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Reika [66]
2 years ago
14

Neeeeeeed helpppppppppppppp

Mathematics
2 answers:
r-ruslan [8.4K]2 years ago
5 0

Step-by-step explanation:

Simplify the expressions

2 {}^{3}  - 2 {}^{1}  = 2 {}^{3}  - 2 = 8 - 2 = 6

2 + 3 = 5

{3}^{2}  = 9

So the expressions from least to greatest is

2 {}^{1}  + 3 {}^{1}  < 2 {}^{3}  - 2 {}^{1}  < 3 {}^{2}

8_murik_8 [283]2 years ago
3 0

Answer:

2^{1} + 3^{1}, 2^{3} - 2^{1}, 3^{2}

Step-by-step explanation:

3^{2} = 9

2^{3} - 2^{1} = 6

2^{1} + 3^{1} = 5

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-\dfrac{1}{3}b=9\ \ \ |multiply\ both\ sides\ by\ (-3)\\\\-3\cdot\left(-\dfrac{1}{3}b\right)=-3\cdot9\\\\\huge\boxed{b=-27}\leftarrow\boxed{C}
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A  viable solution is the ordered pair (0,0)

Step-by-step explanation:

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4 0
2 years ago
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There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
3 years ago
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