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2 years ago

A group of seventh and eighth graders were asked whether they prefer reading fiction or nonfiction.

Mathematics

2 answers:

zloy xaker [14]2 years ago

7 0

7th graders prefer fiction is 0.75 = 75%
8th graders prefer fiction is 0.68 = 68%
7th graders is greater than 8 graders

answer
A. greater than

k0ka [10]2 years ago

5 0

Answer:

Option A. greater than

Step-by-step explanation:

The percent of seventh graders surveyed who preferred fiction = 0.75 or 75%

The percent of eight graders surveyed who preferred fiction = 0.68 or 68%

Now we can say the percentage of seventh graders who preferred fiction is GREATER THAN the percentage of eight graders who preferred fiction.

Therefore option A. is the correct option.

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Somebody... PLEASE help What is the only solution of 2x2 + 8x = x2 – 16?

makvit [3.9K]

Answer:

x= - 4

Step-by-step explanation:

2 * x^2 + 8 * x = x^2 - 16

2 * x^2 - x^2 + 8 * x + 16 = 0

x^2 + 8 * x + 16 = 0

(x + 4) * (x + 4) = 0

x = - 4

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1 year ago

4. At a discount furniture store, Chris offers the salesperson $600 for a couch and a chair. The store charges 8% sales tax. If

Lana71 [14]

Answer:

4. $648

5. $86.40

Step-by-step explanation:

You take the percent (we'll take 4 for example, so 8%) and change it to a decimal: .08. You then multiply the decimal (.08) by your other number (in this case $600) and the resulting number is your answer, but not final. Take whatever the answer was (in this case 48) and add it to the orignal number (600) to get $648.

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2 years ago

Last month, Carla’s neighbors paid her to take care of their cat when they went on vacation.she spent $4 of her earnings on an a

kap26 [50]

Answer:

she got 27 dollars

Step-by-step explanation:

when u add $4 and $14 u get $18.then add the 9 dollars she had left and u will get 27 dollars.

Hope this is a clear description

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2 years ago

9. Evaluate (-7) -10 ÷ (-28)-10 pls solve fast

Juli2301 [7.4K]

Answer:

17/38 or ≈ 0.45

Step-by-step explanation:

(-7) -10 / (-28)-10

= -17 / -38

= 17/38

You can keep the answer in fraction as 17/38 or ≈ 0.45

Hope this help you :3

7 0

2 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.

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