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Scrat [10]
3 years ago
8

Cube X has a volume of x, Cube Y has a volume of 8x. If Cube Y has an edge length of y, what is the edge length of Cube X?

Mathematics
1 answer:
ziro4ka [17]3 years ago
6 0
Hello,

Cube X: volume=x
Cube Y: volume=8x
Y has a volume 2^3 times of the X cube
So,
the side of the Y cube is twice the volume of the X cube.

The side of the X cube is half of the cube Y

Side of X cube=y/2

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Complementary angles have a sum of 90 degrees <br>true or false​
swat32
True would be the answer, sorry if it’s wrong
8 0
3 years ago
I'm not sure what this means, please help me I'll give you 10 points ;)
Karo-lina-s [1.5K]

<u>Answer:</u>

So all the possible solutions are:

  • \bold{523,~400,~379~and,~378.6}

<u />

<u>Solution Steps:</u>

<em>First you need to solve the real inequality to understand how to find the rest of the possible equations. </em>

<u>Add 78 to both sides:</u>

  • <u />78+78= Cancels Out
  • 300+78=378

<em>So now we know the real answer, but it ask for all possible answers. </em>

(Means anything larger than 300 when you plug it into x - 78 > 300.)

<u>Numbers that are greater than 378:</u>

  1.)  377-78=299 > 300  (False)

  2.)  523-78=445>300  (True)

  3.)  0-78=-78>300  (False)

  4.)  -62-78=-140>300  (False)

  5.)  400-78=322>300  (True)

  6.)  379-78=301>300  (True)

  7.)  222-78=144>300  (False)

  8.)  378.6-78=300.6>300 (True)

______________________________

 \bold{Hope~this~helps!}\\\bold{If~you~need~help~with~anything~else,~feel~free~to~ask!}\\\\\bold{~~~~~-TotallyNotTrillex}

5 0
3 years ago
The given triangles are similar. Solve for x.
Maru [420]
The first one is 21.93 rounded up to 22.
8 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
During a sale, a store offered a 15% discount on a TV that originally sold for $420. After the sale, the discounted price of the
wolverine [178]

Answer:

Step-by-step explanation:

Your answer

303.45

Mark it as Brainlist. Follow me for more answer.

7 0
3 years ago
Read 2 more answers
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