Solve the equation 10r – 2 = 28 for r.

Simplify 4radical2plus7radical2minus3radical2

zlopas [31]

Let "radical 2" be represented by "r."

Then you are to simplify 4r + 7r - 3r. This comes out to 11r - 3r = 8r.

The answer is 8 radical 2.

3 0

3 years ago

John bought new headphones originally listed for $ 80.99 . They are \(35\%)| off. Which equation can be used to find the amount

Katarina [22]

Answer:

$28.35

Step-by-step explanation:

The headphones are $80.99 and are 35% off. To find the amount he will save, change the 35% to a decimal by moving it two times to the left, which is 0.35. Multiply 80.99 and 0.35 to get 28.3465. Round to get $28.35.

7 0

2 years ago

The annual rainfall (in inches) in a certain region is normally distributed with = 40 and = 4. What is the probability that star

sdas [7]

Answer:

0.93970

Step-by-step explanation:

Solution:-

- Denote a random variable "X" The annual rainfall (in inches) in a certain region . The random variable follows a normal distribution with parameters mean ( μ ) and standard deviation ( σ ) as follows:

X ~ Norm ( μ , σ^2 )

X ~ Norm ( 40 , 4^2 ).

- The probability that it rains more than 50 inches in that certain region is defined by:

P ( X > 50 )

- We will standardize our test value and compute the Z-score:

P ( Z > ( x - μ ) / σ )

Where, x : The test value

P ( Z > ( 50 - 40 ) / 4 )

P ( Z > 2.5 )

- Then use the Z-standardize tables for the following probability:

P ( Z < 2.5 ) = 0.0062

Therefore, P ( X > 50 ) = 0.0062

- The probability that it rains in a certain region above 50 inches annually. is defined by:

q = 0.0062 ,

- The probability that it rains in a certain region rains below 50 inches annually. is defined by:

1 - q = 0.9938

n = 10 years ..... Sample of n years taken

- The random variable "Y" follows binomial distribution for the number of years t it takes to rain over 50 inches.

Y ~ Bin ( 0.9938 , 0.0062 )

- The probability that it takes t = 10 years for it to rain:

= 10C10* ( 0.9938 )^10 * ( 0.0062 )^0

= ( 0.9938 )^10

= 0.93970

3 0

3 years ago

Whats the slope of (2,-5),(-6,5)

Makovka662 [10]

The slope of the line with the points (2,-5) and (-6,5) is 5/-4 (or 10/-8).

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