What is the measure of ZRMN?

For better clarification I guess

marissa [1.9K]

Answer:

1.) $=\frac{\sqrt{3}}{3}$

2.) $42\sqrt{30}$

Step-by-step explanation:

1.) $\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{3}}{\sqrt{3}}$

2.) $\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{b}=\sqrt{a\cdot b}$

$\sqrt{6}\sqrt{5}=\sqrt{6\cdot \:5}$

6 0

3 years ago

Find a formula for the nth term in this arithmetic sequence. a1=0, a2=0.5, a3=1, a4=1.5

MrRissso [65]

Answer:

nth term is;

0.5n-0.5

Step-by-step explanation:

As we can see, the first term is 0

The common difference is 0.5-0= 1-0.5 = 1.5-1 = 0.5

Formula for nth term of an arithmetic sequence is;

a + (n-1)d

So we have

0 + (n-1) 0.5

= 0.5n-0.5

6 0

3 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

4 years ago

Which of these numbers is prime? Choose 1 answer: 3 30 63 77 E 91

Sav [38]

Answer:

3

Step-by-step explanation:

3 is the only number here with factors of itself (3) and 1

3 0

3 years ago

Read 2 more answers

Using the approximation 3.14 for \pi what is the radius of a circle with circumference 75.4m

Charra [1.4K]

Answer:

The radius of the circle is 12.006 m.

Step-by-step explanation:

Let us assume the radius of the circle = r

Circumference = 75.4 m

Now, CIRCUMFERENCE OF A CIRCLE = 2 π r

⇒ 75 .4 m = 2 π r

Now, putting π = 3.14, we get:

75 .4 m = 2 (3.14) r

⇒ 75.4 = 6. 28 r

or, r = 75.4/6.28 = 12.006

or, r = 12.006 m

Hence, the radius of the circle with circumference 75.4 m is 12.006 m.

3 0

4 years ago