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nikklg [1K]
2 years ago
5

HAI HELP ME ASAP PLEASE

Mathematics
1 answer:
11111nata11111 [884]2 years ago
7 0

Answer:

Y/X = 2/3 x^2 + 16/3

Y= 2/3 x^3 + 16/3 x

Just replace y with y/X and X with x^2

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What is a real life word problem for (-7) + (7). Solve problem. Can you help?
notka56 [123]

Answer:

Step-by-step explanation:

Sally owes 7 dollars to Matthew. She opens her piggy bank and takes out 7 dollars. Now, how much money does she owe to Matthew?

-7 + 7 = 0

-7 + 7 is the same as  7 - 7

You can clearly see that 7-7 is 0.

3 0
3 years ago
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Please help me with this ❤️❤️❤️
tester [92]

Answer: h=P/mg

Step-by-step explanation:

7 0
2 years ago
A 19% discount on a watch saved a shopper $76 . find the price of the watch before the discount
Alik [6]
The watch costed $400 before the discount.

76 * 100 = 7,600
7,600 / 19 = 400
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
Inverse of the function h(x)=\dfrac{3}{2}(x-11)h(x)= 2 3 ​ (x−11)h, left parenthesis, x, right parenthesis, equals, start fracti
blagie [28]

Answer:

The inverse function of h(x) = 3/2*(X-11)  is:  g(x)= (2/3*X) +11

Step-by-step explanation:

h(x) = Y= 3/2*(X-11)

To find the inverse function, the first step is to exchange the position of the variables with each other. So;

X= 3/2*(Y-11)

Now we need to isolate the variable Y from this equation. The final result will be te inverse function.

X=3/2*(Y-11)

2*X=3*(Y-11)

2/3*X=Y-11

2/3*X +11 = Y

Y= (2/3*X) + 11  (Inverse function)

5 0
3 years ago
Read 2 more answers
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