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2 years ago

What is the equation of the line in slope-intercept form? y = _ x+_=

Mathematics

1 answer:

pashok25 [27]2 years ago

4 0

Answer:

y=3/5x+3

Step-by-step explanation:

So find 2 points, in this case it will be

(-5,0) and (0,3)

Use the slope formula---> y2-y1=x2-x1 and enter the points above

= 3-0/0-(-5)

3/5

Slope= 3/5

y=3/5x+b

then enter (0,3)

3=0+b

b=3

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What is the least common denominator for adding the fractions 4/15, 1/12, and 3/8

il63 [147K]

I think you want the least common multipule (exg LCM of 2,3,4 is 12)

so we find the factors and include all of them so like this

15=3 times 5
12=2 times 2 times 3
8=2 times 2 times 2

so to include all of them we include 3 2's, 1 3 and 1 5 or 2 times 2 times 2 times 3 times 5 or 8 times 15 or 4 times 30 or 120

the LCM is 120

so 4/15=32/120
1/12=10/120
3/8=45/120

add 32/120+10/120+45/120=(32+10+45)/120=87/120

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Answer:

x=-16

Step-by-step explanation:

y=-20

y=x-4

-20=x-4

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3 years ago

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In March, April, and June it rained 2 inches. In February, May, and September it rained 1 inch. In August and October, it rained

koban [17]

Answer:

Step-by-step explanation:

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3 years ago

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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.

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.

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3 years ago

Graph the solution set of the system of inequalities or indicate that the system has no solution.

alina1380 [7]

The graph is the intersection of the interior of a circle with a half-plane.

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