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3 years ago

Mount Everest is the highest point on

Mathematics

2 answers:

vlabodo [156]3 years ago

7 0

Due to the earth level C -2,180 would b correct

Dahasolnce [82]3 years ago

6 0

Answer:

19,880 meters

Step-by-step explanation:

Add both ( 11,030+8,850) = 19,880 meters would be your difference!

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago

F(x) = |3x-4| give the domain and range

gulaghasi [49]

Answer:

Domain: -∞≤x≤∞

Range: 0≤F(x)≤∞

Step-by-step explanation:

The x variable can take any value.

In the case of F(x), the function only can take positive values, due to the absolute value function, and the function does not have any other restriction.

3 0

3 years ago

Can someone please explain this problem to me idk what I am doing wrong and can you draw the problem for me thanks

Bogdan [553]

Answer:

you're not doing anything wrong

Step-by-step explanation:

In order for cos⁻¹ to be a function, its range must be restricted to [0, π]. The cosine value that is its argument is cos(-4π/3) = -1/2. You have properly identified cos⁻¹(-1/2) to be 2π/3.

Cos and cos⁻¹ are conceptually inverse functions. Hence, conceptually, cos⁻¹(cos(x)) = x, regardless of the value of x. The expected answer here may be -4π/3.

As we discussed above, that would be incorrect. Cos⁻¹ cannot produce output values in the range [-π, -2π] unless it is specifically defined to do so. That would be an unusual definition of cos⁻¹. Nothing in the problem statement suggests anything other than the usual definition of cos⁻¹ applies.

This is a good one to discuss with your teacher.