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amid [387]
3 years ago
11

Please help me pleaseeee

Mathematics
2 answers:
adelina 88 [10]3 years ago
8 0
The answer is n=18

Hope this helps :))
Amanda [17]3 years ago
5 0

Answer:

18

Step-by-step explanation:

n=(6*42)/14

n=252/14

n=18

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Rewrite (21+31).4 using the disbributive property of multiplication over addition
Mamont248 [21]

Answer: evaluate it is 52

Step-by-step explanation:

7 0
3 years ago
-81= −45−9n i need help please
miss Akunina [59]

-45 - 9n = -81

-9n = -81 + 45

-9n = -36

n = -36/-9

n = 4

Hope this helps you. :)

4 0
3 years ago
Read 2 more answers
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.

7 0
4 years ago
Lisa is now six years older than Cheryl. In two years, Lisa will be twice as old as Cheryl. How old is Lisa now?
Elena-2011 [213]
--------------------------------------------------
Define x :
--------------------------------------------------
Let Cheryl be x years old.
Cheryl = x
Lisa = x + 6

--------------------------------------------------
In two years time:
--------------------------------------------------
Cheryl = x + 2
Lisa = x + 8

--------------------------------------------------
Lisa will be twice as old as Cheryl :
--------------------------------------------------
x + 8 = 2(x + 2)

--------------------------------------------------
Solve x :
--------------------------------------------------
x + 8 = 2(x + 2)
x + 8 = 2x + 4
x = 4

--------------------------------------------------
Find their age :
--------------------------------------------------
Cheryl = x = 4
Lisa = x + 6 = 4 + 6 = 10

--------------------------------------------------
Answer: Lisa is 10 years old now.
--------------------------------------------------
4 0
4 years ago
If there is a 30% chance of sun tomorrow and a 40% chance of wind and no sun, what is the probability that it is windy, given th
slega [8]
A, 57%, because 0.40 divided by all possible outcomes, which is 0.70 because we know it will not be sunny tomorrow, is 0.57.
8 0
3 years ago
Read 2 more answers
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