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sukhopar [10]
4 years ago
7

write a rule that you can use to find the total number of beads in a bracelet when you know the total number of beads

Mathematics
1 answer:
liberstina [14]4 years ago
6 0
Reading the total number ?
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50 POINTS! Plot function h on the graph.<br><br> NO LINKS
tangare [24]

See attachment for the graph of the piecewise function h(x)

<h3>How to plot the function?</h3>

The function is given as:

h(x) = | -4,    x < 3

         | x + 5,    x >= 3

The above function is a piecewise function.

It has 2 separate functions at two domains

This means that we plot the sub-functions in the piecewise function at their respective domain

See attachment for the graph of the piecewise function h(x)

Read more about piecewise function at:

brainly.com/question/27262465

#SPJ1

3 0
2 years ago
It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains
sdas [7]

Answer:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

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

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

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 IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

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

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

6 0
3 years ago
2x-5 <br><br>-------- = 2<br><br> 8<br><br><br>(2x-5/8 = 2)
inysia [295]

Step-by-step explanation:

2×-5=-8 and 2×-5/8=-1.25

4 0
2 years ago
Read 2 more answers
The owner of a local grocery store increased his order of apples from 42 crates to 96 crates. What is the percent of increase in
Sonja [21]
Answer: 64% hope this helped
6 0
3 years ago
What is a line whos slope is -1/4
Ipatiy [6.2K]
Y=-1/4x+5 hope it helps
3 0
3 years ago
Read 2 more answers
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