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3 years ago

What is the policy about blank assignments?

Mathematics

1 answer:

Elan Coil [88]3 years ago

7 0

Answer:

What is all about no assignment policy?

“By ensuring that they complete all assignments and projects in school, the no-homework policy enables our learners to find balance between their academic development and personal growth by having ample time for enjoyable activities with family,” the department said in a statement

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Efectúa las siguientes operaciones, aplicando la propiedad distributiva de la multiplicación. De ser posible simplifique los res

Alchen [17]

Answer:

Step-by-step explanation:

e) 3√2 (-3√6 -2 -6)

3√2 (-3√6 - 8)

-9√12 - 24√2

-9√(2^2 * 3) - 24√2

-18√3 - 24√2

6(-3√3 - 4√2)

f) -5√2 (3 + 2√2 - 6 - 1)

-5√2(-4 + 2√2)

20√2 - 20

20(√2 - 1)

7 0

3 years ago

How do you complete this One-Step equation?<br>5 1/2 + p = 6

dalvyx [7]

5 1/2 + p = 6

subtract 5 /12 from each side

5 1/2 - 5 1/2+ p = 6 - 5 1/2

p = 6 - 5 1/2

borrow from the 6

p = 5 2/2 - 5 1/2

p = 1/2

8 0

3 years ago

Read 2 more answers

Which of the purchases would be the least expensive? Purchase Amount of sale Sales tax rate A $150 5% B $145 7% C $140 8% D $135

Lera25 [3.4K]

Answer:

It is d

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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

An inequality is shown. 27/7n > 4/3 Select all the values of n that make the inequality true.

ankoles [38]

Answer:

Step-by-step explanation:

<h2>Hope it help & study well</h2>

6 0

3 years ago