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4 years ago

5

I was at the store and saw two sizes of avocados being sold. The regular size sold for $0.79 each. For what prices would the lar

ger avocado be a better deal?

1 answer:

Anon25 [30]4 years ago

3 0

Answer:

the price on the larger avocado maybe a little more but at the same time it's better because of the fact yes you're paying a little more but you get a little more than you would with the smaller avocados

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3 years ago

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slava [35]

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Step-by-step explanation:

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3 years ago

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3 years ago

There are 9 performers who are to present their acts at a variety show. How many different ways are there to schedule their appe

nika2105 [10]

For this type of question it is easiest to think about it in terms of a simple multiplication. The way we can do this is by working out how many possibilities there are for each position on the program. This is as follows:
For the first act, there are 9 possible performers we can pick
For the second act, because we have already assigned a performer to the first act, there are only 8 possible performers
For the third act, because we have already assigned a performer to the first two acts, there are only 7 possible performers
This goes on and on until we reach the final act, where we have used all the performers except one.
Therefore we can say:
For the first act there are 9 possibilities
For the second act there are 8 possibilities
For the third act there are 7 possibilities
For the fourth act there are 6 possibilities and so on
If we want to write this mathematically we can just say that there are:
9 possibilities x 8 possibilites x 7 possibilities etc right down until we are at the final act where there is 1 possibility.
To write this out as an equation we just say:
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = the number of different ways
In maths however, we have a much simpler way of saying this, which we call "factorial".
That whole stream of numbers up there can just be said to be "9 factorial", or if we are writing it in an equation "9!". You may find that your calculator has a button which has an "!" on it. If you put in the number of performers and then press the "!" button, it will give you the answer to your question.
Hope this helped you :)

3 0

3 years ago

In 2002, the mean age of an inmate on death row was 40.7 years with a standard deviation of 9.6 years according to the U.S. Depa

marissa [1.9K]

Answer:

The <em>95% confidence interval</em> for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Step-by-step explanation:

The <em>confidence interval</em> of the mean is given by the next formula:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$ [1]

We already know (according to the U.S. Department of Justice):

The (population) standard deviation for this case (mean age of an inmate on death row) has a standard deviation of 9.6 years ( $\\ \sigma = 9.6$ years).
The number of observations for the sample taken is $\\ n = 32$ .
The sample mean, $\\ \overline{x} = 38.9$ years.

For $\\ z_{1-\frac{\alpha}{2}}$ , we have that $\\ \alpha = 0.05$ . That is, the <em>level of significance</em> $\\ \alpha$ is 1 - 0.95 = 0.05. In this case, then, we have that the <em>z-score</em> corresponding to this case is:

$\\ z_{1-\frac{\alpha}{2}} = z_{1-\frac{0.05}{2}} = z_{1-0.025} = z_{0.975}$

Consulting a cumulative <em>standard normal table</em>, available on the Internet or in Statistics books, to find the z-score associated to the probability of, $\\ P(z$ , we have that $\\ z = 1.96$ .

Notice that we supposed that the sample is from a population that follows a <em>normal distribution</em>. However, we also have a value for n > 30, and we already know that for this result the sampling distribution for the sample means follows, approximately, a normal distribution with mean, $\\ \mu$ , and standard deviation, $\\ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$ .

Having all this information, we can proceed to answer the question.

Constructing the 95% confidence interval for the current mean age of death-row inmates

To construct the 95% confidence interval, we already know that this interval is given by [1]:

$\\ \overline{x} \pm z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

That is, we have:

$\\ \overline{x} = 38.9$ years.

$\\ z_{1-\frac{\alpha}{2}} = 1.96$

$\\ \sigma = 9.6$ years.

$\\ n = 32$

Then

$\\ 38.9 \pm 1.96*\frac{9.6}{\sqrt{32}}$

$\\ 38.9 \pm 1.96*\frac{9.6}{5.656854}$

$\\ 38.9 \pm 1.96*1.697056$

$\\ 38.9 \pm 3.326229$

Therefore, the Upper and Lower limits of the interval are:

Upper limit:

$\\ 38.9 + 3.326229$

$\\ 42.226229 \approx 42.23$ years.

Lower limit:

$\\ 38.9 - 3.326229$

$\\ 35.573771 \approx 35.57$ years.

In sum, the 95% confidence interval for the current mean age of death-row inmates is between 42.23 years and 35.57 years.

Notice that the "mean age of an inmate on death row was 40.7 years in 2002", and this value is between the limits of the 95% confidence interval obtained. So, according to the random sample under study, it seems that this mean age has not changed.

7 0

3 years ago