All categories

3 years ago

Ignore the answer I put please help and I’ll give brainliest

Mathematics

1 answer:

Leokris [45]3 years ago

5 0

Answer:

Step-by-step explanation:

3/12 reduces to 1/4.

1/2 alone is greater than 1/4, so when 2/10 is added to 1/2, it is certainly greater than 3/12.

Answer: B.

You might be interested in

Find the range. Write your answer in algebraic notation (with the inequalities).

Svet_ta [14]

Answer:

(-3, infinity)

Step-by-step explanation:

Range the y values, meaning up and down. The highest point is neverending therefore, infinity. The lowest point is -3.

I'm pretty sure this is correct.

4 0

3 years ago

Alberta has 5 shirts and 4 pants. 3 of the shirts are her favorite and 2 of the pants are her favorite. If Alberta chooses an ou

tiny-mole [99]

Answer:

$\frac{3}{10}$

Step-by-step explanation:

In probability, "AND" means "multiplication" and

"OR" means "addition".

We want the probability that she picks FAVORITE SHIRT "AND" FAVORITE PANT.

Probability that she picks favorite shirt = 3/5

Probability that she picks favorite pants = 2/4 = 1/2

Since, "AND", we "multiply" both to get:

3/5 * 1/2 = 3/10

5 0

4 years ago

An art teacher recorded whether the students in three of her classes prefer water color or oil painting. She plans to create

il63 [147K]

Answer:

C. class number and painting preference.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Are there any outliers in the data? What does the outlier mean?

Brrunno [24]

Answer:

<em>44, definition below</em>

Step-by-step explanation:

An outlier is a number that is much larger or smaller than the general populous of numbers. For example, if there was a dot plot with dots clustered at 2, 3, and 5, but there was one or two dots around 10, then 10 would be considered an outlier.
I would consider 44 to be an outlier in the number set.

If I am incorrect in my reasoning, please let me know so that I can plan better for my future answers. Have an amazing day.

5 0

3 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

4 years ago