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3 years ago

Factor out the GCF 4g^3-6g+20g^2

Mathematics

1 answer:

elena55 [62]3 years ago

6 0

Answer:

$\sf 2g(2g ^2-3+10g)$

Step-by-step explanation:

$\sf 4g^3-6g+20g^2$

GFC: 2g

We'll Write the GCF first, then, divide each term by the GCF.

$\sf 2g(\frac{4g^3}{2g}-\frac{6g}{2g} +\frac{20^2}{2g} )$

$\sf 2g(2g ^2-3+10g)$

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240 = 100%
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3 years ago

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1. The correct answer should be A
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dezoksy [38]

Answer:

a. Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.

b. The probability is 0.2029 or 20.29%

c. The probability is 0.2029 or 20.29%

d. The probability is 0.1671 or 16.71%

e. The probability is 0.9975 or 99.75%

Step-by-step explanation:

a. Yes, there are 10 independent trials, each with exactly two possible outcomes, and a constant probability associated with each possible outcome.

b. Let's use the binomial distribution table, this way:

Binomial distribution (n=10, p=0.697)

f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0000 0.0000

1 0.0002 0.0002

2 0.0016 0.0017

3 0.0095 0.0112

4 0.0384 0.0496

5 0.1059 0.1555

6 0.2029 0.3584

7 0.2668 0.6252

8 0.2301 0.8553

9 0.1176 0.9729

10 0.0271 1.0000

The probability is 0.2029 or 20.29%

c. If 69.7% of 18-20 years old consumed alcoholic beverages in 2008, therefore, 30.3% did not and the binomial distribution table is:

Binomial distribution (n=10, p=0.303)

f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0271 0.0271

1 0.1176 0.1447

2 0.2301 0.3748

3 0.2668 0.6416

4 0.2029 0.8445

5 0.1059 0.9504

6 0.0384 0.9888

7 0.0095 0.9983

8 0.0016 0.9998

9 0.0002 1.0000

10 0.0000 1.0000

The probability is 0.2029 or 20.29%

d. Let's use the binomial distribution table, this way:

Binomial distribution (n=5, p=0.697)

f(x) F(x) 1 - F(x)

x Pr[X = x] Pr[X ≤ x]

0 0.0026 0.0026

1 0.0294 0.0319

2 0.1351 0.1671

3 0.3109 0.4779

4 0.3576 0.8355

5 0.1645 1.0000

P(0) + P(1) + P (2) = 0.0026 + 0.0294 + 0.1351

The probability is 0.1671 or 16.71%

e. Using the same binomial distribution table we used in d. we have:

P(1) + P (2) + P(3) + P(4) + P (5) = 0.0294 + 0.1351 + 0.3109 + 0.3576 + 0.1645

The probability is 0.9975 or 99.75%

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