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miss Akunina [59]
3 years ago
12

HELP ASAP!!

Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0
Form: y=ax+b
4a + b = 7 \\ 0a + b = 7
b=7
a=0
y=0x+7
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5x – 19 &lt;1<br> OR<br> - 4x + 3 &lt; -6<br> Solve for x
astra-53 [7]

Answer:

x < 4

OR

x > - 2.25

= 4 > x > - 2.25

Step-by-step explanation:

5x - 19 < 1

5x < 1 + 19

5x < 20

x < 20/5

x < 4

OR

- 4x + 3 < - 6

- 4x < - 6 - 3

- 4x < - 9

x > 9/- 4

x > - 2.25

7 0
3 years ago
How could you use coordinate geometry to prove that segment BC is congruent to segment AD?
liberstina [14]

Answer:

  Prove the lengths are the same

Step-by-step explanation:

When we say segments are congruent, we mean their lengths are the same.

__

Let's see if they are congruent.

  AD = √((3-(-3))² +(2-2)²) = 6

  BC = √((6-0)² +(6-6)²) = 6

AD ≅ BC . . . . their lengths are the same

8 0
3 years ago
Read 2 more answers
Sin^2x=1/4. Find all solution on the interval [0,2pi).
steposvetlana [31]
I really don’t know I’m just answering questions that I know sorry
4 0
3 years ago
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
If 300 words or typed in 6 minutes how many words will be typed in 17 minutes
Ierofanga [76]

Answer:

850 words can be typed in 17 minutes

Step-by-step explanation:

Divide 300 by 6 and multiply it by 17 and that is your answer, 850.

5 0
2 years ago
Read 2 more answers
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