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vaieri [72.5K]
2 years ago
11

In the coordinate plane, the points A(-8, -6), B(-9, 11), and C(3. - 7) are reflected over the y-axis to the points A, B, and C'

, respectively. What are the coordinates of A' B', and C'?​
Mathematics
1 answer:
QveST [7]2 years ago
6 0

Answer:

A' = (8,-6)

B' = (9,11)

C' = (-3,-7)

Step-by-step explanation:

When reflecting a point over the y axis, multiply the point's x coordinate by -1. Do this for each point and you have your answer.

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PtichkaEL [24]
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2 years ago
A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur
Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.1 = 0.9, so Z = 1.28.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28\frac{1.5}{\sqrt{305}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

8 0
3 years ago
Come here if you can solve this highly complicated question~ :0
Natasha2012 [34]
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3 years ago
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Can someone think of a poem which starts with "Today I feel ...."<br>pls
GuDViN [60]
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Today I feel glad...
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7 0
3 years ago
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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
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