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3 years ago

What is the slope of the line

Mathematics

2 answers:

Zina [86]3 years ago

6 0

Your slope is -3/2
Hope that helps!

noname [10]3 years ago

6 0

Let's choose 2 points:
(-2, 3) and (2, -3)
slope = difference in y / difference in x
slope = (3 --3) / (-2 -2)
slope = 6 / -4
slope = -3 / 2

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Which shows the products ordered from least to greatest

alexgriva [62]

Answer:

Can you add a picture to your question?

Step-by-step explanation:

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2 years ago

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Of the coffee makers sold in an appliance store, 4.0% have either a faulty switch or a defective cord, 2.5% have a faulty swi

lukranit [14]

Answer: Hence, Probability that a coffee maker will have a defective cord is 0.376=37.6%.

Step-by-step explanation:

Since we have given that

Let A be the event of getting a faulty switch.

Let B be the event of getting a defective cord.

Here, P(A∪B) = 4% = 0.04

P(A∩B) = 0.1% = 0.001

P(A) = 2.5% = 0.025

We need to find P(B):

As we know that

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\0.04=P(A)+0.025-0.001\\\\0.04=P(A)+0.024\\\\P(A)=0.04-0.024=0.376$

Hence, Probability that a coffee maker will have a defective cord is 0.376=37.6%.

6 0

3 years ago

Can anyone help me with this?

Illusion [34]

Answer:

Step-by-step explanation:

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2 years ago

What is the solution to the inequality? −4x−8>−20

iogann1982 [59]

Answer:

see work attached and shown

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2 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

2 years ago