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julsineya [31]
2 years ago
13

"cube has a volume of 125 cubic inches. What is the length of each edge?"

Mathematics
1 answer:
s2008m [1.1K]2 years ago
6 0
5. Each side is equal and you need to find what number times itself 3 times gets 125, which in this case is 5. 5x5x5= 125, which means 5 is the correct answer.
You might be interested in
ALGEBRA 2 PLEASE HELP! QUICKLY!! I AM TERRIBLE AT ALGEBRA AND I HAVE LOADS OF STUFF TO Do TODAY CAN YOU PLEASE HELP ME WITH THIS
Genrish500 [490]
A) the posible roots of the function f(x)=x⁴-3x³-5x²+21x+22 are: 
=-1,+1,-2,+2,-11,+11,-22+22.

b)
one zero of the funtion can be: x=-1 ⇒y=0   (-1,0)

c)
other zero of the funtion is: x=-2 ⇒y=0  (-2,0)

d)
if x=0 then y=22   (0,22)

e) In the file is the graph. 


4 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
I need help 2(x + 1) = 2x + 2.
Ksju [112]
2(x + 1) = 2x + 2\\
2x+2=2x+2\\
0=0\\
x\in\mathbb{R}
7 0
3 years ago
SURFACE AREA HELP ASAP!!
Norma-Jean [14]

Answer:

Step-by-step explanation:

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6 0
3 years ago
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Lucy has block letters that are in the shape of cubes with sides 2 3 foot long. She has a box which is 1 1 3 foot long, 2 3 foot
MrMuchimi

Complete question :

Lucy has block letters that are in the shape of cubes with sides 2 3 foot long. She has a box which is 1 1 3 foot long, 2 3 foot wide, and 2 3 foot tall. Find the number of block letters that will fit in the box.

Answer:

1/2

Step-by-step explanation:

Number of block letters that will fit in :

Volume of box ÷ volume of each letter cube

Volume of each letter cube = (side length)³ = (2/3)³ = 8/27 ft³

Volume of box = length * width * height

Volume of box = 1 1/3 * 2/3 * 2/3

Volume of box = 4/3 * 2/3 * 2/3 = 16 / 27 ft³

Therefore ;

8/27 ÷ 16/27

8/27 * 27/16

= 8/16

= 1/2

7 0
3 years ago
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