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Tema [17]
2 years ago
15

Solve 18>12+x. Graph the solution.

Mathematics
1 answer:
Dmitry [639]2 years ago
8 0

Answer:

x < 6

Step-by-step explanation:

Step 1: Subtract 12 from both sides.

  • x+12-12 < 18-12
  • x < 6

Therefore, the answer is x < 6.

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Date : n _<br>5+ [ 4 +(-7) -{ 6 ×(5+ 1x4)}]<br><br>​
s2008m [1.1K]

Answer:

−28−6^2x4

Step-by-step explanation:

6 0
2 years ago
How your day been? Mines been kinda great.​
garri49 [273]

Answer:

kinda bad

Step-by-step explanation:

but overall good

6 0
2 years ago
Natalie has 28 erasers. She divided some of her erasers equally among 3 friends. Natalie has 10 erasers left. How many erasers d
Korvikt [17]
Hey there! :D

Subtract 10 from 28. 

28-10= 18

Now, divide that by 3. (for her three friends) 

18/3= 6

Each friend got 6 erasers. 

I hope this helps!
~kaikers
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
Determine if each ratios or rates are equivalent.
maxonik [38]

Answer:

no these are not equivalent

Step-by-step explanation:

these aunt equivalent because 12+12 =24 and 3+3=6 not 7 or 28 so no ccx these are not equivalent

4 0
3 years ago
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