Can the sides of a triangle have lengths 7, 15, and 15?

g On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any

Ksju [112]

Answer:

0.249579

Step-by-step explanation:

P(rain) = 0.3

P(no rain) = 1 - 0.3 = 0.7

The event of rain falling a second time within the next 5 days is possible in these ways

1. Rain on days 1 and 2

2. Rain on days 1 and 3; none on day 2

3. Rain on days 1 and 4; none on days 2 and 3

4. Rain on days 1 and 5; none on days 2, 3 and 4

5. Rain on days 1 and 6; none on day 2, 3, 4 and 5

$P(\text{second rain within 5 days}) = 0.3^2+0.3^2\times0.7+0.3^2\times0.7^2+0.3^2\times0.7^3+0.3^2\times0.7^4 = 0.3^2(1+0.7+0.7^2+0.7^3+0.7^4)= 0.09\times2.7731=0.249579$

5 0

4 years ago

Which set of ordered pairs represents a function?

KATRIN_1 [288]

Answer:

{(3, -1), (7, 1), (-6, -1), (9, 1), (2, -1)}

Step-by-step explanation:

Any relation having repetitive x-coordinates, is NOT a function.

3 0

3 years ago

The table below shows the price for different numbers of notebook:

levacccp [35]

So if 2 notebooks cost 12 divide that
12/2 =6 $ each
So to check if its true the 3notebooksx $6 dollars each should =$18 for 3 notebooks
5 notebooks at $6 each = $30 for 5 notebooks
so What is 4notebooks x$6 each ? =4x6 = $24 for 4 notebooks

8 0

3 years ago

this year, the winner of the new york marathon ran the 8 kilometer event in 24 minutes. what is the runners unit rate

DedPeter [7]

Based on the given situation above, the given values are 8 kilometers and 24 minutes. To simplify these values, we will come up with the result of 1 km in 3 minutes. Therefore, the winner of the New York marathon ran 0.33 kms per one minute so the runners unit rate is 0.33kms/minute.

3 0

3 years ago

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