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2 years ago

11. Solve for x: x3 + 3x2 + 3x + 2 = 0

Mathematics

1 answer:

sineoko [7]2 years ago

5 0

Answer:

x = -2, (-1 + √3 i ) / 2 , (-1 - √3 i) / 2.

Step-by-step explanation:

x^3 + 3x^2 + 3x + 2 = 0

f(-2) = -8 + 3*4 - 6 + 2

= -8 + 12 - 6 + 2

= 0

So (x + 2) is a factor . Dividing:

x^2 + x + 1

---------------------------------

x + 2) x^3 + 3x^2 + 3x + 2

x^3 + 2x^2

----------------

x^2 + 3x

x^2 + 2x

-----------

x + 2

x + 2.

----------

Solving x^2 + x + 1:

x = [ - 1 +/- sqrt(1 - 4*1*1) ] / 2

= (-1 + √3 i ) / 2 , (-1 - √3 i) / 2.

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It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell

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a) 75

b) 4.33

c) 0.75

d) $3.2 \times 10^{-13}$ probability that no one in a simple random sample of 100 young adults owns a landline

e) $6.2 \times 10^{-61}$ probability that everyone in a simple random sample of 100 young adults owns a landline.

f) Binomial, with $n = 100, p = 0.75$

g) $4.5 \times 10^{-8}$ probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.

This means that $p = 0.75$

(a) On average, how many young adults do not own a landline in a random sample of 100?

Sample of 100, so $n = 100$

$E(X) = np = 100(0.75) = 75$

(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33$

(c) What is the proportion of young adults who do not own a landline?

The estimation, of 75% = 0.75.

(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?

This is P(X = 100), that is, all do not own. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}$

$3.2 \times 10^{-13}$ probability that no one in a simple random sample of 100 young adults owns a landline.

(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?

This is P(X = 0), that is, all own. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}$

$6.2 \times 10^{-61}$ probability that everyone in a simple random sample of 100 young adults owns a landline.

(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?

Binomial, with $n = 100, p = 0.75$

(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?

This is P(X = 50). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}$

$4.5 \times 10^{-8}$ probability that exactly half the young adults in a simple random sample of 100 do not own a landline.

8 0

2 years ago