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Colt1911 [192]
2 years ago
13

Express the product of 2x^2+6x-8 and x+3 in standard form

Mathematics
1 answer:
Radda [10]2 years ago
7 0
The answer is  2x³ + 12x² + 10x - 24

(2x² + 6x - 8)(x + 3) = x(2x² + 6x - 8) + 3(2x² + 6x - 8) =
                                = 2x³ + 6x² - 8x + 6x² + 18x - 24 =
                                = 2x³ + (6x² + 6x²) + (-8x + 18x) - 24 =
                                = 2x³ + 12x² + 10x - 24
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Find the area of the circle. Round your answer to the nearest tenth.
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Step-by-step explanation:

area of a circle

A = pi x r^2

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3 years ago
The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma
nata0808 [166]

Answer:

a) \sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b) Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

c) P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d) P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e) P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

X \sim N(110000,40000)  

Where \mu=110000 and \sigma=40000

If we select a sample size of n =35 the standard error is given by:

\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

P(X > 112000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)

And we can use the complement rule and we got:

P(Z>0.354) = 1-P(Z

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

P(X > 100000)

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)

And we can use the complement rule and we got:

P(Z>-1.768) = 1-P(Z

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

P(100000

And we can use the z score given by:

z= \frac{\bar X  -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(100000< \bar X

And we can use the complement rule and we got:

P(-1.768

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mixas84 [53]

Answer:

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6 0
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lukranit [14]

Answer:

Step-by-step explanation:

1.5x^2+6x+.... so that it can be rewritten in the form a(x+b)^2

Let .... be C

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Compare above with 1.5(x+2)^2 = 1.5(x^2+4x+4)

C/1.5=4

C=6

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It can be rewritten to 1.5(x+2)^2

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Vitek1552 [10]
5x+128 should be it. 65+75=140. 140-12=128. you can't do anything to the x because they're not common terms so you leave it at that.
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3 years ago
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