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3 years ago

15

A sewing instructor asked his students how many spools of knitting yarn they carried in their sewing kits. The table shows the d

ata he recorded.

1 answer:

const2013 [10]3 years ago

4 0

I think the answer is mean. (The average) it could also be “mean average deviation” but I’m not sure what that one means...

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Can you please help me out this is DUE IN 5 MINUTES please.

Rina8888 [55]

Answer

You flip it, so

Step-by-step explanation:

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3 years ago

Let [n] = 1 if n is odd and 0 if n is even, for all positive integer n. if [n] * [n+8] = 0, then what is one possible value of n

astra-53 [7]

$[n][n+8]=0$

if either or both of $[n]$ and $[n+8]$ are 0, i.e. if one or both of $n$ and $n+8$ are even. But both will be even if $n$ is even, and odd otherwise, so any even $n$ will be a solution, e.g. $n=2$ .

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3 years ago

Find the measure of the indicated angle

Paul [167]

Add them together and find x :)

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2 years ago

Suppose a simple random sample of size n is drawn from a large population with mean mu and standard deviation sigma. The samplin

Sidana [21]

Answer:

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

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

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

$\mu_{\bar X} = \mu$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

Step-by-step explanation:

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".

Solution to the problem

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

$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

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

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

$\mu_{\bar X} = \mu$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

6 0

3 years ago

Read 2 more answers

ANSWER PLZ QUICK AND FAST WHAT THE ANSWER QUICK

pav-90 [236]

The answer is B
Because when x=0, y=-4 and when y=0, x=-1. Hope I was able to help

5 0

3 years ago