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Arturiano [62]
3 years ago
10

(2x+5y-z)+(-6x-4y+7z)

Mathematics
1 answer:
11111nata11111 [884]3 years ago
8 0

Answer:

-4x + y + 6z

Step-by-step explanation:

2x + 5y - z + (-6x - 4y + 7z) < distribute positive 1 to (-6x - 4y + 7z)

2x + 5y - z - 6x - 4y + 7z < combine like terms to get the answer

2x - 6x = -4x

5y - 4y = y

-z + 7z = 6z

our new expression is:

-4x + y + 6z we cannot simplify this any further, so this is our answer

You might be interested in
Which equation best matches the graph shown below?
n200080 [17]

Answer:

y= 0.2(x+1)^2+5 (Bottom left answer)

Step-by-step explanation:

Hope this helps :)

6 0
2 years ago
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
3 years ago
How can you tell that the data in the table is function or not?
stealth61 [152]

that is not a function, because there are two outputs for three. two inputs can have one output, but one input cant have two outputs.

6 0
3 years ago
Is a function or not
Anna71 [15]

Answer:

that is indeed a function

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Please help me with this homework
snow_lady [41]

Answer:

10.44 m

Step-by-step explanation:

Solve this as a triange. The base is 3m and the perpendicular is 10m. The angle between the base and wall is 90° so it will be a right angled triangle.

Solve this by Pythagoras theorm:

let length of ladder is c

base is b

And perpendicular is a

Acc to Pythagoras theorem:

C² = a² + b²

C² = 10² + 3²

C² = 100 + 9

C² = 109

√C² = √109

C = 10.44m

8 0
2 years ago
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