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omeli [17]
2 years ago
5

What is the value of expression for 5.75 - 1 2 (20 ÷ 2.5) ÷ 2 + 6

Mathematics
1 answer:
Korolek [52]2 years ago
4 0

Answer:

-36.25

Step-by-step explanation:

5.75 - 1 2 (20 ÷ 2.5) ÷ 2 + 6

5.75 - 1 2 (8) ÷ 2 + 6

5.75 - 96 ÷ 2 + 6

5.75 - 48 + 6

= -36.25

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lakkis [162]

Answer:

hope it helps...any queries comment me!!!

4 0
2 years ago
(a) An angle measures 43°. What is the measure of its supplement
Inga [223]
The answer is 137° because supplementary means the 2 angles must equal 180°
8 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
Hello Help Plis…………..
Lostsunrise [7]

Answer:

7. 20 + 5 + 3/10 + 6/100

8. 7 + 4/10 + 2/100 + 5/1000

Step-by-step explanation:

We are writing two numbers in expanded form, to help, we can use the guide above both numbers. Expanded form is all about breaking apart a number by place value and putting them in an additional sequence.

Let's look at 7, 25.36.

There are the tens, ones, tenths, and hundredths place being used, therefore :

20 + 5 + 3/10 + 6/100

We can do the same for 8, 7.425.

There are ones, tenths, hundredths, and thousandths being used, therefore :

7 + 4/10 + 2/100 + 5/1000

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2 years ago
Which statement best describes the effect on the graph of y = (x – 9)2 if the equation is changed to y = (x + 9)2?
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Answer:

You won't be in the negative but postive side of the graph

Step-by-step explanation:

6 0
3 years ago
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