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Lemur [1.5K]
3 years ago
14

X=4 what is f(x)? what point is on the graph of f

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
8 0
You already have the x
F(x)=4
(4,0)
You might be interested in
Which notation best represents the phrase “no more than 200”
Lilit [14]

Answer:

X<_200

Step-by-step explanation:

We know that if the number connot be more than 200 then the number is less than or equal to 200

7 0
3 years ago
After surveying 100 parents, researchers concluded that 90 percent of babies can walk by the age of 10 months. Why is this an in
Maru [420]

This is an inappropriate conclusion because of <u>D. At 10 months</u>, ninety percent of toddlers have not reached the developmental milestones to walk.

<h3>What is an inappropriate conclusion?</h3>

An inappropriate conclusion is one arrived at without a proper research design.  An inappropriate conclusion can also be reached when the interpretation of the research result is not based on the findings.

Our conclusion implies that there is a wrong interpretation of the research findings because of the researcher's bias.

Thus, the research conclusion is <u>inappropriate</u> because of Option D.

Learn more about research conclusions at brainly.com/question/24542637

3 0
3 years ago
Is education related to programming preference when watching TV? From a poll of 80 television viewers, the following data have b
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
Lin's family has complete 70% of a trip. They have traveled 35 miles how far is the trip
Bogdan [553]

Answer  

50 miles:) I'm positive  but there may be a chance it's 200

4 0
3 years ago
If l is parallel to m, and m is parallel to a third line n, then l is __________ parallel to n.
sveticcg [70]
The appropriate choice is ...
  A. always

_____
For lines, the property of being parallel is transitive. A||B and B||C means A||C.
8 0
3 years ago
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