Which of the following inequalities is false?

What value for n makes the equation true? 8n - 2 = 2

kiruha [24]

Answer:

n = 1/2

Step-by-step explanation:

The answer is 1/2

I hope this helps you

3 0

3 years ago

Read 2 more answers

You made a deal with Ms.Williams to read 100 pages every night. You know that it takes you 32 minutes to read 16 pages. How long

AnnyKZ [126]

Put the problem into a proportion to make it easier to solve.

32 x
=
16 100

You can use cross products to simplify this proportion. Multiply 32 by 100, and set that number equal to 16 *x.

Your equation should now look like this
3200=16x

You are trying to get the variable alone on one side, so you should divide both sides by 16.

If this is performed correctly, then your final answer should be

x=200.

It will take you 200 minutes (3 hours 20 mins) to read the pages that you promised Ms. Williams

5 0

3 years ago

What is the definition of mean???

Scilla [17]

Mean, in terms of math, is the total added values of all the data in a set divided by the number of data <em>in</em> the set. Make sense? If not, here' an example...

Let's say this is my data set:
1, 2, 5, 4, 3, 8, 7, 4, 6,10

To find the mean...
Step 1: Add all of them together.
1+2+5+4+3+8+7+4+6+10 is what? 50. Now that you have this number...
Step 2: Divide by the amount there are. Basically, count up all of the numbers. How many are there? There are 10. Finally...
Step 3: Divide. 50/10 is 5, so the mean of this data set would be 5. Get it? I sure hoped this helped :)

3 0

3 years ago

Jeff earns $27,800 per year. What is his bi-monthly gross pay?

taurus [48]

$27,800/6=$4,633.33 if thats every two months if its twice a month them its $1158.33

6 0

3 years ago

Read 2 more answers

The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and

natali 33 [55]

Answer:

a) $\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03$

$p_v = P(\chi^2_{4} >17.03)=0.0019$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Step-by-step explanation:

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".

Assume the following dataset:

Quality management Excellent Good Fair Total

Excellent 40 35 25 100

Good 25 35 10 70

Fair 5 10 15 30

Total 70 80 50 200

Part a

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{70*100}{200}=35$

$E_{2} =\frac{80*100}{200}=40$

$E_{3} =\frac{50*100}{200}=25$

$E_{4} =\frac{70*70}{200}=24.5$

$E_{5} =\frac{80*70}{200}=28$

$E_{6} =\frac{50*70}{200}=17.5$

$E_{7} =\frac{70*30}{200}=10.5$

$E_{8} =\frac{80*30}{200}=12$

$E_{9} =\frac{50*30}{200}=7.5$

And the expected values are given by:

Quality management Excellent Good Fair Total

Excellent 35 40 25 100

Good 24.5 28 17.5 85

Fair 10.5 12 7.5 30

Total 70 80 65 215

And now we can calculate the statistic:

$\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(3-1)=4$

And we can calculate the p value given by:

$p_v = P(\chi^2_{4} >17.03)=0.0019$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent Ex=Reputation of company excellent

G= Quality Management good Gx=Reputation of company good

F= Quality Management fait Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

7 0

3 years ago