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3 years ago

What is a complex fraction?

2 answers:

7 0

A complex fraction is a fraction with a mixed number, or in it's numerator(Top number) or denominator(Bottom number) or both.

zalisa [80]3 years ago

5 0

A complex fraction is a fraction that includes another fraction for example, 1/2
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6

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Help fast asap! Please

const2013 [10]

A circle is 360°. So 360/12= 30°. Hope that helps.

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3 years ago

Read 2 more answers

The graph below shows Roy's distance from his office (y), in miles, after a certain amount of time (x), in minutes: Graph titled

Marina CMI [18]

Answer:

Felix

Step-by-step explanation:

The graph contains 3 segments,

first one is for the first 4minutes,

second one is for the next 2 minutes (standing still)

third one is for the last 2 minutes.

Only Felix has it right, the other students use absolute time in their statements, in stead of the difference between start and end. (e.g., from 4 to 6 is 2 minutes).

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3 years ago

the ishango bone eas found 20000years ago in the congo. why do archaeologist bellieved that the scratxhes on this bine was from

Lerok [7]

Answer:

Step-by-step explanation:

The Ishango bone is a bone tool, dated to the Upper Paleolithic era. It is a dark brown length of bone, the fibula of a baboon, with a sharp piece of quartz affixed to one end, perhaps for engraving. It was first thought to be a tally stick

3 0

2 years ago

Find dy, dx if f(x) = (x + 8)3x. 3xln(x + 8) the product of the quantity 3 times the natural log of the quantity x plus 8 plus 3

boyakko [2]

F(x) = 9x^2(x + 8)^2

8 0

2 years ago

Read 2 more answers

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:

$P(-1.768$

8 0

3 years ago