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ArbitrLikvidat [17]
3 years ago
11

I don't understand these problems. 1. 2.4 divided by 6

Mathematics
2 answers:
ch4aika [34]3 years ago
4 0
2.4 divided by 6 = 0.4 
Nadya [2.5K]3 years ago
3 0
1) 1 divided by 6 is 0.16666666666 so you can round that into 0.2
2) 2 divided by 6 is 0.33333333333 so you can round that to 0
3) 4 divided by 6 is 0.66666666666 so you can round that to 0.1
4) !!!***PLEASE give this a THANKS!!! :) :D :)) ALSO PLEASE give this the BRAINIEST ANSWER/CROWN! TYSM! :)***!!!
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Select the correct description for the expression 5(x-7)
Anettt [7]

Answer:

d

Step-by-step explanation:

i got it

3 0
3 years ago
Read 2 more answers
A geometric sequence is defined by the explicit formula an=5(-4)^n-1 what is the recursive formula for the nth term of this sequ
Furkat [3]
Hello,

a_{n}=5*(-4)^{n-1}\\
 a_{n-1}=5*(-4)^{n-2}\\\\
\dfrac{a_{n}}{a_{n-1}}=(-4)^{n-1-(n-2)}=-4\\\\

Formula\ is\ :\\
 a_{n}=-4*a_{n-1}\\
a_{1}=5\\



4 0
4 years ago
Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of
alexandr402 [8]

Answer:

1) H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

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

3) \chi^2 = \frac{(21-33.143)^2}{31.143}+\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) p_v = P(\chi^2_{2} >19.72)=0.00005222

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated 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:

                    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 significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

Part 2

The statistic to check the hypothesis is given by:

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

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:

                    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

Part 3

And now we can calculate the statistic:

\chi^2 = \frac{(21-33.143)^2}{31.143}+\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)=(2-1)(3-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.72)=0.00005222

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

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

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

8 0
3 years ago
Find the Area of the figure below, composed of a parallelogram and one semicircle. Rounded to the nearest tenths place
jolli1 [7]
The answer is248.5 units^2
4 0
3 years ago
Read 2 more answers
4 points
Natalija [7]

Answer:

1575 cubic feet

Step-by-step explanation:

3 0
3 years ago
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