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3 years ago

What is the sample space for a 5 sided number cube?

Mathematics

1 answer:

grigory [225]3 years ago

3 0

Answer:

{1, 2, 3, 4, 5}

Step-by-step explanation:

Sample space is the set of all possible outcomes. Supposing that the 5 sided number cube has numbers one to five on its sides, the possible outcomes are the numbers that can be rolled, then its sample space is: {1, 2, 3, 4, 5}

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A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose pe

soldier1979 [14.2K]

Using the normal distribution, it is found that:

a) 0.8599 = 85.99% probability that x is more than 60.

b) 0.1788 = 17.88% probability that x is less than 110.

c) 0.6811 = 68.11% probability that x is between 60 and 110.

d) 0.0643 = 6.43% probability that x is greater than 125.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 87, thus $\mu = 87$ .
The standard deviation is of 25, thus $\sigma = 25$ .

Item a:

This probability is 1 subtracted by the p-value of Z when X = 60, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 87}{25}$

$Z = -1.08$

$Z = -1.08$ has a p-value of 0.1401.

1 - 0.1401 = 0.8599

0.8599 = 85.99% probability that x is more than 60.

Item b:

This probability is the p-value of Z when X = 110, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{110 - 87}{25}$

$Z = 0.92$

$Z = 0.92$ has a p-value of 0.8212.

1 - 0.8212 = 0.1788.

0.1788 = 17.88% probability that x is less than 110.

Item c:

This probability is the p-value of Z when X = 110 subtracted by the p-value of Z when X = 60.

From the previous two items, 0.8212 - 0.1401 = 0.6811.

0.6811 = 68.11% probability that x is between 60 and 110.

Item d:

This probability is 1 subtracted by the p-value of Z when X = 125, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{125 - 87}{25}$

$Z = 1.52$

$Z = 1.52$ has a p-value of 0.9357.

1 - 0.9357 = 0.0643.

0.0643 = 6.43% probability that x is greater than 125.

A similar problem is given at brainly.com/question/24863330

7 0

3 years ago

If 3x - 6y = 8, then x - 2y =

Fiesta28 [93]

Answer:

8/3 or 2.6666

Step-by-step explanation:

You have divided the first equation by 3 -- all except the 8. So you must divide the 8 by 3 giving you 8/3 which is 2 2/3 or 2.66666

Answer (x - 2y) = 8/3

8 0

3 years ago

Read 2 more answers

A local pizza shop has a membership program for frequent buyers. The membership costs $25 per month and members get a discounted

AlladinOne [14]

Answer:

Step-by-step explanationcvm

8 0

3 years ago

Read 2 more answers

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

3 0

3 years ago

Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit. Total

Andrews [41]

Jill had a total score of 40 on 13 quizzes, so her average was
40/13 ≈ 3.08

_____
1 quiz @ 1 = total of 1
3 quizzes @ 2 = total of 6
4 quizzes @ 3 = total of 12
4 quizzes @ 4 = total of 16
1 quiz @ 5 = total of 5
total score = 1 + 6 + 12 + 16 + 5 = 40
total quizzes = 1 + 3 + 4 + 4 + 1 = 13

8 0

3 years ago