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Nuetrik [128]
3 years ago
7

Factor completely 12x4 + 6x3 + 18x2.

Mathematics
1 answer:
Semmy [17]3 years ago
8 0

Answer:

B)3x^2 (4*x^2 + 2x + 6)

Step-by-step explanation:

Step 1: Find the Greatest common factor of the given expression.

12x^4 + 6x^3 + 18x^2

The above expression can be written as .

= 2*2*3*x^4 + 2*3*x^3 + 2*3*3*x^2

Here 3x^2 is prime factor


Step 2: Let's take out the 3x^{2} and write the remaining terms in the parenthesis.

= 3x^2 (2*2*x^2 + 2x + 2*3)

= 3x^2 (4x^2 + 2x + 6)

Therefore, the answer is B)3x^2 (4*x^2 + 2x + 6)

Thank you.

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The value 5 is a natural number.

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Give the rule for the translation if the coordinates of the preimage are A(2,-4) B(5,-4) C(5,-2) D(4,-2) E(4,-1) F(3,-1) G(3,-2)
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  (x, y) ⇒ (x -7, y)

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3 years ago
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A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

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H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

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  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

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  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

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Hey
Here it is!!

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