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3 years ago

Identify the examples as primary sources or secondary sources.

Mathematics

1 answer:

algol133 years ago

5 0

Answer:

Primary : letters, artifacts, contracts

Secondary : documentaries, encyclopedias, biographies

Step-by-step explanation:

(I'm not too sure about contracts, so you might want to check yourself for that one)

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Juliana wants to find the mass of a watermelon which unit should she use

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I would use grams, because in my country watermelons are small, but if they were huge I would use kilograms

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3 years ago

Three potential employees took an aptitude test. Each person took a different version of the test. The scores are reported below

sergejj [24]

Answer:

Kerri

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Kerri = 1.7

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Vincent = 1

Step-by-step explanation:

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2 years ago

Is the function in the table linear or nonlinear and why? x y 0,−100 and 1,-50 and 2,0 and 3,100 and 4,150. (A)The function is n

Studentka2010 [4]

Answer:

The correct option is c.

Step-by-step explanation:

Linear function: The rate of change of a linear function is always constant.

Non-Linear function: The rate of change of a non-linear function is not constant.

From the given coordinate pairs it is noticed that the function is passing through the points (0,-100), (1,-50), (2,0), (3,100) and (4,150).

$m=\frac{y_2-y_1}{x_2-x_1}$

The slope of function for points (0,-100) and (1,-50) is

$m_1=\frac{-50-(-100)}{1-0}=50$

The slope of function for points (2,0) and (3,100) is

$m_2=\frac{100-0}{3-2}=100$

Since the slopes of function are different, therefore the given function is non-linear.

The function is not linear because the rate of change is not constant and option c is correct.

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3 years ago

Read 2 more answers

What is the range of the function f(x) = 2x +4 when the domain is {-2, 0, 3}?

Vladimir79 [104]

Look at the attached picture..

Hope it will be helpful to you...

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3 years ago

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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