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3 years ago

What are the two main branches of statistics?

Mathematics

2 answers:

Korolek [52]3 years ago

7 0

Answer:

descriptive statistics and inferential statistics

Step-by-step explanation:

Pavel [41]3 years ago

3 0

Descriptive and inferential

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Which value is the best estimate for x? A=24.4 B=27.0 C=63.0 D=65.6

svp [43]

I think it’s C. 63.0

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3 years ago

What is the domain and range of the following graph?

puteri [66]

Using it's concepts, the domain and the range of the graph are given as follows:

Domain: all real values except x = -1.

Range: All real values.

<h3>What are the domain and the range of a function?</h3>

The domain of a function is the set that contains all the values of the input. In a graph, it is given by the values of x, which is the horizontal axis of the graph.

The range of a function is the set that contains all the values of the output. In a graph, it is given by the values of y, which is the vertical axis of the graph.

In this graph, have that the function is defined for all values of x except x = -1, and assumes all real values, hence the domain and the range of the graph are given as follows:

Domain: all real values except x = -1.

Range: All real values.

More can be learned about the domain and the range of a function at brainly.com/question/10891721

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6 0

2 years ago

25/160 into a decimal??

Andrej [43]

Answer:

0.15625

Explanation:

Divide:

25 divided by 160

Which Equals by:

0.15625

Step-by-step explanation:

Hope this Helps!!

7 0

3 years ago

Read 2 more answers

Sunshine bank offers 6% simple interest on deposits. Tangerine bank offers 5% interest compound annually. State the difference i

Arada [10]

B) $35 Hope it helps :)

5 0

4 years ago

Read 2 more answers

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

3 years ago