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gregori [183]
3 years ago
15

I WOULD APPRECIATE IT A LOT IF SOMEONE COULD ANSWER THESE PROBLEMS!!! THANK YOU IF YOU CAN!

Mathematics
1 answer:
romanna [79]3 years ago
8 0
10. C.

11. D.

12. C.

13. C.

14. A.

15. D.
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Find (F/g)(x)<br> F(x) = sqrt x^2-1<br> g(x) sqrt x-1
Margaret [11]

Answer:

Option A. √(x + 1)

Step-by-step explanation:

Data obtained from the question include:

f(x) = √(x² – 1)

g(x) = √(x – 1)

(f/g) (x) =..?

(x² – 1) => difference of two square

(x² – 1) => (x – 1)(x + 1)

f(x) = √(x² – 1)

f(x) = √(x – 1)(x + 1)

(f/g) (x) = f(x) /g(x)

f(x) = √(x – 1)(x + 1)

g(x) = √(x – 1)

(f/g) (x) = √(x – 1)(x + 1) / √(x – 1)

(f/g) (x) = √[(x – 1)(x + 1) / (x – 1)]

(f/g) (x) = √(x + 1)

7 0
3 years ago
7th GRADE WORK HELPPP!!! ILL BRAINLIST!!!
Ksenya-84 [330]

Answer: Rachel's prediction is faster than the actual unit rate, and Theadore's prediction is slower.

6 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
Express the following numbers as a ratio between an integer and a natural number.
Julli [10]

Answer:

Step-by-step explanation:

a) 7/5, 3/10, -13/4,-27/1,0

b)36/1,-45/1,21/5, -4/5, 91/6,-2/9

5 0
3 years ago
Solve x^2 – 7x = -13.
lara31 [8.8K]

Answer:

No solutions I think

Step-by-step explanation:

8 0
3 years ago
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