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2 years ago

Please answer correctly !!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Mamont248 [21]2 years ago

7 0

<h2><em>If you put and label the three points in a line, you will see that RS + ST = RT. </em></h2><h2><em> </em></h2><h2><em>Then you only need to substitue with the expressions given for RS and ST to find RT. </em></h2><h2><em> </em></h2><h2><em>RT = x +1 + 2x - 2 = 3x - 1 </em></h2><h2><em> </em></h2><h2><em>Also, RT = 5x - 5 </em></h2><h2><em> </em></h2><h2><em>Then, 3x - 1 = 5x - 5 </em></h2><h2><em> </em></h2><h2><em>5x - 3x = 5 - 1 </em></h2><h2><em> </em></h2><h2><em>2x = 4 </em></h2><h2><em> </em></h2><h2><em>x = 4/2 = 2 </em></h2><h2><em> </em></h2><h2><u><em>X=2</em></u></h2>

viktelen [127]2 years ago

3 0

X Would Be 2 I Hope That Helped .

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10 x 200 is how many tens

melisa1 [442]

10 x 200 is 2000
therefore there are 200 tens .

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3 years ago

Read 2 more answers

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.

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3 years ago

What is the slope in this equation?

Elena L [17]

Answer:

3/2 looks right to me.....

6 0

3 years ago

Read 2 more answers

What is the surface area of a die in which each edge has a length of 15 millimeters? (Hint: On a cube, the length, width, and he

oee [108]

In order to find the surface area of a cube, you need to find the area of one of the sides.

15*15 = 225

this would then be multiplied by 6 to account for all of the sides of the cube.

225*6 = 1350 millimeters ^2

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3 years ago

Consider the diagram below

MissTica

Sbsnsjsuhxhxjxsii t amamandjdmx

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3 years ago