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3 years ago

Pleaseee Help. Don't answer if you aren't going to take it serious.

1 answer:

6 0

<span>Stem-and-leaf plots are used for showing the frequency of which certain classes of values occur. use a stem-and-leaf plot and let the numbers themselves to show pretty much the same information. In short, they are meant to show frequencies of which numbers or values occur. Not for colors or objects.

Yes, a frequency table is a perfect way to solve the problem! These are meant to help you find the number of something that occurs, and is a simple and clean way to show your word.</span>

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∠PQR and ∠SQR form a linear pair. Find the sum of their measures. ____

sdas [7]

A linear pair means the adjacent angles form a line...meaning they equal 180 degrees

6 0

3 years ago

What do the medians and ranges of two dot plots tell you about the data

Tatiana [17]

That both (along with the mean) help the data be described the data's spread, which can be the slope if only two points.

5 0

3 years ago

When you multiply a function by -1 what is the effect on its graph?

vlada-n [284]

Answer:

Step-by-step explanation:

Depends on what you mean by multiplying by - 1. I assume you are not going to multiply the y or f(x) term by - 1.

If that is so, take an example. Suppose you have a graph that is y=x^2

That's a parabola that opens upwards and it has a line going through its focus which is a point on the +y axis.

When you multiply the right hand side by - 1, the graph you get will be y = - x^2.

That opens downward and the focus is on the - y axis.

That means that the effect of the graph is that it flips over the x axis, which I think is the third answer.

5 0

3 years ago

Math question please help

almond37 [142]

Answer:

y = -7x +8

Step-by-step explanation:

The slope-intercept form of the equation of a line is ...

y = mx +b

where m represents the slope, and b represents the y-intercept. The y-intercept is the value of y when x=0. Here, you have m=-7 and b=8. Your equation is ...

y = -7x +8

7 0

2 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

3 years ago