cherry, P(C) = 3/13 = 0.231

A bell, P(Be) = 1/13 = 0.0769

a) Probability of getting 3 lemons = (4/13) × (4/13) × (4/13) = 256/2197 = 0.1165

b) Probability of getting no fruit symbol

On each slot, there are 4+3 = 7 fruit symbols.

Probability of getting a fruit symbol On a slot = 7/13

Probability of not getting a fruit symbol = 1 - (7/13) = 6/13 = 0.462

Probability of not getting a fruit symbol On the three slots = 0.462 × 0.462 × 0.462 = 0.0983

c) Probability of getting 3 bells, the jackpot = (1/13) × (1/13) × (1/13) = 1/2197 = 0.000455

d) Probability of not getting a bell on the 3 slots

Probability of not getting a bell on one slot = 1 - (1/13) = 12/13 = 0.923

Probability of not getting a bell on the 3 slots = (12/13) × (12/13) × (12/13) = 1728/2197 = 0.787

e) Probability of at least one bar is a sum of probabilities

Note that Probability of getting a bar = 5/13 and probability of not getting a bar = 8/13

1) Probability of getting 1 bar and other stuff on the 2 other slots (this can happen in 3 different orders) = 3 × (5/13)×(8/13)×(8/13) = 960/2197 = 0.437

2) Probability of getting 2 bars and other stuff on the remaining slot (this can also occur in 3 different orders) = 3 × (5/13)×(5/13)×(8/13) = 600/2197 = 0.273

3) Probability of getting 3 bars on the slots machine = (5/13) × (5/13) × (5/13) = 125/2197 = 0.0569

Probability of at least one bar = 0.437 + 0.273 + 0.0569 = 0.7669 = 0.767

5 0
3 years ago
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