If f(x)=square root 4x+9+2, which inequality can be used to find the domain of f(x)?

PLEASE HELP 10 POINTS

yuradex [85]

Answer:

27

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Look at this please i dont need help im just trying to tell everybody something

ch4aika [34]

Answeri did it

3Step-by-step explanation:

8 0

3 years ago

The height of a triangle is 5 m less than its base. The area of the triangle is 42 m².

marysya [2.9K]

A=bh/2, h=b-5 so

A=(b/2)(b-5)

A=(b^2-5b)/2 and A=42 so:

(b^2-5b)/2=42

b^2-5b=84

b^2-5b-84=0

b^2-12b+7b-84=0

b(b-12)+7(b-12)=0

(b+7)(b-12)=0 and since b>0

b=12m

5 0

3 years ago

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There are a total of 71 students in the cafeteria.

Vsevolod [243]

Answer:

49

Step-by-step explanation:

you can make the information into equations them solve simultaneously:

b + g = 71

b = 2g + 5

7 0

2 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