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2 years ago

What is the measure of Angle N M R?

Mathematics

2 answers:

hichkok12 [17]2 years ago

7 0

Answer:

The correct answer is 63.

Step-by-step explanation:

irga5000 [103]2 years ago

5 0

Answer:

I took the test!

Step-by-step explanation:

hope this helps. :D

You might be interested in

Jonathan put ice cream in the freezer, and the ice cream remained frozen.

Phantasy [73]

Answer:

C- Based on the data, this is an example of causation

Step-by-step explanation:

This is because by him putting the ice cream into a freezing area, the ice cream remained frozen. Imagine if he put in in a warm area, it would melt. Hope this helps!

8 0

2 years ago

Read 2 more answers

Which of the following is equivalent to x^-1/2 1) -1/2x 2) -√x 3) 1/√x 4) -1/2x

irina [24]

Answer:

Option (3) is correct.

Step-by-step explanation:

We need to find the expression that is equivalent to $x^{-1/2}$ .

The property is : $x^{-a}=\dfrac{1}{x^a}$

Here, a = 1/2

So,

$x^{-1/2}=\dfrac{1}{x^{1/2}}$

We know that, $x^{1/2}=\sqrt{x}$

$\dfrac{1}{x^{1/2}}=\dfrac{1}{\sqrt{x} }$

So, $x^{-1/2}$ is equivalent to $\dfrac{1}{\sqrt{x} }$

Hence, the correct option is (3).

3 0

2 years ago

At a school fair, students were challenged to hit one of the small congruent circles on the large rectangular board with a ball.

shtirl [24]

Answer:

~4.7%

Step-by-step explanation:

Area of the rectangle: 35*52=1820

Area of 1 small circle: A=πr2=π·32≈28.27433

28.27433*3=84.82299

84.82299/1820=0.04660603846

Therefore, about 4.7%

6 0

2 years ago

HELP PLS PLS PLS PLS

solmaris [256]

Answer:

The probability of a person going to work by car is 48

And the probability of different way is 52

Step-by-step explanation:

As it's represented in the table above, there are 48 percent employees use cars as their transportation to work, and to find the rest sum all the probability of other transportation together which will give you 52 percent.

5 0

1 year ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence 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:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

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

The statistic to check the hypothesis is given by:

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

Part b

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:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

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

"=1-CHISQ.DIST(33.75,2,TRUE)"

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

7 0

2 years ago