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aleksklad [387]
3 years ago
13

Can someone explain the snake method, I don't understand how to do it

Mathematics
2 answers:
gavmur [86]3 years ago
8 0
<h3>Answer: <em>x ∈ [−2; 1] ∪ 3.5</em></h3>

Step-by-step explanation:

storchak [24]3 years ago
6 0

Answer:

  • x ∈ [−2; 1] ∪ 3.5

Step-by-step explanation:

<h3>Given</h3>

<u>Inequality</u>: (x-1)(x+2)(2x-7)≤0

<h3>Solution: </h3>

<u>If we solve the corresponding equation (x-1)(x+2)(2x-7)²= 0, we get roots </u>

  • x =  -2, 1, 3.5

<u>We need to consider the following 4 intervals: </u>

  • (−∞; −2), [−2; 1], (1; 3.5), (3.5; ∞)

<u>1st interval</u> (−∞; −2)

  • The expression (x-1)(x+2)(2x-7)² is positive as two of the multiples are negative and one is always positive (square number), and therefore  does not satisfy the inequality.

<u>2nd interval</u>  [−2; 1]

  • The expression is negative as only one of the multiples is negative, and therefore the interval (−1; 2) satisfies the inequality.

<u>3rd interval</u> (1; 3.5)

  • The expression is positive as all the multiples are positive. Therefore, the interval (1; 3.5) also does not satisfy the inequality.

<u>4th interval</u>

  • The expression is positive as above, and therefore also does not satisfy the inequality.  

<u>So, the answer to the inequality is: </u>

  • x ∈ [−2; 1] ∪ 3.5
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Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of
frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

3) \chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

4) df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association 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:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married                     21                              37                            58                116

Not Married              59                             63                            42                164

Total                          80                             100                          100              280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is \alpha=0.05

Part 2

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

Part 3

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{80*116}{280}=33.143

E_{2} =\frac{100*116}{280}=41.429

E_{3} =\frac{100*116}{280}=41.429

E_{4} =\frac{80*164}{280}=46.857

E_{5} =\frac{100*164}{280}=58.571

E_{6} =\frac{100*164}{280}=58.571

And the expected values are given by:

                    Diag. Alcoholic   Undiagnosed Alcoholic    Not alcoholic    Total

Married             33.143                       41.429                        41.429                116

Not Married     46.857                      58.571                        58.571                164

Total                   80                              100                             100                 280

And now we can calculate the statistic:

\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

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Answer:

8

Step-by-step explanation:

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